Most mathematicians agreed that infinity multiply by zero is not a zero because infinity is not a natural number. However, in my theory, that is wrong.
When I explain multiplication, I explain multiplication as "A zero value sum to the value of the first number a number of times that is defined by the value of the second. If there is a third number then a zero value sum to the equation result of the first two numbers a number of times that is defined by the third number."
I will skip the fourth, the fifth number, and so forth but the explanation above is understandable. If you put the above in context if an infinity number multiply by a value of zero then the equation will produce a zero value as the result value. From the context of my explanation, the equation will be something like this: "a zero value sum to an infinite value 0(A number of times that is defined by the second number) times". From that context, if Zero sum to anything zero times then the result of that can only be a zero value. It does not matter if the number had one, two, three, or an infinite amount of digits. In this circumstance, if we put zero as the first number then a zero value will sum to a zero value an infinite amount of times. In that case, it does not matter how many times that equation goes on, in the end, the result can only be a zero value.
Something similar will also apply to any numbers(including infinity) multiply by an infinite amount of zeroes. In that case, if we are sure that the infinity of zero number only holds the digits zero then it does not matter how many zeroes there are, the value of all the zeroes combined can only be a zero value. continue reading
]]>The source code is very basic currently, but if you like to use it in your project feel free to. Any bug can be reported directly here. I do not mind contribution to the source code either. The link to the project is here https://github.com/kevinhng86/fai_geoIP, I might occasionally update the source code in the futures.
I will come back to continually release the infinity subject. It just currently I have to finish up with the traffic counter platform. To see my traffic counter in action, you can visit http://ozlu.org or http://codercentral.net, they are all operated from one centralized location. The ad server that I am building and debugging also serving ads from one centralized location. continue reading
]]>This is just a short release for my readers to not get bored while waiting for the finishing of the infinity formula. I halt the editing on the infinity formula due to the development of an automation program for advertisement distribution.
These three functions are a parser that will read in any text file or any string. When reading of the string, any code in between of {{PHP}} {{/PHP}} will be run as php code. Any others text will be echo by the echo command. There can be more than one set of {{PHP}}{{/PHP}}. I can't go into detail for this article right at this moment, but this will change into an article that will show you in detail how the function work. The benefits for this function is you can centralize all your php code on one domain in a text file, and run it off from another server. This is good for multiple domains that are not on the same server. You can always use .htaccess file to limit which IP can access the text folder that stored the php codes. Nevertheless, this is eval code so please do understand that there is a security risk involved if the text file can be read by unauthorized personnel or can be altered by anyone.
To use the code all you have to do is have the three function together, pass any text string or text file to writer($data) and voila it will work. The php code inside the tag will be executed first before anything else is being echoed. You can modify this as desired. Nevertheless, do take security measurement when running eval from a clear readable text. If the php code inside the text file is broken, this function will exit with an error. I programmed this in PHP 7.0. That is the default result of what would happen when the php code that passed into an eval code is broken. We can always check the code by parsing it into a php parser before parsing the code to the eval() function. Just a quick reminder, text in between of {{php}} {{/php}} will be treat as php code. Do not nest the {{php}}{{/php}} tags, doing so will create an error. Please do not change the start tag and end tag, that will also create an error. If you want to customize the start tag and end tag you do need to modify the code. Where the plus 7 is, that is the string length of the start tag. The plus eight is the string length of the end tag. When choosing the start tag and end tag, it is best to choose set of word that you will never use or most likely never be used in writing or in code, otherwise, sometimes, you can grab the wrong text then the code will go in error. There must also be both the start tag and the end tag. If the set of start and end is missing one tag, the script will also go in error. In the final version of this article, I will show you how to build the function to check for both, the start tag and the end tag. Also, the php code, including the start and end tag will be removed from the input text before the text being echo by the function.
When the eval code is return, the return eval function will always return an array. Nevertheless, that can be customize to return string when one tag is found, and when there is more than one set of tag then an array would be return. Else that, everything is easy as this is just a basic demonstration of something that will be more complex when putting in real life pratical examples.
[code lang="php"]
function runEval($invalue){
if (gettype($invalue) === "string") eval($invalue);
if (gettype($invalue) === "array"){
for( $i = 0; $i < count($invalue); $i++){
eval($invalue[$i]);
}
}
}
function returnEval($a, $starttag, $endtag){
$tempstr = "";
$codeArray = array();
while( strpos($a, $starttag) || strpos($a, $endtag) ){
$posa = 0;
$posb = 0;
$strlen = strlen($a);
$posa = strpos($a, $starttag);
$from = $posa + 7;
$to = $strlen - $posa;
$a = substr($a, $from, $to);
$posb = strpos($a, $endtag);
$strlen = strlen($a);
$from = $posb + 8;
$to = $strlen - $from;
$tempstr = substr($a, 0, $posb);
$a = substr($a, $from, $to);
array_push($codeArray, $tempstr);
}
return (count($codeArray) > 0? $codeArray : 0) ;
}
function writer($data){
$st = "{{php}}";
$et = "{{/php}}";
$code = returnEval($data, $st, $et);
if($code !== 0){
runEval($code);
if (gettype($code) === "string") $data = str_replace($st . $code . $et, '', $data);
if (gettype($code) === "array"){
for( $i = 0; $i < count($code); $i++){
$data = str_replace($st . $code[$i] . $et, '', $data);
}
}
}
echo $data;
}
[/code]
This code below is the update and will run the php exactly where it is. This code below will most likely be the code for this article if I ever have time to write this article.
[code lang="php"]
$st = "{{php}}";
$et = "{{/php}}";
$data = 'I want to echo the word world. Let try it: {{php}}echo "world";{{/php}}
Cool, I just echo that. Let do something else.
{{php}} for($i = 0; $i < 10; $i++){
echo $i . "\n";
}
{{/php}}
Now let check to see weather I can echo something else more cool like what I look like if I can grab a variable that is my parent.
{{php}}
echo $parent;
{{/php}}
';
$parent = "I am a parent variable";
function returnEval($a, $starttag, $endtag){
$tempstr = "";
$codeArray = array();
while( strpos($a, $starttag) || strpos($a, $endtag) ){
$posa = 0;
$posb = 0;
$strlen = strlen($a);
$posa = strpos($a, $starttag);
$from = $posa + 7;
$to = $strlen - $posa;
$a = substr($a, $from, $to);
$posb = strpos($a, $endtag);
$strlen = strlen($a);
$from = $posb + 8;
$to = $strlen - $from;
$tempstr = substr($a, 0, $posb);
$a = substr($a, $from, $to);
array_push($codeArray, $tempstr);
}
return (count($codeArray) > 0? $codeArray : 0) ;
}
function runEval($invalue){
if (gettype($invalue) === "string") eval($invalue);
if (gettype($invalue) === "array"){
for( $i = 0; $i < count($invalue); $i++){
eval($invalue[$i]);
}
}
}
$code = returnEval($data, $st, $et);
if($code !== 0){
if (gettype($code) === "string") $data = str_replace($st . $code . $et, "<?php " . $code . " ?>", $data);
if (gettype($code) === "array"){
for( $i = 0; $i < count($code); $i++){
$data = str_replace($st . $code[$i] . $et, "<?php " . $code[$i] . " ?>", $data);
}
}
}
$evalStr = "?>" . $data;
eval($evalStr);
?>
[/code] continue reading
I do like articles from ghacks.net and find that they are good articles. I would always love to have ghacks.net's articles on this site whether if they are in full or not. Nevertheless, when I imported ghacks.net articles into my website, it was through automation feed reading and ghacks.net does not use feed excerpt and left their content in full. Currently, I do not have time to modified the old posts that coming from ghacks.net due to too many tasks on hand at this moment. Just like everyone else, I do really respect the intellectual copyright of other, therefore, the old ghacks.net's articles will be made unavailable at the current moment until I can edit them.
ghacks.net does have quality articles in regard to windows and technology news, I would recommend going to them to read on some of the latest news. Currently, I do not import articles from ghacks.net until I can ask ghacks.net on how many words in an article that they would allow for post republishing. continue reading
]]>After analyzing the properties of infinity numbers, I came to a conclusion. From the conclusion, I define that it is possible to calculate a mathematical equation's result value for any infinity numbers. For some instance, we may never be able to get the result value for the entire equation because the numbers are infinite in length. Nevertheless, we can still produce part of the result value. In other instance, it is possible to actually get a result value that would be very similar to the actual result value of an entire equation of infinity numbers. From my conclusion, I compare infinity to time and from there I wrote the infinity formula.
