Regarding the subtraction and borrowing a digit from upper digits, I know how that works for more than one digit numbers. However, I can not figure out for one digit numbers! It is an obvious thing since we were in school. But I need a mathematically explanation.


What is the result of 21-13?

 2 (10^1)       1 (10^0)
-1 (10^1)      -3 (10^0)

For the first digit, we know that 1<3, so we have to borrow from the next digit. The next digit is 2*10 and the borrowed value will be 10. Therefore, the first number is changed to 1 (10^1) (10+1) (10^0). Now we are able to subtract 3 from 11 and then 1 from 1. The results will be 8


Now assume, we want to find the result of 1-3 with the above mentioned methodology. We know that

 0 (10^1)        1 (10^0)
-0 (10^1)       -3 (10^0)

We have to borrow a 10 from the second digit but it is 0. There is also no more non-zero digit in the left (all left digits are zero means 0...000001).

So how we reach 1-3=-2??


According to the comments and answers, borrowing is not a universal method and may fail if there is no 1 in the left. The question now is, what is the universal method for subtraction?

  • $\begingroup$ The usual way is just to recognize that $1-3 = -(3-1)$ so we can always just subtract the bigger number (in absolute value terms) from the smaller and then add a negative sign at the end if needed. $\endgroup$ – user275324 Sep 30 '15 at 17:02
  • $\begingroup$ The borrowing rule simply does not apply when you want to make a subtraction $x-y$ with $x<y$. $\endgroup$ – Crostul Sep 30 '15 at 17:04
  • $\begingroup$ Simple: you don't use this method for single digits. This method isn't exactly a sophisticated mathematical algorithm or anything- it's just a heuristic taught in middle school. For $1-3 = -2,$ the same teachers who taught us that same heuristic also taught us to move "one unit right" on the 1-D number line, and then "3 units left" to land on -2. $\endgroup$ – daOnlyBG Sep 30 '15 at 17:05
  • $\begingroup$ @Crostul That's not accurate. Borrowing still works. It just can't be applied to this problem. $\endgroup$ – Akiva Weinberger Sep 30 '15 at 17:05
  • $\begingroup$ If you think about it, the borrowing method gives the answer as $-10+8$. Which is correct, it's just not in simplest terms. $\endgroup$ – Akiva Weinberger Sep 30 '15 at 17:21

There's more than one way to get to an answer. The borrowing method is just one method.

As it happens, the borrowing method isn't particularly useful for $1-3$. (We have nothing to borrow from!) So we use another method.

On thing we can do is notice that: $$(1-3)+(3-1)=0$$ (Just remove the parentheses. Everything cancels.) Now, any two things that add up to zero are opposites of each other! That means that $(1-3)=-(3-1)$, or $1-3=-(2)=-2$.

In general, $x-y=-(y-x)$. If you ever have to subtract a large number from a smaller one, just switch them around and put a minus sign in front.

  • $\begingroup$ (Some of you may notice how I've been influenced by group theory in my second-to-last paragraph.) $\endgroup$ – Akiva Weinberger Sep 30 '15 at 17:24
  • $\begingroup$ This answer is OK. But you said in the comments that borrowing still works for one digits. $\endgroup$ – mahmood Sep 30 '15 at 17:54
  • $\begingroup$ @mahmood I said, "Borrowing still works. It just can't be applied to this problem." $\endgroup$ – Akiva Weinberger Sep 30 '15 at 17:55
  • $\begingroup$ OK I saw your last comment in the OP $\endgroup$ – mahmood Sep 30 '15 at 17:58
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    $\begingroup$ I'll say again: "If $x\ge y$, compute $x-y$ (using borrowing). If $x<y$, compute $-(y-x)$ (use borrowing for $y-x$ and stick a minus sign in front)." $\endgroup$ – Akiva Weinberger Sep 30 '15 at 18:19

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