How would you go about proving these:
For all integers n, n is even if and only if n -1 is odd
if d|(a+b) and d|a, where a,b and d are integers, d is not equal to 0
Suppose n is even then n = 2m for some m.
therefore n = 2m
obviously n-1=2m - 1 is odd
if n- 1 is odd then (n-1) + 1 = is n which is even for n-1 is odd.
we can also say that for n-1 being odd implies em-1 is odd where n = 2m for some m.
thus (n-1) = (2m-1)-1 = 2m-2 = 2(m-1) which is even.
Hence the proof