When we first talk about an infinity number, the first question would come up to mind is can we classify a number that is infinitely in length as a number? In my opinion, whether if the number is infinitely in length or not, we can always classify the number as a number. The closest evidence that I came up with to back up this theory is time. I would use time for a comparison in this case.
As when we are looking at time as the subject, time is also infinitely in measurement. No one would know if time would flow infinitely or will time meet a wall and will end. For the property of time, we know that we can see events that occurred in the past but not the present nor the future. We may be able to predict the future bases on statistical data but would not be able to see events that will occur in the future.
Also, we, ourselves can never see events that are occurring in the present. The events of now that is occurring right in front of our eyes seem like the present, and our brain perceives it that way. Nevertheless, what we are seeing in our eyes are actually the past. It would take sometimes for light to hit an object and then reflect off the object into our eyes. When our eyes receiving the light, our brain would have to process the lights into images. Because of the speed of light and because of how fast our brain process data, the process of translating an event that occurs in front of our eyes into an image would be really fast and would be completed in less than a microsecond or a nanosecond. The fast speed of the process would let our brain perceive the event that occurred in front of us as if we are actually seeing the present. However, that event actually occurred less than a nanosecond in the past. In my opinion, in a true equation, whether if the event occurred one year ago or one nanosecond ago it would still be considered as an event of the past and not the present.
When we take this to a micro scale, even if we are standing in front of a mirror and looking at ourselves, the images that we are seeing in the mirror are still the images of the past and not the images of the present. On a larger scale, if we are looking at a planet that is one billion light year distance away from us, we are not seeing that planet present. We are actually seeing the past of that planet. It would take one billion years for light to travel from that planet to our planet, Thus, if that planet was destroyed or turned into a planet that habitable by life forms half a billion years ago, we would not know until half a billion years later. This same principle also applies to our closest star, which would be the Sun. If there are any changes on the surface of the Sun, we would not be able to detect the changes until at least eight minutes later.
Not just our eyes that can't see events that occurred in the present, the pain that we feel or the feeling of touching an object are also events that occurred in the past. When an object touched our body, it would take a very small amount of time for our nerve to be able to send a signal to our brain, and with that signal from the nerve, our brain would be able to process the signal into what we can describe as feelings.
Even when we are playing games on the computer or typing something on the computer. To us, it almost seems like when we are typing something on the computer, the words would be displayed so quickly that it is actually the present. Nevertheless, we, in fact, pressed the key on the keyboard some microseconds or nanoseconds ago.
Is there a method for us to be able to perceive events of the present? In my opinion, at the current state, I do not believe that we can perceive events that occur at their exact moments. Nevertheless, I do believe that to be able to perceive events that occur right at the exact moments of when they occurred is to be able to move forward in time and look back at the events. Only then, we would be able to perceive events that occur at their exact moment. In short, to be able to see the present is to be able to see the future and look back in time. Will technology or will our body develop to give us the ability to perceive events of the present by looking forward into the future? I would never know.
When comparing time to an infinity number, I would treat time as a mathematical equation. What events that we have seen that already occurred in the past would be similar to seeing numbers that we had already seen and might have applied a calculation to. Thus, when we are asking the question if an infinity number is a number would be similar to asking if the infinity time is time. Time is infinite in the amount and in my opinion, whether if the events in time have happened yet or not, they would still be considered as time. Therefore in principle, if we were to receive an infinite string of digits, the digits that we have seen are numbers and the digits that we haven't seen are still numbers. In my opinion, it just because the numbers are infinitely in length does not mean that they are not a number. They are still numbers, just that they are numbers that are infinitely in length.
For an example, my birthday that happened last year would be considered as an event of time that had already occurred. My birthday of the future or how many birthdays would I have left would still be considered as events of time. These time events, however, have not occurred but are time events of the future. If my birthday of the future does not happen, another event of time would replace the time event of what suppose to be my birthday. Nevertheless, these events would still be considered as an event that occurs in time or an event of time.
In my conclusion, I concluded that a number, whether infinitely in length or not should still be considered as a number. Therefore, an infinity number is still a number, it just that the infinity number is infinitely in length.
When looking at the length of infinity numbers, I asked myself, can one infinity number be longer than another infinity numbers in terms of the amount of digit one number can have? With some analysis, in my opinion, infinity numbers are always the same length. I drew my conclusion bases on one simple principle, and that principle is if the numbers are infinitely in length that means there is no limit on the amount of digits one number can have, and that also means, any number that is infinitely in length will never have an ending. In another explanation, if we are using the term infinity for a length then infinity is the largest available length in any type of measurement. This is because infinity will never end, and due to that, there would not be another unit of measurement that is larger than infinity as a unit of measurement. Since infinity as a length is the largest available length, there would not be another length that is larger than the infinity length. There will be other lengths that are smaller than the infinity length, nevertheless, infinity length can neither be smaller than itself. With that perspective in mind, the infinity length can't be larger or smaller than itself.
To put the perspective above in context, I will demonstrate some examples. For the first example, we have two numbers and each number contained 1 trillion digits. If we were to discuss in regard to the length property of those numbers, then we know that they are both 1 trillion digits in length, and if we were to ask which number is the longer or the shorter number, then we know that both numbers are the same in length. The reason of why the numbers are the same in length is because both numbers are 1 trillion digits long. Now, if one number is 1 trillion digits long and the other number is 1 trillion and 1 digits long, then the numbers are now different in length. Thus, in this instance, if we were to measure the length of the numbers, the longer number would be the number that has 1 trillion and 1 digits. It may seem that the length of 1 trillion digits is a superbly huge length, nevertheless, the 1 trillion digits length is not endless and is calculable. When this principle is applied to infinity lengths, the property would be similar. If both numbers can never run out of digits then either number can't be the larger number bases on their length. Also, either number can't be the shorter number bases on how many digits one number had.
For another more complicated example, I will use a real life example. In this example, we have two cars, one is black and one is white. The white car will travel a route that is 1000km long, and the black car will travel another route that is 1000km long. Both cars will be driven at the same speed. The white car started its route 5 hours before the black car. Now, if we were to ask which car finish their route first then it is obvious that the white car will. Nevertheless, if we were to ask which car will travel the longer or the shorter distances then we know that the answer is both cars will travel the same distance. This is because they will both travel a route that is 1000km long. Now, if the white car was to route through an additional route that is just 1km long, then it is obvious that the white car will travel the longer distance. Nevertheless, when coming back to the 1000km route, 1000km can't be smaller or larger than 1000km. This is because 1000km is the same length as 1000km. If we were to use a distance of infinity instead of using the distance of 1000km, then both car will still, in fact, travel the same distances.
For the final example, let imagine that we have two pieces of paper that are so large that they are infinitely in size. If we were to write down one random digit onto each piece of paper at the same time and at the rate of one digit per second for an infinite amount of second, at any moment during the process, both of our pieces of paper will always have the same amount of digits, and the amount of digits that each piece of paper is going to have is going to be endless.
When it comes to a fractional part value for infinity numbers, the same in length property would also attach to the numbers. If we are going to say that there is an infinite amount of digit before and after the decimal for both numbers. That means there are an infinite amount of digits before the decimal in both numbers. With that, it also means, both numbers contained an unlimited amount of digits after the decimal. Thus, both infinity numbers with a fractional part are also equal to one another as of lengthwise.
