Originally Posted by
CaptainBlack 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