Consider the statement: For all x, y, z ∈ Z. At least one of x− y, x− z and y − z is even
I want to check if my proof is correct via contradiction:
Suppose(for a contradiction) that all of x-y, x-z, and y-z are odd:
x - y = 2k+1 for some k ∈ Z
rearranging for x:
x=2k+1+y
y - z = 2n+1 for some n ∈ Z
rearranging for z:
z=y-2n-1
taking x-z and subbing in
x - z =2k+1+y-(y-2n-1)
= 2k+2n+2
=2(k+1+n)
therefore x-y is even which is a contradiction therefore the original statement is true
On
That's fine. ANd you proof is good.
Alternatively:
By pigeon hole principle, if we are given three items, each with two possible options, then at least two of the items must have the same option, as for them all to be different options would require at least three options. That's why if you flip three coins two of them must agree on heads or tails or if you pick three socks from a drawer of red and blue two must match.
Likewise given $x,y,z$ each either odd of even, then two of them are the same parity. (Because there you can't have all three be three different parities!)
And lemma: If $a,b$ are the same parity then $a-b$ is even.
Pf: If $a=2m$ and $b =2n$ are even then $a-b = 2(m-n)$ is even. If $a=2m-1$ and $b=2n-1$ are odd then $a-b = (2m-1)-(2n-1)= 2(m-n)$ is even.
And that's it. Of $x,y,z$ two of them are the same parity and depending on which of the two $x-y$ or $y-z$ or $x-z$ is even. (If $x,y$ are same parity $x-y$ is even; if $x,z$ are same parity then $x-z$ is even; if $y,z$ are same parity then $y-z$ is even.)
Yes your proof looks fine.
Another way to think about this is to notice that the sum $$ ( x - y) + (y - z) + (x-z) = 2 (x-z) $$ is even, so the numbers $(x-y), (y-z), (x-z)$ can't all be odd.
If you want to continue posting here you should consider learning LaTeX. It will make your equations easier to read.