For a conclusion, numbers are always the same length if they are infinity numbers.
In my understanding, the general formula for calculating infinity numbers would be "Infinity equation Infinity = Infinity". For the case of this article, I would write the general formula as the general infinity formula for calculating a mathematical multiplication equation. Thus, the general formula for calculating infinity numbers in this example would be written as "infinity x infinity = infinity".
General Formula: ∞ x ∞ = ∞
For my doctrine of infinity calculation, I have two formulas for calculating mathematical equations that involve infinity numbers. The first formula is for when we are calculating infinity numbers that contain a variety of digits. This means the infinite numbers are numbers that do not only contained only one single repeating digit that runs infinitely in length. For my first formula, the formula is written as "Infinity Calculation is equal to Infinity Assumption and is equal to Infinity Alteration". In short, I write my formula as "infC = infA = infAL".
Infinity Numbers That Contains Varieties Of Digits: infC = infA = infAL
My second formula is a formula for calculating infinity numbers in the circumstance that the digits in the infinity numbers do not change. In another explanation, I classified this as infinity numbers that are made up by only one digit and that one digit will be repeating itself infinitely. For example, an infinite amount of nine multiply by another infinite amount of nine. To put this in context, we are assuming that we are multiplying two numbers together and each number only contains the digit nine and the digit nine will repeat itself indefinitely. For my second formula, I write the formula as "Infinity Number equation Infinity Number equal to Infinity Pattern". For a short version, I write this formula as "infN e infN = infP".
Infinity Numbers With A Single Repeating Digit: infN e infN = infP
The first formula: "infC = infA = infAL".
When we are applying mathematical equations to infinity numbers, we would not be able to get the true final result value for the entire equation. This is because the infinity numbers will never have an ending and because the numbers will never have an end, we would not be able to get the final result value for the entire equation. Nevertheless, in my opinion, we can always get the result value for the part of the numbers where we had already seen. Thus, I labeled the result values that are produced by an equation of what we had already seen in the infinity numbers as the "State Of Infinity Result Value".
To put this in context, if we were to receive two strings of digits that are infinite in length and we are applying a mathematical equation to the strings of digits, we would be able to get the result value for the digits that we had already received but not the result value for the entire two strings of digits. Let assume that we are receiving the same amount of digit each time for both strings. At each time when we received a new set of digits, we would be able to calculate the result value for all the digit that we had received. For a similar comparison, this would be similar to an event in time. If time was a mathematical equation, then we are experiencing the result value for what had occurred but we had not experienced or been able to see the result value for what had not occurred. In short, a "State of Infinity Result Value" is the equation's result value of the part that we had already received and calculable within the infinite numbers.
The first formula is not just only usable for calculating numbers that are infinite in length. The first formula is applicable to circumstances where numbers are not infinite in length but are of an unknown length. In other words, very large numbers that we can't read and apply a mathematical equation to in one time.
In a perspective, I described formula one as a formula for dealing with "Infinity Numbers That Contains Varieties Of Digit".
For the properties of formula one, we would have two properties, and they are "Infinity Assumption" and "Infinity Alteration". For "Infinity Assumption", we would also have two sub-properties and they are "Infinity Post-Assumption" and "Infinity Pre-Assumption". With "Infinity Post-Assumption", we have one property that only applies to a mathematical addition or a mathematical subtraction equation. The property is "Infinity Larger-Assumption".
Property Map of Infinity Assumption:
Infinity Assumption:
The amount of "State of Infinity Result Value" that we are required to produce in a mathematical equation at each part of the infinity numbers depends on the type of mathematical equation that we are dealing with.
For the "Infinity Alteration" property, "Infinity Alteration" will have one sub-property and that property is "Infinity Unchanged". I also classified "Infinity Unchanged" as "Eternal Unchanged". Therefore, sometime I may use the name interchangeably for this property. "Infinity Unchanged" is a property that is found within a subtraction, an addition and a multiplication equation of infinity numbers. Nevertheless, the "Infinity Unchanged" property is not found from a division equation of infinity numbers that are not made up by only one repeating digit.
Property Map of Infinity Alteration:
Infinity Alteration:
When we are calculating infinity numbers, often time, we would have to deal with carry-over values. The only equation that we do not have to deal with a carry-over value is a division equation. How about a subtraction equation? In the general subtraction formula, the carry-over/regrouping values would be replaced with the borrowing/regrouping values. Nevertheless, I wrote a subtraction formula to accommodate my infinity formula. In my subtraction formula, I do not use the borrowing/regrouping value but would use a carry-over value instead. For a reference to my subtraction formula, the full article for the formula can be found at Kevin's subtraction formula.
For a summary of my subtraction formula, I use absolute values for calculating the result value at each set of digits. I do not swap the position of the numbers when the first number is the lower number but I keep the numbers in their original position. In my subtraction formula, I do not borrow a value from the next set of digits in the first number. Rather, I either add or subtract a value of one to the next set of digits in the first number, and whether to subtract or to add is bases on whether if the first number is the larger or the smaller number. Thus, in my infinity formula, carry-over values are applied to an addition, a subtraction, and a multiplication equation but not to a division equation.
When we are calculating the seeable and calculable part of infinity numbers to get the "State of Infinity Result Value" and when it comes to the carry-over values. It is very important when there is a carry-over value that produced from the equation. This carry-over value that I mentioned here is not the carry-over value from applying the equation to each set of digits within the seeable part of the infinity numbers. But is the final carry-over value from the equation of the entire seeable part of the infinity numbers.
For example, from the equation of ∞ 3787 + &infni; 7018 and we are considering that there is an infinite amount of digits that will be added to the beginning of both of the infinity numbers. We would not know what digits are going to be added to the beginning of both of the infinity numbers. Nevertheless, the seeable part of the first number is 3787 and the seeable part of the second number is 7018, and when we added the two seeable parts together, we would have an answer of 10,805. Since we added four digits to four digits and we have not completed the equation, our result value at this time should only contain four digits. Anything beyond four digits would be a carry-over value. Since infinity numbers are numbers that are infinite in length, we would not be able to be at the last digits set in the infinity numbers and the equation is an endless equation. Therefore, in the example of ∞ 3787 + ∞ 7018, when added the two seeable parts of the infinity numbers together, we would have a value of one as the final carry-over value for the equation before we are obtaining more digits from the infinity numbers. The main question here is when should we write or when should we not write this carry-over value into the "State of Infinity Result Value"?
When it comes to a true addition equation or a multiplication equation, we are always increasing the digit value of the numbers that involve in the equation. In terms, the equations would always produce equal to or larger digit values on top of the digit value of the largest numbers out of the two numbers that involved in the equation. In this type of equation, because we are always adding digit values on top of digit values, writing down the carry-over values into the "State of Infinity Result Value" for any part of the equation without applying the carry-over values twice will not alter the next "State of Infinity Result Value" in a way that the value will be an incorrect value. For example, if I was to add 1357 + 1701 and I only want to add 357 + 701 before adding the two 1s together, I would get a result value of 1058. I can write the result value as is 1058 and then add this result value to the result value of the two one. Which can also be written as 1000 + 1000 + 1058, of which, would be equal to 3058. When this principle is applied to equations of infinity numbers, the same property would also apply to the "State of Infinity Result Value". In the example of ∞ 357 + ∞ 701, at this part in the numbers, it would make more sense if we were to write the "State of Infinity Result Value" as 1058. Rather than writing the value as 058. Therefore, in a conclusion, the final carry-over value of the current part in the infinity numbers during an addition and a multiplication equation should be written down into the "State of Infinity Result Value".
When I describe in regard to a true addition equation, I define a true addition equation as an addition equation that in final will always increase the values of the numbers that involved, and is not an addition equation where the equation turned into a subtraction equation due to the differential in the negative or the positive base of the numbers. This can be similar to when I describe in regard to a true subtraction equation, I describe a true subtraction equation as a subtraction equation that truly reduces the value of the first number to the value of the second number, and is not a subtraction equation that turned into an addition equation due to the differentials of the negative or the positive base between the two numbers.
