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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??

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Obligatory xkcd comic. – J. M. Dec 6 '10 at 13:43
up vote 3 down vote accepted

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 $b$.

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HINT $\rm\ \ \ a+b-c\ <\ a,b\ \iff\ a,b\ <\ c$

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