1. ## Proof by contrapositive

I am having a terrible time understanding this one:

Prove that for all x E r, and y E R, if x is rational and y is irrational, then x + y is irrational. use proof by contradiction known as contrapositive.

OK I understand that contrapositive (~p --> ~q) so

Proof: if x is not rational then y is not irrational

Now I have no idea what to do next?

Please help, I have a test coming up in two days and I need to grasp this stuff.

2. Originally Posted by robasc
I am having a terrible time understanding this one:

Prove that for all x E r, and y E R, if x is rational and y is irrational, then x + y is irrational. use proof by contradiction known as contrapositive.

OK I understand that contrapositive (~p --> ~q) so

Proof: if x is not rational then y is not irrational

Now I have no idea what to do next?

Please help, I have a test coming up in two days and I need to grasp this stuff.
So assume the negation of the conclusion.

Original conclusion: x + y is irrational
Negation of it: x + y is rational

So under this assumption you can write x + y = r, where r is in Q (rationals). Now what happens when you solve for y?

3. I have no idea, I am assuming that the answer will be false because a rational number and a irrational number will conclude to be an irrational number but how do I prove that?

4. Originally Posted by robasc
I have no idea, I am assuming that the answer will be false because a rational number and a irrational number will conclude to be an irrational number but how do I prove that?
What does "answer will be false" mean?

x + y = r with r in $\mathbb{Q}$, by assumption
$\implies y = r - x$

recall that $\mathbb{Q}$ is closed under + and -. Thus y is rational. Contradiction.