When we come to the carry-over value for a true subtraction equation, the property of the carry-over value does change. In a true subtraction equation that involves two numbers, we are reducing the digit value of one number to another number. From my formula, the carry-over value that produced from a sub-equation of a digit sets in the numbers has to be added to or subtracted from the next set of digits in the first number. Thus, we can not write the final carry-over value from a subtraction equation of the current seeable part of the infinity numbers into the "State of Infinity Result Value". This is because we have not seen the next set of digits to able to apply the carry-over value to. This property would also apply in the circumstance where we were to use the general subtraction formula to calculate the infinity numbers. We can't write down the borrowing value either, simply because we can't borrow from what we can't see or a nil value. Therefore, in conclusion, in a subtraction formula of infinity numbers, the final carry-over value of the current seeable part of the infinity numbers can not be written into the "State of Infinity Result Value". This is because the value is to be applied to the next set of digits, or to the equation of the next set of digits and is not a true result value when applying alone. Applying the carry-over value to the "State of Infinity Result Value" without any set of digits will produce an incorrect result value. This is simply because we are carrying a value to a value that we have not seen.
Subtraction is also the one part of the infinity formula that seems to not make sense at first glance. Nevertheless, by looking at an example we can understand more. For example, if we were to use my subtraction formula for an equation of 3384 subtract to 2554 and I only want to subtract 384 to 554 before subtracting the final digits set of 3 from the first number to the final digits set of 2 in the second number. We would get a result value of 830 and a carry-over value of 1. If we were to write down the result value for this state as 1830, it would not make sense. Also, if we wrote the result as 1830, we would not be able to apply the result value to next set of digits. If we didn't write down the final carry-over value, we can write the equation as (3000 - 1000) - 2000 + 830, of which will give us the correct answer of 830. Now, if we were to write down the final carry-over value into the result value from the last equation before we process the last set of digits, we would have 3000 - 2000 + 1830. From which, would give us 2830 as a result value for the equation. Nevertheless, that is not the correct result value and if we were to write the equation as 3000 - (2000 + 1830), we would get -830 as the result value for the equation. Of which, wouldn't be the correct value either.
If we were to treat the above example as infinity numbers, we can write the equation as ∞384 - ∞554. In this circumstance, if we are assuming that the first number is going to be the larger number, we would have ∞830 as the "State of Infinity Result Value" or simply 830. On the other hand, if we are going to assume that the second number is going to be the larger number, we would have ∞170 as the "State of Infinity Result Value" or simply 170. In both of our assumptions and whether if the first number is the larger number or not, we would still have a carry-over value that we would have to carry to the equation of the next seeable part of the numbers. To put this theory into a practical application, with the example of the first number being the larger one, We can have 18,384 - 3,554, of which, would give us 14,830 as a result value. For an example of the second number being the larger number, we can have 18,384 - 38,554 and the result value of the equation would be -20,170.
When it is a division equation, we wouldn't have a carry-over value. This is because in a division equation, whether at any part in the numbers, we would have to divide all the digit from the first number to all the digit from the second number to get a result value. Therefore, in an infinity division equation, at each time when we received a new set of digits that increase the seeable part of the infinity numbers. We would have to divide all the seeable digits in the first number to all the seeable digits in the second number. We wouldn't be able to divide part of the numbers to each other and somehow manage the result value from one part of the numbers to the result value from another part of the numbers to produce a correct result value for the entire equation. Therefore, during an infinity division equation, at any time we are obtaining a new set of digits in the infinity numbers, it is going to be a completely new equation. Hence, there wouldn't be any value that is going to be applied to the later or the previous equations. However, there is a remainder value. Nevertheless, the remainder value and how many decimal places are we going to use is up to our decision at the beginning instance of the equation, and this would be different between each case.
continue reading
From all the web addresses below, if you go onto the website using a browser, then view the source code of the page. Unless stated, within the source code, search for the word escort. In most browser, when viewing the source code of a webpage, you can just press Ctrl + F to search for a phrase or word in the source code. There will be some link that links to porn sites that would be hidden from the browser rendering engine by having the 0 font size, out of area display or with hidden HTML attribute. This is done by the same group as I stated on my list of hacked government and education websites.
*** This list is live and may update from time to time. If a website on this list become clean and you know please e-mail me at admin@iblog.website, thank you. ***
----- April 17 -----
----- April 18 -----
----- April 19 -----
----- April 25 -----
*** Note: If there is a "< <<" then search for what contained inside the " ". If there is not, just search for "escort" ***
---- April 26 -----
m4m
----- April 27 -----
Note: This list had become too big, I will not be adding more website to this list, but will only update the hacked websites' status when I know that the hacked website is fixed. In a month from now, I will launch a platform that will mainly be use as a service for listing and report currenly hacked website. Nevertheless, I will program the platform and not using anyone else premade platform. Thus, will take sometimes. Nevertheless, thank you for the support of reading this list and helping me contact the website owners.
*** You do not need to go into the source code for sites below. The sites below are different languages but they are holding the same links and setup with a very similar pattern. Just look through them for a bit and you can see the pattern. ***
The three tools that aided me in this investigation are Moz Site Explorer, SemRush Backlink Checker and Ahrefs.com Backlink Checker. Besides the tools, I also use Google Search Engine for aiding and tracking down websites that been hacked. Tools I mentioned may not be able to tracked link from .gov and .edu domains, nevertheless, Google could.
To know how much websites these hackers once hacked I took a snapshot from my investigation. I used one of the domains that I suspected to be associated with this hacker group and check for the history of their backlink using ahref.com tool. The domain is www.ankaralikizlar.com.
As you can see in this snapshot below, before January the 8th of 2016, there are some domains that referring to ankaralikizlar.com but that numbers were below 400 domains. Just on January 3th of 2016, the amount of domain that referring to ankaralikizlar.com went down to 137 domains. Now, just on January the 8th of 2016, ankaralikizlar.com's referring domain counts went up to 2,067 referring domains. That mean on January the 8th, there are 2,067 domains that had at least one link pointing to ankaralikizlar.com. Those numbers do eventually drop because websites' owners found out and reinstall their server or data. Nevertheless, when going through the list of websites that been hacked, I found it sad that some websites did get banned and it is not their fault.
At the height of this hack scheme for this single domain, bases on backlink history record from ahref.com, there could have been 2300+ websites that been hacked and carried ankaralikizlar.com's link unknowingly. Take into consideration that ahref.com does not regularly crawl country's domains extension that is not well known, including others country's .gov and .edu domains. This below is the four snap of this lone domain from ahref.com. As of today, ahref.com report the amount of domain pointing to ankaralikizlar.com is a bit under 200 domains, nevertheless, combine with others country's TLD, there are still 300+ domains that are pointing to ankaralikizlar.com at the time of this writing. The majority of sites that were hacked were either running a CMS system and bases on ahref.com data, a large percentage were running a version of Wordpress.
]]>Less than 2 months ago, one of my websites got hacked, hackers were injecting porn links as a none display HTML division code block into my HTML code. I have been trying to figure out who doing this for sometimes. I have suspicion over two groups of people that could involve or know these hackers but I am not certain, because, it is not solid evidence but just circumstantial. Now, I am following the hacker trails for days and found these websites that got hacked.
This list is now dedicated as a listing of only governmental and educational websites that are currently being hacked. For the list of none governmental and educational websites that are currently being hacked, please refer to my other list at http://kevinhng86.iblog.website/2017/04/16/hacked-website-list-of-none-govermental-and-none-educational-website/. The only section in this list that contained none governmental or educational sites is the section of websites that I contacted and some small exceptions. Some of these sites are not in English, therefore, I am asking for help from people who speak these site language to contact them. Anytime a website is being listed on to this list, at the time of listing, the website was still being hacked. If its come to my attention that any website on this list is fixed, I will put the word fixed next to the website name.
