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? why three integer ? two is enough

Well, the question has been put up like that....

yeah, u hv given a trimmed version of what i did
basically u hv used Euclid's div lemma only

oh cool, i didnt know if i remembered it or not
you are probably correct then

Therefore, taking b as 2 we get a = 2q + r and \[0\le r < 2\]

So, if r = 0, then a = 2q + 0 = 2q
If r = 1, then a = 2q + 1

Hence all positive integers are of the form 2a and 2q+1

right and if a number can be represented in the 2q form it can be said it is divisible by 2

Yes, so for my question I hv two possibilities
1) 2q, 2q+1 and 2q
2) 2q+1, 2q, 2q+1

correct

Thanks for confirming..U get a medal !!