Show that the integer n is even if and only if n - 2[n/2] = 0
Thanks very much.:)
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Show that the integer n is even if and only if n - 2[n/2] = 0
Thanks very much.:)
Does n - 2[n/2] = 0 means $\displaystyle n - 2\frac{n}{2} = 0$?Quote:
Originally Posted by suedenation
Why LaTex isn't shown?!Quote:
Originally Posted by OReilly
The software driving MHF has been upgraded and LaTeX has notQuote:
Originally Posted by OReilly
yet been reinstalled.
RonL
It should mean that there exists a k in N, such that:Quote:
Originally Posted by OReilly
n=2k
It could be either the floor, ceiling of some sort of nearest integer function,
as these are n/2 if this is an integer, and something else if not an integer.
RonL
1. Suppose n is even, then it may be written n=2k, for some k in N.Quote:
Originally Posted by suedenation
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
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.Quote:
Originally Posted by CaptainBlack
-Dan
No. If and only if means that if n - 2[n/2] = 0, then n is even, and ifQuote:
Originally Posted by topsquark
n is even then n - 2[n/2] = 0.
It is a short hand for bi-implication.
Also as a consequence if n is not even then n - 2[n/2] != 0, similarly
if n - 2[n/2] != 0 then n is not even.
RonL