From all the web addresses below, if you go onto the website using a browser, then view the source code of the page. Within the source code, search for the word escort. In most browser, when viewing the source code of a webpage, you can just press Ctrl + F to search for a phrase or word in the source code. There will be some link that links to porn sites that would be hidden from the browser rendering engine by having the 0 font size, out of area display or with hidden HTML attribute. This is really hard to track in the beginning because first I thought site owner trying to sell PageRank, then it is obvious when it's come to education and government sites.
After I got hacked, I reinstall everything but keep the old data as a backup on my office computer for evidence. If you are a government official or have the authority for investigation and if you need the information, please contact me and obtain the data for investigation.
The following are sites that got hacked and I divide them into two categories. The one that I contacted and the one that I couldn't contact. You can check them both out. For the one that I contacted, you could also try to contact them if they didn't fix the problem. Some of the sites I contacted was not in English but I gave English a try. Last I checked they still didn't fix it. If I found any more site that hacked I will update this page.
Check out the website that I contacted. Even the government website of Lebanon got hacked and they still not fix it at this time of writing and I did contact them.
*** This list is live and may update from time to time. If a website on this list become clean and you know please e-mail me at admin@iblog.website, thank you. ***
*** If your website is on this list, you can ask me to remove your web address from this list, one it is fixed, please use your e-mail @ your domain to contact me at admin@iblog.website ***
-------- April 12 | Un Contacted--------
----- April 13 | Un Contacted -----
----- April 14 | Un Contacted---
----- April 15 | Un Contacted---
----- April 17 | Un Contacted -----
----- April 18 | Un Contacted -----
----- April 19 | Uncontacted -----
*** Some of these websites of today don't have .edu in them but they are actually belong to the educational sector ***
----- April 25 | Uncontacted -----
*** Note: If there is a "< <<" then search for what contained inside the " ". If there is not, just search for "escort" ***
----- April 26 | Uncontacted -----
---- April 27 | Uncontacted -----
Note: This list had become too big, I will not be adding more website to this list, but will only update the hacked websites' status when I know that the hacked website is fixed. In a month from now, I will launch a platform that will mainly be use as a service for listing and report currenly hacked website. Nevertheless, I will program the platform and not using anyone else premade platform. Thus, will take sometimes. Nevertheless, thank you for the support of reading this list and helping me contact the website owners.
----- April 12 | Contacted -----
** Each line below can be search in Google **
I also wrote a subject on how to calculate Infinity Numbers. If you are interested in learning infinity calculation, click here.
When explaining in regard to a subtraction equation, the first thing comes to mind is the rate of digits that is applied at a time by the equation. I called this rate as the specified amount of digit that the equation is being applied at a time. In most cases for an explanation of the formula, the explanation would most likely stick with the rate of one digit at a time. In my explanation, the specified amount of digit the equation is applying at a time can also be called the set of digits or the digits set.
When it comes to a subtraction equation, the general knowledge formula is the borrowing method. The borrowing method is applied when the digits set in the first number is lower in digit value than the second digits set in the second number.
In the general formula, the subtraction equation alway started from the rightmost digits on the right side of the numbers and then process to the leftmost digits of the numbers. Before applying the subtraction equation to any digits set, if the digit value of the digits set in the first number is lower than the digit value of the digits set of the second number then a value has to be borrowed from the next digits set of the first number. Also, the borrowing value for a subtraction equation that only involves two numbers would always be a value of one. This value of one is then placed to the left of the leftmost digit of the current set of digits from the first number. For example, if we classified our digits set to only contain one digit. Then, if our digits set in the first number is a digit 2, and we need to borrow a value of one from the next digits set. Then we would place the one that we borrowed in front of the 2, and now we would have 12 as a value. In another example, if our digits set contained two digits and the digit within is 52, and we need to borrow a value of one from the next digits set. Then we would place a borrow value of one in front of the 52. After the placement, the 52 would now become 152.
If needed, the digits set that the current set of digits would always borrow from, would always be the next digits set that is on the left of the current set of digits. The borrowing digit value would be subtracted from the digits set where the borrowing value was borrowed from. The borrowing value would most often be called the borrowing/regrouping value. In my opinion, the coined term borrowing/regrouping is because the digit that was borrowed would be placed at the beginning of the digits set that needed that borrowing value. Thus, what was borrowed would be regrouped with digits set that needed the value by placing the borrowed value at the beginning or to left of the leftmost of the digits set.
Borrowing/regrouping value is only needed when the value of the digits set of the first number is lower than the value of the digits set of the second number. If the digits set in the first number is higher in value than the second number's then we would not need a borrow value, and a normal subtraction procedure would be applied to the sets of digits from both numbers.
To demonstrate the general formula, I would apply the formula to a couple of examples and we are subtracting one digit at a time.
Example 1: 25 -17 Step 1: 5 can't subtract to 7, therefore, we have to borrow a value of one from the next set of digits which is the digit two. Thus, the equation for the first digits set is: 15 - 7 = 8 Step 2: Since we borrowed a value of one from the digit two, we reduce the digit two by a value of one. The equation for the second set of digits is: 1 - 1 = 0 Answer: 25 -17 -- 08, or 8 is the answer of 25 - 17
Example 2: 851 -758 Step 1: 1 can't subtract to 8, therefore, borrow a value of 1 from five. current equation: 11 - 8 = 3 Step 2: a value of 1 was borrowed from 5, therefore, the digit 5 would now become the digit 4. 4 can't subtract to 5, therefore, borrow a value of 1 from 8. current equation: 14 - 5 = 9 Step 3: a value of 1 was borrowed from 8, therefore, the digit 8 would now become the digit 7. It is possible to subtract seven to seven, hence, no need to borrow. Note: We can never borrow from the last set of digits because there is nothing to borrow from. current equation: 7 - 7 = 0 Answer: 851 -758 ----- 093 or 93 is the answer of 851 - 758.
The general formula works perfectly fine when the first number is the number with the larger value. Nonetheless, the general formula would not be able to produce an answer in the event of when the first number is smaller than the second number. When applying the general formula and the second number is the bigger number, instead of having the first number on top and subtract to the second number, we would have to place the second number on top and subtract to the first number. In an event of when the second number is the larger number, the result value of the equation would be in the opposite negative or positive base from the two numbers. For example, when positive A is smaller than positive B and we are subtracting A to B then the result value of the equation would be a negative result value. For another example, when negative A is smaller than negative B and we are subtracting A to B then the result value of the equation would be a positive result value.
There is a reason of why we can't subtract a smaller number to a larger number without a reversal of the equation. The reason is that of the last set of digits. In the last set of digits, there isn't another digits set for the equation to borrow from. The equation neither can borrow from a zero value. If we were to borrow from a zero value or a nil value then, in fact, we are producing an additional value that is not supposed to be there. Which would obviously render the equation to produce an incorrect answer. The reason for why reversing or swapping the position between the larger number and the smaller number would work is because of absolute values.
For absolute values, I have an example. In the example, we have an equation of 5 - 2. In this equation, the equation would produce the result value of 3. If we were to reverse the equation and have an equation of 2 - 5 then we would have a result value of -3. Although the result values from both equations differ in the negative or the positive base, nevertheless, their absolute value is the same, and that absolute value is a value of 3. Thus we can say that the equation of 5 - 2 and 2 - 5 would produce the same absolute value. To put the example in principle, if we were to have an equation of 78 - 98 then the result value of the equation would be the negative value of the result value of the equation of 98 - 78, which is the reverse order of the first equation. In another explanation, the result value of the equation of 78 - 98 would be the negative value of the absolute value of either equation 78 - 98 or 98 - 78.
