Prove the irrationality of ar+s or ar-s using proof by contradiction

[INdeWATERS]

Use proof by contradiction to prove the following: Let a be an irrational number and r a nonzero rational number. Prove that if s is a real number, then either ar+s or ar-s is irrational.

I am stuck with this proof. Here's what I have so far,

Proof Suppose, by way of contradiction, that this is not true. Then there exists a real number s such that it is not the case that ar+s or ar-s is irrational. By DeMorgan's Law we have ar+s and ar-s are both irrational. So choose s such that...??

(a) I have proved in class that an irrational number times a nonzero rational number (ar) is irrational, so no need to include that proof in the proof.
(b) Do I need to choose an s so that ar+s or ar-s is irrational? Would it suffice to let s be a rational number?

Thanks for the help!

 

[Dick]

The contradiction of "either ar+s or ar-s is irrational" is "both ar+s and ar-s are rational". Try it from there.

 

[INdeWATERS]

The contradiction of "either ar+s or ar-s is irrational" is "both ar+s and ar-s are rational". Try it from there.

Proof: Suppose, by way of contradiction, that this is not true. Then there exists a real number s such that both ar+s and ar-s are rational.

I still don't see how to derive a contradiction. What should I choose s to be?

Thanks for you help!

 

[Dick]

Proof: Suppose, by way of contradiction, that this is not true. Then there exists a real number s such that both ar+s and ar-s are rational.

I still don't see how to derive a contradiction. What should I choose s to be?

Thanks for you help!

You don't have to choose s to to be anything in particular. If ar+s and ar-s are rational, then their sum is rational, isn't it?