Math 2200 Notes on 2.6
Summary
In Exercise 2.6.1, (b), (e), (f), and (g) are false, and the rest are true. In Exercise 2.6.2, (c) and (d) are false,
while (a) and (b) are true. I’ll prove a few of the hardest problems below.
I cannot overemphasize the value of thinking about whether a claim is plausible before plunging into a
proof! I will throw plenty of false claims at you during the course of the semester, so be ready to find them
and dispose of them by supplying counterexamples.
Exercise 2.6.2 (b)
Prove or disprove the following claim:
Claim. Let r and s be real numbers. If r is rational and s is irrational, then r + s is irrational.
I said in class that a claim stated in the form “if ... then” will never require a proof by contradiction
since a proof by contrapositive will amount to the same thing. I was wrong, and this claim is an example.
You can’t get very far with the contrapositive, because all you have to work with is the assumption that
r + s is rational. But a proof by contradiction gives you three assumptions: r is rational, s is irrational, and
r + s is rational, with which you can find a contradiction in the same way as in Exercise 2.4.2 (a). Here goes:
Proof. Suppose r is rational and s is irrational. Assume for contradiction that r + s is rational. Then there
are integers a and b with b != 0 such that r + s = ab . Since r is rational, there are integers c and d such that
r = dc . Then s = ab − dc = ad−bc
is rational, contradicting the fact that s is irrational.
bd
Exercise 2.6.2 (d)
Prove or disprove the following claim:
Claim. Let r be a real number. Then r is rational if and only if r2 is rational.
It wouldn’t be completely fair of me to include this claim in a learning √celebration
√ because although
the claim is false, in order to give a counterexample you need to know that
2
(or
3, etc.) is irrational,
√
which we haven’t proven in class. Don’t worry too much about proving 2 is irrational. Still, the proof
showcases the power of proof by contradiction, so I’ve included it below for your enjoyment. I also strongly
recommend that you attend Dan Ciubotaru’s talk on irrational numbers in the Undergraduate Colloquium
(see http://www.math.utah.edu/ugrad/colloquia.html for details.) Here’s a way to disprove the above claim:
This claim is false because ( ⇐= ) is false! Counterexample: let r =
irrational.
√
2. Then r2 = 2 is rational but r is
(Note that ( =⇒ ) is actually true, but the claim is still false because ( ⇐= ) is false.)
1
Exercise 2.6.2 (c)
Claim. If r and s are irrational, then rs is irrational.
This claim is easy to disprove once you know a counterexample for (d), because the same counterexample
works! Here is what you could write:
This claim is false! Counterexample: let r = s =
√
√
2. Then r and s are irrational, but rs = 2 is rational.
2 is Irrational
The proof is a bit more complicated than the proofs we’ve been doing and requires some facts we haven’t
mentioned. I don’t expect you to be able to prove things that are this difficult (yet!), but I will present the
proof anyway, because it is a classic example of proof by contradiction:
√
Claim. 2 is irrational.
√
Proof.
Assume for contradiction that 2 is rational. Then there are integers a and b with b != 0 such that
√
2
2 = ab . Moreover, we may choose a and b to have no common factors larger than 1. Then 2 = ab2 , so
a2 = 2b2 . Thus a2 is even, which implies that a is even. So there exists an integer k such that a = 2k. Then
2b2 = (2k)2 = 4k 2 , so b2 = 2k 2 . Thus b2 is even, which implies that b is even. But we have shown that both
a and b are even, which contradicts the fact that a and b have no common factors larger than 1.
2