MATH 308 — Lec 9 — Existence & Contradiction

Lecture 9 — Existence Proofs and Proof by Contradiction
MATH 308  Bridge to Advanced Mathematics  Chartrand §5.2, 5.4

Goals for today. By the end of this lecture you should be able to:

• prove an existence statement “$\exists x\, P(x)$∃xP(x)” by exhibiting a witness (a constructive proof);

• recognize a nonconstructive proof — one that shows something exists without producing it;

• write a proof by contradiction: assume the negation, derive an impossibility, conclude;

• prove the two landmark theorems this method was made for: $\sqrt{2}$2​ is irrational, and there are infinitely many primes.

Existence proofs (§5.4)

Back in Lecture 4 we met the existential quantifier: “$\exists x \in S,\ P(x)$∃x∈S, P(x)” asserts that at least one $x$x in $S$S makes $P(x)$P(x) true. Such a statement is called an existence statement. There is a delightfully direct way to prove one.

Definition 1. A proof of “$\exists x \in S,\ P(x)$∃x∈S, P(x)” is constructive if it exhibits a specific $x_0 \in S$x0​∈S and verifies $P(x_0)$P(x0​). That $x_0$x0​ is called a witness. A proof is nonconstructive if it establishes that such an $x$x exists without producing one.

To prove “there exists,” you do not need every $x$x to work — you need one. So the whole game is: find a witness, then check it.

Example 1 (Constructive). There exist integers $x,y$x,y with $7x + 5y = 1$7x+5y=1. Take $x_0 = 3$x0​=3 and $y_0 = -4$y0​=−4. Then $7(3) + 5(-4) = 21 - 20 = 1$7(3)+5(−4)=21−20=1. $\square$□

We did not solve an equation or explain how we found $(3,-4)$(3,−4); a proof only has to present a witness and confirm it works. (Scratch work stays on scratch paper — compare the “read it backwards” remark of Lecture 8.)

Example 2 (Constructive). There is a prime number that is even. The witness is $2$2: it is prime, and it is even. Notice this single example also disproves “every prime is odd” — the seed of counterexamples, our topic next time (Lecture 10).

Sometimes we can prove a thing must exist even though we cannot lay hands on it. The showpiece is the following, and it borrows one fact we will nail down in §2: $\sqrt{2}$2​ is irrational.

Theorem 1 (A nonconstructive existence proof). There exist irrational numbers $a$a and $b$b such that $a^{b}$ab is rational.

Proof. Consider the number $\sqrt{2}^{\,\sqrt{2}}$2​2​. It is either rational or irrational.

• If $\sqrt{2}^{\,\sqrt{2}}$2​2​ is rational, take $a = b = \sqrt{2}$a=b=2​. Both are irrational (§2), and $a^{b} = \sqrt{2}^{\,\sqrt{2}}$ab=2​2​ is rational. Done.

• If $\sqrt{2}^{\,\sqrt{2}}$2​2​ is irrational, take $a = \sqrt{2}^{\,\sqrt{2}}$a=2​2​ and $b = \sqrt{2}$b=2​. Both are irrational, and

  $a^{b} = \left(\sqrt{2}^{\,\sqrt{2}}\right)^{\!\sqrt{2}} = \sqrt{2}^{\,(\sqrt{2}\cdot\sqrt{2})} = \sqrt{2}^{\,2} = 2,$ab=(2​2​)2​=2​(2​⋅2​)=2​2=2,

which is rational. Done.

One of the two cases certainly holds, so a suitable pair $(a,b)$(a,b) exists. ◻

Remark 1. This is genuinely nonconstructive: the argument never tells us which case is the true one, so at its end we still cannot name a specific witness pair. We have proved existence without exhibiting an example — a perfectly valid, and slightly unsettling, kind of proof.

