Let  and  be real numbers. If  is irrational, then  is irrational or  is irrational Let  and  be real numbers. If  is irrational, then  is irrational or  is irrational.
Prove by contrapositive
I believe the contrapositive is if x is rational or y is rational, then xy is rational
How do I go about proving this?
 A: The contrapositive of ($xy$ is irrational $\implies x$ is irrational or $y$ is irrational) is:
$x$ is rational and $y$ is rational $\implies xy$ is rational.
Pf:  Let $x = \frac ab; a,b\in \mathbb Z$  And $y = \frac cd; c,d\in \mathbb Z$.
The $xy = \frac ab \frac cd = \frac {ac}{bd}$.  And $ac\in \mathbb Z$ and $bd\in \mathbb Z$ so $\frac {ac}{bd}$ is rational.
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FWIW
It's worth having the following under your belt:
$rational \times rational = rational$.  Pf:  $\frac mn \frac jk = \frac {mj}{nk}$.
$\underbrace{rational}_{\text{not equal to zero}} \times irrational = irrational$.  Pf:  $\frac ab\times irrational=?????? \implies irrational = ????? \times \frac ba=?????\times rational$.  If $?????$ is rational then we have $irrational = rational\times rational$ which we just proved was impossible.
$irrational \times irrational = impossible\ to\ tell$.  Knowing that what each of the multiplicands can't be doesn't tell us what the product can or cant be.  Example $\sqrt 2 \times \sqrt 8=\sqrt{16} =4$.  But $\sqrt 2\times \sqrt 3 = \sqrt 6$.
A: Actually you got the contrapositive slightly (but significantly) wrong.
The negation of $x$ irrational OR $y$ irrational is (both) $x$ AND $y$ are rational.
Can you proceed from this point? Just let $x = \frac ab, y = \frac cd$.
A: Your contrapositive should have and, not or, between "$x$ is rational" and "$y$ is rational". Otherwise it's correct, and the proof is direct: let $x=a/b,y=c/d$ and multiply.
