Math – Kevin’s Subtraction Formula

Before going into the description of my formula for a subtraction equation, I would first give the summary description of the general formula for a subtraction equation that would most likely be used on a daily basis.

I also wrote a subject on how to calculate Infinity Numbers. If you are interested in learning infinity calculation, click here.

General Mathematical Subtraction Formula

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.

My Mathematical Subtraction 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.

The formula for when the first number is larger or A > B

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 

The formula for when the first number is smaller or A < B

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

Summary And The Easy Explanation

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.
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This post was written by Kevin and was first post @ http://kevinhng86.iblog.website.
Original Post Name: "Math – Kevin’s Subtraction Formula".
Original Post Link: http://kevinhng86.iblog.website/2017/04/10/math-subtraction-formula/.

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