- #1
INdeWATERS
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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!
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!