What you are trying to prove is FALSE:
$\sqrt{2} + (-\sqrt{2}) = 0$
and 0 is certainly rational.
Prove the statement: If x is irrational and y is irrational, the x + y is irrational.
Contrapositive: If x + y is rational, then x is rational or y is rational.
My best guess at an approach is to prove the contrapositive. This is how I would start:
Proof:
Suppose x + y is rational. Then, by defn., x + y = p/q for some intergers p,q where q ≠ 0...
I'm not sure what to do from here. How do I deduce that x is rational or y is rational?
About the only "non-trivial" assertion we have along these lines is:
If $x$ is rational, and $y$ irrational, then $x+y$ is irrational.
Proof: (by contradiction)
If $x$ is rational and $y$ is irrational, but $x+y$ is rational, we then have $y = (x+y) - x$ is rational (since the rationals form an abelian group under addition), contradiction.
Yea, I have to prove that one too. But I was told to prove the contrapositive here. That is, assume x + y is rational and x is rational and then duduce that y is rational. A little less intuitive but I suppose it still gets the job done.
It's the same proof:
If $x+y$ is rational, and $x$ is rational, then $-x$ is rational
(for if $x = \dfrac{p}{q}$, then $-x = \dfrac{-p}{q}$ and $-p$ is an integer if $p$ is).
Then, surely, $y = (x+y) + (-x)$ is rational, because the sum of any two rational numbers is rational:
let $r_1,r_2$ be rational, so:
$r_1 = \dfrac{a}{b}, a,b,\in \Bbb Z, b \neq 0$
$r_2 = \dfrac{c}{d}, c,d,\in \Bbb Z, d \neq 0$
then $r_1 + r_2 = \dfrac{ad + bc}{bd}$ and clearly $ad + bc$ is also an integer, and so is $bd$ and furthermore, $bd \neq 0$.
In general, any proof by contradiction (of an implication) can be turned into a "straight proof" of the contrapositive, and vice-versa.