Proof by contradiction (§5.2)

Suppose we want to prove a statement $R$R. Proof by contradiction (Latin: reductio ad absurdum) turns the job inside out:

$\text{assume } \sim\! R \text{ is true, reason until you reach an impossibility, and conclude } R.$assume ∼R is true, reason until you reach an impossibility, and conclude R.

The “impossibility” is a contradiction: a statement of the form $C \wedge (\sim\! C)$C∧(∼C) — something that cannot be true (recall negation, Lecture 3). Since our reasoning was valid, the only thing that could have gone wrong is the assumption $\sim\! R$∼R; so $\sim\! R$∼R is false, i.e. $R$R is true.

The shape of a proof by contradiction: the assumption $\sim\! R$∼R is the only input, so a contradiction convicts it.

A gentle first example fixes the pattern.

Proposition 2. There is no largest integer.

Proof. Suppose, for contradiction, that there is a largest integer; call it $N$N. Then $n \le N$n≤N for every $n \in \mathbb{Z}$n∈Z. But $N + 1 \in \mathbb{Z}$N+1∈Z and $N + 1 > N$N+1>N, so $N$N is not an upper bound after all. This contradicts “$n \le N$n≤N for every $n$n” (take $n = N+1$n=N+1). Hence no largest integer exists. ◻

Irrationality of $\sqrt{2}$2​

A real number is rational if it equals $\tfrac{a}{b}$ba​ for integers $a,b$a,b with $b \ne 0$b=0, and irrational otherwise. Every rational number can be written in lowest terms: $\tfrac{a}{b}$ba​ with $a,b$a,b sharing no common factor bigger than $1$1 (in particular, not both even). We need one small fact, proved — as promised in Lecture 5 — by contrapositive.

Lemma 3. For every integer $n$n: if $n^{2}$n2 is even, then $n$n is even.

Proof. We prove the contrapositive: if $n$n is odd, then $n^{2}$n2 is odd. Suppose $n$n is odd, say $n = 2k+1$n=2k+1 with $k \in \mathbb{Z}$k∈Z (Lecture 5). Then

$n^{2} = (2k+1)^{2} = 4k^{2} + 4k + 1 = 2\,(2k^{2}+2k) + 1,$n2=(2k+1)2=4k2+4k+1=2(2k2+2k)+1,

which is odd. By contraposition, $n^{2}$n2 even $\Rightarrow n$⇒n even. ◻

Theorem 4 (Pythagoras). $\sqrt{2}$2​ is irrational.

Proof. Suppose, for contradiction, that $\sqrt{2}$2​ is rational. Then we may write

$\sqrt{2} = \frac{a}{b}, \qquad a,b \in \mathbb{Z},\ b \ne 0,\ \text{in lowest terms (so $a,b$ are not both even).}$2​=ba​,a,b∈Z, b=0, in lowest terms (so a,b are not both even).

Squaring gives $2 = \dfrac{a^{2}}{b^{2}}$2=b2a2​, hence

$a^{2} = 2b^{2}. \tag{$\ast$}$a2=2b2.(∗)

By $(\ast)$(∗), $a^{2}$a2 is even, so by Lemma 3 $a$a is even; write $a = 2c$a=2c with $c \in \mathbb{Z}$c∈Z. Substituting into $(\ast)$(∗),

$(2c)^{2} = 2b^{2} \ \Rightarrow\ 4c^{2} = 2b^{2} \ \Rightarrow\ b^{2} = 2c^{2}.$(2c)2=2b2 ⇒ 4c2=2b2 ⇒ b2=2c2.

So $b^{2}$b2 is even, and again by Lemma 3 $b$b is even. But now $a$a and $b$b are both even — contradicting our choice of $\tfrac{a}{b}$ba​ in lowest terms. Therefore $\sqrt{2}$2​ is not rational; it is irrational. ◻

The diagonal of a unit square has length $\sqrt{1^2+1^2}=\sqrt{2}$12+12​=2​. Theorem 4 says this length is incommensurable with the side: no common unit divides both — the discovery that shook the Pythagoreans.

