Consider the statement: For all $x, y, z ∈ Z$. At least one of $x−y$, $x−z$ and $y−z$ is even. Prove this statement by contradiction.
So the contradictory statement would be that neither one of $x−y$, $x−z$ and $y−z$ is even, eg they're all odd. I know how to do the proof, but with proof by contradiction, would I be allowed to let $x$,$y$, and $z$ equal actual integers, eg $1,2,3$ or would I have to make them equal eg. $2k+3$, $2m+6$ and $2n+8$ to complete this proof?
On
Assume WLOG that $x-y$ and $y-z$ are both odd (if at least one is even, we are done). However, this means that $(x - y) + (y - z) = x - z$ is even. $\blacksquare$
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With regards to the question "Would I be able to let $x,y,z$ equal actual integers?":
No. One example of the converse being right is sufficient to disprove a claim, however one example of the converse being wrong is not sufficient to prove it.
You must assume that $x-z, y-z, x-y$ are arbitrary odd numbers, and find a way to show that this makes no sense.
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Alternative approach : pigeonhole principle.
Given integers, $a,b$ then $2 | (a-b) \iff a,b$ have the same parity (odd or even).
There are only two parities : odd or even. Given three #'s, $x,y,z$ then two of them must have the same parity.
On
Fix $x,y,z\in\mathbb{Z}$ and suppose all of $x-y, x-z,y-z$ were all not even. Then they must all be odd. We know that an odd integer plus or minus an odd integer returns an even integer, so we will check some cases.
$(x-y)-(x-z)=x-y-x+z=z-y$ which is even. However, because we know $-(y-z)=z-y$, then we have that an even integer is equal to an odd integer, a contradiction.
You can check through the other possible cases to show that none of the configurations will hold.
By the way of contradictio, assume that $x-y$, $z-x$ and $y-z$ are odd numbers.
By adding all three, you get: $(x-y)+(z-x)+(y-z)=0$. But the sum of three odd numbers must be odd. And here you have a contradiction.
MOREOOVER. This proof says that it is also impossible to have exactly 1 odd numbers among them.