Show that the integer n is even if and only if n - 2[n/2] = 0

Thanks very much.:)

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- Sep 11th 2006, 11:12 PMsuedenationNumber theory proof
Show that the integer n is even if and only if n - 2[n/2] = 0

Thanks very much.:) - Sep 12th 2006, 03:28 AMOReilly
- Sep 12th 2006, 03:29 AMOReilly
- Sep 12th 2006, 03:51 AMCaptainBlack
- Sep 12th 2006, 03:52 AMCaptainBlack
- Sep 12th 2006, 05:42 AMCaptainBlack
1. Suppose n is even, then it may be written n=2k, for some k in N.

Then n-2[n/2]=2k-2[k], but k is in N so [k]=k, therefore n-2[n/2]=2k-2k=0.

Which is half of the proof, we have proven that if n is even n-2[n/2]=0

2. Suppose n-2[n/2]=0, now n is in N, so may be written n=2k+r, where

r is in {0,1}. Then n/2=k+r/2, so [n/2]=k (assuming that [x] denotes the

greatest integer less than x, make your own arrangements for other

intended meanings), therefore n-2[n/2]=2k+r-2k, but this is zero, therefore

r=0, and so n is even.

Which is the other half of the proof, as we have proven that if n-2[n/2]=0

then n is even.

RonL - Sep 12th 2006, 09:08 AMtopsquark
Perhaps I'm not seeing something here, but shouldn't we also be showing that if n is odd then n - 2[n/2] is not 0? (It's not hard to do, it just seems necessary.) 2. already shows that if n - 2[n/2] = 0 then n is even so we don't have to cover any other case.

-Dan - Sep 12th 2006, 10:31 AMCaptainBlack