Infinitely many primes

Recall that an integer $p \ge 2$p≥2 is prime if its only positive divisors are $1$1 and $p$p; an integer $\ge 2$≥2 that is not prime is composite. We use one structural fact.

Lemma 5. Every integer $n \ge 2$n≥2 has at least one prime divisor.

Proof. Among the divisors of $n$n that are $\ge 2$≥2 there is a smallest one (there is at least one such divisor, namely $n$n itself); call it $p$p. We claim $p$p is prime. If not, $p$p would have a divisor $d$d with $1 < d < p$1<d<p; but $d \mid p$d∣p and $p \mid n$p∣n give $d \mid n$d∣n (transitivity, Lecture 7), so $d$d is a divisor of $n$n with $2 \le d < p$2≤d<p — contradicting the minimality of $p$p. Hence $p$p is prime, and $p \mid n$p∣n. (That a smallest such divisor exists is the Well-Ordering Principle, made rigorous by strong induction in Lecture 12.) ◻

Theorem 6 (Euclid). There are infinitely many prime numbers.

Proof. Suppose, for contradiction, that there are only finitely many primes. List them all:

$p_1, p_2, \dots, p_k.$p1​,p2​,…,pk​.

Form the integer

$N = p_1 p_2 \cdots p_k + 1.$N=p1​p2​⋯pk​+1.

Since $N \ge 2$N≥2, Lemma 5 gives a prime $p$p with $p \mid N$p∣N. As our list contains every prime, $p = p_i$p=pi​ for some $i$i, so $p \mid p_1 p_2 \cdots p_k$p∣p1​p2​⋯pk​ as well. A common divisor of two integers divides their difference (Lecture 7), so

$p \ \Big|\ \bigl(N - p_1 p_2 \cdots p_k\bigr) = 1.$p ​ (N−p1​p2​⋯pk​)=1.

But then $p \le 1$p≤1, contradicting $p \ge 2$p≥2 (every prime is at least $2$2). Hence the assumption fails: there are infinitely many primes. ◻

The primes $2,3,5,7,11,13,\dots$2,3,5,7,11,13,… thin out but never stop. Euclid’s proof builds, from any finite list, a number forcing a prime outside it.

Remark 2. Euclid’s argument is often mis-stated as “$N = p_1\cdots p_k + 1$N=p1​⋯pk​+1 is itself prime.” That is false ($2\cdot3\cdot5\cdot7\cdot11\cdot13 + 1 = 30031 = 59 \cdot 509$2⋅3⋅5⋅7⋅11⋅13+1=30031=59⋅509). What is true, and all we need, is that $N$N has a prime factor, and that factor cannot be on the list. Read the proof again: it never claims $N$N is prime.

Homework problems

1. Give a constructive proof: there exist integers $x,y$x,y with $3x + 11y = 1$3x+11y=1. (Exhibit a witness and check it.)

2. Prove there exists a positive integer that is both a perfect square and one more than a multiple of $5$5. (Find a witness.)

3. Prove by contradiction that $\sqrt{3}$3​ is irrational. (Adapt Lemma 3: if $3 \mid n^2$3∣n2 then $3 \mid n$3∣n.)

4. Prove by contradiction: there is no smallest positive rational number.

5. Prove by contradiction: if $n \in \mathbb{Z}$n∈Z and $n^2$n2 is odd, then $n$n is odd. (Compare with proving it by contrapositive — which do you prefer, and why?)

6. Is the number $2\cdot3\cdot5\cdot7 + 1 = 211$2⋅3⋅5⋅7+1=211 prime? If so, what does that illustrate about the misreading of Euclid’s proof in the Remark above?

Suggested textbook exercises (Chartrand §5.2, 5.4): 5.2, 5.6, 5.10, 5.14, 5.18, 5.22, 5.26.