One property of borrowing a value of one from the next set of digits is in regard to borrowing from the lowest value. If we are borrowing from the next set of digits which happen to be a value of zero then the next set of digits would automatically become the largest value that can be contained within the digits set, and the equation would also automatically borrow a value of one from the next digits set that is on the left of the digits set of where we just borrowed from. For example, if we were to process the equation at the rate of one digit at a time and a value of one is borrowed from a zero digit then that digit of zero would become the digit nine and we are automatically borrowing a value of one from the digit that is on the left of the digit zero, which had now become the digit nine. For another example, if the equation were to be processed at the rate of two digits at a time and a value of one is borrowed from the set of digits that contained double zeroes. Then the set of digits that contained the two zero digits would then become 99, and we are automatically borrowing a value of one from the next set of digits that is on the left of where we just borrowed from.
In this type of explanation, at first glimpse, it may seem that we might eventually borrow twice from the lowest value digit. However, double borrowing can never occur because after borrowing a value, the zero set of digits would contain the highest value. With the highest value, there isn't a possibility that the set of digits that was once contained the zero digit would require any additional value to be able to subtract to another set of digits that is the same size to itself by the amount of digit.
One shortcut to a subtraction equation is when both numbers are equal in digit value. When this occurs, a zero result value would be produced. This is because when a number is reduced to a value of itself then there would not be a value left.
Another property of a subtraction equation that is not usually mentioned is when two numbers to three numbers are subtracting to each other. In this scenario, the equation would always reduce or does not alter the digit value of the largest number. In this type of equation, the equation would never increase the digit value of the larger number or produce a result value that is larger than the largest number that involves with the equation. Thus, the digit value of a result value from this type of subtraction equation would always be smaller than or equal to the digit value of the largest number from the same equation. This property does not apply if there are more than three numbers that involve in the equation. Also, this property is only applied to a true subtracting equation where the equation is an actual subtracting of value type of subtraction equation and is not an equation where the subtraction equation turned into an addition equation due to the numbers are different in the negative and positive base.
For example:
1. 9 - 0 = 9
2. 9 - 8 - 8 = -7
3. 9 - 9 - 9 = -9
Nevertheless, when we added a fourth numbers to the equation:
1. 9 - 9 - 9 - 9 = -18
2. 9 - 8 - 8 - 7 = -14
The above description is for when subtracting both numbers that are in the same negative or positive base. For example, subtracting positive A to positive B or subtracting negative A to negative B. When subtracting two numbers that are different in the negative or the positive base then the equation is defined as an equation where we are adding the first number's digit value to the second number's digit value and result value would have the same negative or positive base as the first number. For example, +N1 - -N2 = +R, or -N1 - +N2 = -R. A subtraction equation that turned into an addition equation would be carried out with the same principles and procedures as an addition equation.
While programming with programs that dealt mainly with mathematical equations that involving huge numbers that can't be processed in one equation and the numbers have to be split into chunks or set of digits, I came up with my own formula for a mathematical subtraction equation. The main key point I solved with my formula is to never have to swap the position of the numbers due to the first number being the smaller number. Nevertheless, my formula can be a bit complicated due to my formula dealt with absolute values or negative values at the micro level.
In my subtraction formula, when both numbers are equal in digit value a zero result value would be produced. This is similar to the general formula. This is because when the first number is reduced to by a value of itself, there would not be any value left. Another property that my formula and the general formula have in common is: in a subtraction equation that is a subtraction equation that truly subtracting values that only involves two to three numbers then the equation would always produce a result value that is smaller than or equal to the largest number in the equation.
In my formula, I do not use the term borrowing/regrouping value. This is due to the property of my formula. My formula does not always borrow a value from the next set of digits. In my formula, there are two conditions that the formula bases on to produce a value at each time a set of digits from the first number is being subtracted to the set of digits in the second number. The first condition is if the digits set from A can be subtracted to the digits set from B. The second condition of the formula is bases on is if A is the larger or the smaller number. Evaluating bases on the first condition, the equation would either produce a value or not. This value would then be subtracted from or added to the next digits set bases on the second condition, which is if A is the larger or the smaller number. Thus, I would not call this value the borrowing/regrouping value. I would call this value, the carry-over value. The reason of why I use the term carry-over value is because the value would be carry-over and apply to the equation of the next digits set. Since the formula does not borrow and then place the borrowed value to the front of the set of digits, the coined term of regrouping would not apply to my formula.
My formula would also calculate the subtraction equation starting from the digit on the rightmost of the numbers and processing toward the leftmost digit of the numbers. Before going into explaining of the formula, let first look at the subtraction formula for when the first number is larger than the second number.
Step 1: Subtract a set of digits from A to a set of digit from B. If there is a carry-over value and the set of digits from A is zero in value then turn the set of digits from A to become maximum in value by the amount of digit that can be contained in the digits set. Also, keep a value of one for the carry-over value. If there is a carry-over value and the set of digits from A is not zero in value, reduce the digit value of the digits set from A to the carry-over value. Step 2a: If the digits set from A is subtractable to the set of digits from B, process the equation and keep the result value. Step 2b: If the digits set from A is not subtractable to the digits set from B, find the absolute value for the equation of the digits set from A subtract to the digits set from B. Then take a value that is one value larger than the maximum value that can be contained within the digits set, and subtract that value to the absolute value of the equation. Keep the result value of this procedure as the result value for the equation. Also, assign a value of one to the carry-over value. Step 3a: If number B does not have anymore digit and number A still has more digit(s) and there is a carry-over value. Use a value of zero in lieu of the digits set from B. Step 3b: If number B ran out of digit before number A and there isn't a carry-over value then copy all the digits from A that hasn't been processed by the equation in the order of they are to the left of the result value and move onto the final step of the formula. Step 4: Continually repeat the previous steps until the last set of digits in A. Final step: If A and B are negative in value, append the minus sign to the front of the result.
With the formula above in perspective and to simplify the explanation, this explanation for the formula was written for when we are subtracting the first number to the second number by calculating the equation at the rate of one digit at a time. In another term, we can say that the specified amount of digit in the digits set is one digit. When examining step one from the above formula, it can seem to be written in a complicated format. This simply because the equation just started. Nevertheless, the formula was written in a way that step one can be utilized whether if the equation is being applied to the first set of digits or the last set of digits.
At the start of the equation, we would first take the first rightmost digit from A and subtract that digit to the first rightmost digit from B. If the digit from A is subtractable to the digit from B then we write the equation result value down as the result value for the current procedure. If the digit from A is not subtractable to the digit from B, we find the absolute value of the equation. We then take 10 and subtract to the equation's absolute value. The result value from this sub-procedure is kept as the result value for the main procedure. When we can't subtract the digit from A to B without going into the negative value then we would also assign a value of one to the carry-over value. This carry-over value will be applied to the procedure of the next digit.
When defining the absolute value and how to obtain an absolute value of an equation, we should first examine the property of what is the absolute value. An absolute value is a distance value of how far a value is away from zero. In my opinion, zero is an absolute point of no value and zero is neither a negative or a positive value. Thus, in my explanation, I define the absolute value as the distance of how far away a value is from the absolute point of no value. In my opinion, a no value can also be defined as a nil value. When we are looking at the explanation, we know that a positive five would be five distance values away from zero. If we do not take into consideration the fractional part then it would take five whole values to travel from the absolute point of no value to get to the value of five. In a reversal order, it would also take five whole values to travel from the value of five back to the absolute point of nil value.
The property of the absolute value of a negative number would be the same as the property of the absolute value for a positive number. For example, the absolute value for a negative value of five would be a value of five. This is because it would take five whole values distance to travel from the absolute point of no value to get to the value of negative five. It would also take five distance values to travel from the negative value of five to get to the absolute point of nil value. Thus, the absolute value of any number is alway the positive value of that number. Hence, the absolute value of any equation is always the positive version of the equation's result value.
When it comes to a subtraction equation and absolute values, the result value from the equation of 8 - 3 and 3 - 8 would produce the same digit value as a result value. Nonetheless, the result values from the two equations are not the same in the negative and the positive base. However, the absolute values of the two result values from the two equations are exactly the same. Thus, if we want to find the absolute value of digit A subtracts to digit B without getting into the negative value then we can always subtract the digit B to the digit A in the event where the digit B is larger than the digit A.
