Can somebody prove that when we add 2 bit integers and if there is an overflow then the result always will be lesser than the 2 operands used??
Asked
Active
Viewed 155 times
2
-
3Obligatory xkcd comic. – J. M. ain't a mathematician Dec 06 '10 at 13:43
2 Answers
3
Suppose $a$ and $b$ are the numbers you are adding. Suppose that $0 \le a,b < 2^n$. Then we must have that $a+b - 2^n < a$ and $a+b - 2^n < b$. Hence if overflow occurs the result is smaller than $a$ or $b$.
J. J.
- 9,432
