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In binary subtraction i have seen some tutorial and some sites that during 0-1 process borrow value from next 1s.

Let's say one of tutorial is this

http://courses.cs.vt.edu/csonline/NumberSystems/Lessons/Subtraction/

It's first rule

To compute the first column, we need to borrow a 1 from the next column. Recall that two 1s generated a carry in addition. If we reverse this process, we can borrow a 1 from the second column and mark two 1s in the first column.

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Why carry two 1s generated a carry in addition ? Can anyone please explain ?

Anik Islam Abhi
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  • You should post the example that is causing you trouble, so that others don't have to watch the video and in case of link rot. Reading its first "rulse" (is that rule?) it sounds like borrowing in base 10 subtraction-you take one from the next place and replace it with 2 or 10 (or whatever the base is) in the current place. This is the concept they are trying to teach. – Ross Millikan Apr 07 '15 at 03:52
  • To compute the first column, we need to borrow a 1 from the next column. Recall that two 1s generated a carry in addition. If we reverse this process, we can borrow a 1 from the second column and mark two 1s in the first column. I just wanna to know why carry two 1s generated in addition – Anik Islam Abhi Apr 07 '15 at 03:55
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    In decimal, you generate a carry any time the sum exceeds $10$. In binary, you generate a carry any time the sum exceeds $2$. As the only bits are $0$ and $1$, you can only generate a carry from $1+1$ if you are adding two numbers. If you are adding a long column, you can generate many carries, just like in base 10. – Ross Millikan Apr 07 '15 at 03:58

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It looks like you are doing $10_2 - 1_2 = 1_2$ Remember what the positional notation means in base $2$: Each place is twice the previous one. Expressedin base $10$, this is $2-1=1$. When you do $$\begin {align}10_2&\\ \underline{-\quad1_2}&\\ 1_2&\end {align}$$ you recognize that the $1$ in the twos place in the top line represents two in the ones place (but you don't have a symbol for that in base $2$). It is exactly the same as $$\begin {align}10_{10}&\\ \underline{-\quad6_{10}}&\\ 4_{10}&\end {align}$$

Ross Millikan
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