When dealing with the carry-over value, in my formula for when the first number is larger than the second number is to subtract the carry-over value from a digit in A that is next to the digit of where the carry-over value was produced. When applying a carry-over value to a digit, if the digit in A is a nil value, or in other words, a zero value, then we simply increase the value of the digit in A to become the digit 9, which is the maximum value of what our digits set can hold in this circumstance of where we are processing the equation at the rate of one digit at a time. In a circumstance where we are processing the equation at the rate of two digits at a time, then a nil value digits set can be 00, and applying a carry-over value to that digits set would turn that digits set to become 99.
With my formula, when it comes to the last digit in the first number, the equation can never produce a carry-over value. This is because, if the first number is truly larger than the second number then at some point in time before the final digit or is the final digit, even when being applied to by a carry-over value, a digit from the first number would be larger than or equal to the value of digit in the same position from the second number. For example, in an equation of 81111 - 79999, the equation would produce plenty of carry-over values. Nevertheless, in the end, the digit eight neutralized all the carry-over value. For another example, in this equation of 78211111 - 78199999, in the first five equations for the first five digits, each equation would produce a carry-over value. Nonetheless, when it comes to the digit two, the digit two would neutralize the streak of carry-over values, and from that point to the last digit in the equation, the equation does not produce any more carry-over value.
In my formula, when the lower value number ran out of digits and there are still digits left in the higher value number, and there isn't a carry-over value from the previous equation then we simply move all the digits that hasn't been process by the equation from the higher value number to the left of the result value. For example, in this equation of 78298527 - 8527, after processing the first four digits that are 8527 - 8527, which would get us a result value of 0000, we simply moved the last four digits in the larger number that hasn't been processed by the equation to produce the final result value of 78290000 as a result value for the entire equation.
In this case of when the first number is the larger number and it comes to determining if the result value is going to be a positive value or a negative value, the result value would have the same negative or positive base as the two numbers. This is because, when the first number is larger than the second number, there is no possibility for the equation to cross the absolute point of no value and turned the result value into the opposite base as the two numbers.
To simplify this method, I wrote a second version of my subtraction formula. The simplified version can be found at the end of the article. The below are some example of how to apply my subtraction formula for when the first number is the larger number in a subtraction equation.
Example 1: 3558 - 1949 Step 1: 8 - 9 = 10 - |(8 - 9)| = 9 >>>> carry-over = 1 >>>> result = 9 Step 2: (5 - 1) = 4 - 4 = 0 >>>> carry-over = 0 >>>> result = 09 Step 3: 5 - 9 = 10 - |(5 - 9)| = 6 >>>> carry-over = 1 >>>> result = 609 Step 4; (3 - 1) = 2 - 1 = 1 >>>> result = 1609 Answer: 3558 - 1949 = 1609 Example 2: 78501 - 498 Step 1: 1 - 8 = 10 - |(1 - 8)| = 3 >>>>> carry-over = 1 >>>>> result = 3 Step 2: carry-over(0) = 9 - 9 = 0 >>>>> carry-over = 1 >>>>> result = 03 Step 3: (5 - 1) = 4 - 4 = 0 >>>>> carry-over = 0 >>>>> result = 003 Step 4: 78 to the front of 003 >>>>> result = 78003 Answer: 78501 - 498 = 78003
Step 1: Subtract a set of digits from A to B. If there is a carry-over value and the set of digits from A is a maximum value that can be contained in the digits set, then turn the set of digits to the minimum value that can be contained by the digits set. Also, produce a value of one for the carry-over value. If there is a carry-over value and the set of digits from A is not a maximum value that can be contained in the digits set then increase the digit value of the set of digits from A to the carry-over value. Step 2a: If the digits set from A is subtractable to the set of digits from B but does not produce a zero result value. Process the equation then take a value that is one value larger than the maximum value that can be contained within the digits set and subtract to the result value. Keep the result value of the procedure as the equation's result value. Also, assign a value of one to the carry-over value. Step 2b: If the digits set from A is not subtractable to the digits set from B or if subtract the digits set from A to the digits set from B would produce a value of zero. Then keep the absolute value for the equation of the digits set from A subtract to the digits set from B as the result value for the equation. Step 3a: If number A ran out of digits before number B and if there is a carry-over value, use a value of zero in lieu of the digits set from A. Step 3b: If A ran out of digit before B and there isn't a carry-over value, move all digits in B that hasn't been processed by the equation to the left of the result value in the order of they are, and then move onto the final step of the formula. Step 4: Repeat the previous steps until the equation is applied to all the set of digits in B. Final step: If A and B are positive in value, append the negative sign to the result value.
For my subtraction formula, when it comes to the second number is the larger number in the equation. There are only three differences for when we are comparing this formula to the formula for when the first number being the larger number. The first difference between the two formula is when do we keep the result value from a set of digits equation within the subtraction equation. The second difference between the two formula is regard to when does the equation produce a carry-over value. The last difference between the two formula is how does the carry-over value apply to the equation.
For when the second number is the larger number, if the digit from the first number subtracts to the digit from the second number produce a value of zero or less, we keep the absolute value of the equation as the result value. If the digit from the first number subtracts to the digit from the second number produce a result that is larger than zero, we would take 10 and subtract to the equation's result and the result value from this procedure is the result value, also, the equation would produce a value one for the carry-over value.
For dealing with the carry-over value, when the second number is larger in value, we would add the carry-over value to the next digit in the first number. This is the opposite of when the first number being the larger number, when the first number is larger, we would instead minus the digit in the first number to the carry-over value.
When we apply the carry-over value to the digit in the first number and the first number's digit value becomes larger than the maximum value the digits set can hold then we turn that value into a value of zero and assign one to the carry-over value. In other words, if we were to applied a carry-over value to the digit 9, then the digit 9 would become the digit 0, and we would assign a value of one to the carry-over value. In another circumstance of when we are processing the subtraction equation at the rate of two digits at a time and we are applying a carry-over value to the digits set of 99. Then the 99 would become 00 and the equation would produce a value of one as the carry-over value for next equation.
Almost similar to when the first number is the larger number. When the second number is the larger number and in the event that the first number ran out of digits first and there isn't any carry-over value. Then we simply move all the digit from the second number that hasn't been processed by the equation in the order of they are to the left side of the result value. When it comes to determining if the result value is a negative result value or is a positive result value. If both numbers are positive in value, then the result value would be a negative value. If both numbers are negative in value, then the result value would be a positive value. This is because, when the first numbers subtract to something larger than itself, then the result value would cross the absolute point of nil value, and therefore, the result value would be in the opposite negative or positive base as the two numbers.
The examples below is to demonstrate the formula in an equation and we are applying the equation at the rate of one digit at a time.
Example 1: 199 - 211 Step 1: 9 - 1 = 8 ; 10 - 8 = 2 >>>>> carry-over = 1 >>> result = 2 Step 2: carry-over(9) = 0 ; 0 - 1 = |(0 - 1)| = 1 >>>>> carry-over = 1 >>> result = 12 Step 3: carry-over(1) = 2 ; 2 - 2 = 0 >>> result = 012 Answer: 199 - 211 = -012 or is -12 Example 2: -35 - -12343 Step 1: 5 - 3 = 10 - 2 >>>>> carry-over = 1 >>> result = 8 Step 2: carry-over(3) = 4 ; 4 - 4 = 0 >>>>> carry-over = 0 >>> result = 08 Step 3: Since there isn't a carry-over value move 123 to the front of 08. >>> result = 12308 Answer: -35 - -12343 = 12308
For both formulas of when the first number is larger or smaller in value, the shortcut property is, whenever the first number's set of digits subtract to the second number's set of digit produce a result value of zero, we would not need to modify the result value nor would the equation produce any carry-over value. With that in mind, we can simplify some explanation for easy remembering. For example, when A > B, if A's digit subtracts B's digit is a positive value, we keep the result value. If A's digit subtracts B's digit is a negative value, then 10 - |(An - Bn)|, and we add one to the carry-over value. With the explanation of zero just right above, we do not need to remember what to do when the result from the equation of two lone digits subtracting to each other is a zero value. This is because whether if A is larger or B is larger, we simply just write zero as the result value and do not need to modify the value or produce any carry-over value. The only time the equation need to produce a carry-over value under this circumstance is when we applied a carry-over value to a type of digits set that would automatically produce a carry-over value for the next equation.
