Write down careful proofs of the following statements:
a) sqrt(6)- sqrt(2) > 1
b) If n is an integer such that n^2 is even, then n is even.
c) If n= m^3- m for some integer m, then n is a multiple of 6
The Attempt at a Solution
I will rely on P - > Q and not Q - > not P to try and prove the above statements.
a) I think I cant use direct proof (P - > Q) here, because no matter how true Q is, it wont say anything important about P which is the statement sqrt(6)- sqrt(2) > 1. Instead, I will use contradiction.
Contradiction: not P - > Q(false), not Q - > P
sqrt(6)- sqrt(2) ≤ 1
6- sqrt(24) + 2 ≤ 1
2sqrt(12) ≤ 3 (false statement). Negating this statement makes sqrt(6)- sqrt(2) > 1 true.
b) n^2 - > n= 2m
If we assume not P, then n^2 can be assumed to be odd: (1+ 2m)^2= 1+ 4m+ 4m^2= 1+ 4(m+m^2) which is false because we are given that n^2 is even.
From here it's obvious that n should be even to avoid contradiction.
c) not P: 1+ 6m = m^3- m where n is not a multiple of 6.
1+ 6m = m^3- m
0= m^3- 7m -1 ( I will assume that this result is false because solving for x introduces imaginary numbers which I am not sure if I am allowed to work with).
Obviously, n is a multiple of 6.