The below is the explanation of the formula in a simpler format. Nevertheless, the below formulas use negative and positive value as per digit equation rather than the term absolute value. In the formula below, I write the formula as a subtraction equation that is being applied at the rate of one digit at a time.
Formula: A > B (Starting from right to left) Step 1: Subtract a single digit from A to B. Step 2: If the result value is a positive value or is a value of zero, keeps the result value. Step 3: If the result value is a negative value. Convert the result value to a positive value. Then take ten and subtract to the result value. Also, assign one to the carry-over value. Step 4: Write down the result value starting from right to left. Step 5: Move to the next number. If there is any carry-over value, subtract A's digit to the carry-over value. If the result value from this step is a negative value, convert the result value to a positive value and assign a value of one to the carry-over value. Step 6: If the digit in B ended before A and when there isn't a carry-over value, copy all the digit from A that hasn't been processed by the equation in the order of they are to the left of the result value then move on to step eight. If there is a carry-over value, substitute a digit of zero in lieu of the digit from B. Step 7: Repeats the previous steps until there is no more digit in both A and B. Step 8: If (A, B) = negative then the result value is a negative result value.
Formula: A < B (Starting from right to left) Step 1: Subtract a single digit from A to B. Step 2: If the result value is a negative value or is a value of zero, convert the result value to a positive value and keep the result. Step 3: If the result value is larger than zero, subtract ten to the result value and assign a value of one to the carry-over value. Step 4: Write down the result value starting from right to left. Step 5: Move to the next number. If there is any carry-over value, add A's digit to carry-over value. If A's digit becomes zero in this step, assign one to the carry-over value. Step 6: If A ran out of digit before B, and there isn't a carry-over value, move all the digits from B that hasn't been processed by the equation in the order of they are to the left of the result and move onto step 7. If there is a carry-over value then use a value of zero in lieu of the digit from A. Step 7: Repeats the previous steps until there is no more digit in both A and B. Step 8: If (A, B) = positive then append a negative sign to the front of the result value.]]>
When we first talk about an infinity number, the first question would come up to mind is can we classify a number that is infinitely in length as a number? In my opinion, whether if the number is infinitely in length or not we can always classify the number as a number. The closest evidence that I can come up with to back up this theory is time. I would use time for comparison in this case.
As when we are looking at time as the subject, time is also infinitely in measurement. No one would know if time would flow infinitely or will time meet a wall and will end. For the property of time, we know that we can see events that occurred in the past but not the present nor the future. We may be able to predict the future bases on statical data but would not be able to see events that occurred in the future.
We, ourselves can never see events that occur in the present. The events that occur right in front of our eyes seem like the present, and our brain perceives it that way. Nevertheless, what we are seeing in our eyes are actually the past. It would take sometimes for light to hit an object and then reflect off the object into our eyes. When our eyes receiving the light, our brain would have to process the lights into images. Because of the speed of light and because of how fast our brain process data, the process of translating an event that occurs in front of our eyes into an image would be really fast and would be completed in less than a microsecond or a nanosecond. The fast speed of the process would let our brain perceive the event that occurred in front of us as if we are actually seeing the present, however, that event actually occurred less than a nanosecond in the past. In my opinion, in a true equation, whether if the event occurred one year ago or one nanosecond ago it would still be considered as an event of the past and not the present.
When we take this to a micro scale, even if we are standing in front of a mirror and looking at ourselves, the images that we are seeing in the mirror are still the images of the past and not the images of the present. On a larger scale, if we are looking at a planet that is one billion light year distance away from us, we are not seeing that planet present. We are actually seeing the past of the planet. It would take one billion years for light to travel from that planet to our planet, Thus, if that planet was destroyed or turned into a planet that habitable by life forms half a billion years ago, we would not know until half a billion years later. This same principle also applies to our closest star, which would be the Sun. If there are any changes on the surface of the Sun, we would not be able to detect the changes until at least eight minutes later.
Not just our eyes that can't see events that occurred in the present, the pain that we feel or the feeling of touching an object are also events that occurred in the past. It would take a very small amount of time for our nerve to be able to send a signal to our brain when an object touched our body, and with that signal from the nerve, our brain would be able to process the signal into what we can describe as feelings.
Even when we are playing games on the computer or typing something on the computer. To us, it almost seems like when we are typing something on the computer the words would be displayed so quickly that it is actually the present. Nevertheless, we, in fact, pressed the key on the keyboard some microseconds or nanoseconds ago.
Is there a method for us to be able to perceive event of the present? In my opinion, I do not believe at the current state we can perceive events that occur at their exact moment. Nevertheless, I do believe to be able to perceive events that occur right at the exact moment of when they occurred is to be able to move forward in time and look back at the events. Only then, we would be able to perceive events that occur at their exact moment. In short, to be able to see the present is to be able to see the future and look back in time. Would technology or would our body develop to give us the ability to perceive events of the present in the future? I would never know.
When comparing time to an infinity number, I would treat time as a mathematical equation. What events that we have seen that already occurred in the past would be similar to seeing numbers that we had already seen. Thus, when we are asking the question if an infinity number is a number would be similar to asking if the infinity time is time. Time is infinite in the amount and in my opinion, whether if the events in time have happened yet or not, they would still be considered as time. Therefore in principle, if we were to receive an infinite string of digits, the digits that we have seen are still numbers and the digits that we haven't seen are still numbers. In my opinion, it just because the numbers are infinitely in length does not mean that they are not a number. They are still numbers, numbers that are infinitely in length.
For an example, my birthday that happened last year would be considered as an event of time that had already occurred. My birthday of the future or how many birthdays would I have left would still be considered as events of time. These time events, however, have not occurred but are time events of the future. If my birthday of the future does not happen, another event of time would replace the time event of what suppose to be my birthday. Nevertheless, these events would still be considered as an event that occurs in time or an event of time.
In my conclusion, I concluded that a number, whether infinitely in length or not should still be considered a number. Therefore, an infinity number is still a number, it just that the infinity number is infinitely in length.
In my understanding, the general formula for calculating infinity numbers would be "Infinity equation Infinity = Infinity". For the case of this article, which would be the multiplication equation, the general formula for calculating infinity numbers would be "infinity x infinity = infinity".
General Formula: ∞ * ∞ = ∞
I have two formulas for calculating mathematical equations for infinity numbers. The first formula is for when we are calculating numbers that will vary in digits. Each digit in the string of infinity number would differ from one another. In my first formula, "Infinity Calculation is equal to Infinity Assumption and is equal to Infinity Alteration". In short, I write my formula as "infC = infA = infAL". continue reading
]]>I have been long not wanting to disclose the hackers' IPs that tried to attack the front http server and the back-end hosting. Nevertheless, everything will be disclosed, and all logs will be posted. When did and why did? I have a couple of stories to tell, I actually keep a log of all events. Nevertheless, it does not mean that those people did it. It just means a great coincidence. Also, hackers were getting free Amazon AWS Server to mask them as Google bot. That just tell you how cheap hacker now a day become. Digitalocean's hosting IP was also on the list. There are a little bit more than 30 IPs that guarantee hacker and around 100+ IP that could be hacking attempts. Nevertheless, these could be coming from only a handful of the individual.
Nevertheless, enough evidence is collected, it is almost the time to disclose the evidence and share the evidence to get some advice.
Currently, the Infinity subject is being delay for released until the end of April, 2017. This is because I am currently researching on how serious the hacking scheme was. --- The red text was written On April 18, 2017 continue reading
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