MATH 308 — Lec 10 — Counterexamples & Uniqueness

Lecture 10 — Counterexamples and Uniqueness
MATH 308  Bridge to Advanced Mathematics  Chartrand §5.1, 5.4

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

• disprove a universally quantified statement by exhibiting a single counterexample;

• write the logical shape of a disproof as the negation $\sim\bigl(\forall x,\,P(x)\bigr) \equiv \exists x,\,\sim P(x)$∼(∀x,P(x))≡∃x,∼P(x);

• recognize that a pattern verified in many cases is not a proof, and hunt for the case that breaks it;

• prove a uniqueness statement by assuming two objects with the property and deducing they are equal.

Disproof by counterexample (§5.1)

Most theorems in this course are universally quantified: they claim “for every $x$x in some set, $P(x)$P(x) holds.” To prove such a statement we must argue about an arbitrary $x$x (Lecture 5). To disprove it, we need much less — a single offending instance.

Recall from Lecture 4 how the negation of a quantified statement is formed:

$\sim\bigl(\forall x \in S,\ P(x)\bigr) \;\equiv\; \exists x \in S,\ \sim P(x).$∼(∀x∈S, P(x))≡∃x∈S, ∼P(x).

So to show “$\forall x \in S,\ P(x)$∀x∈S, P(x)” is false, we prove its negation “$\exists x \in S,\ \sim P(x)$∃x∈S, ∼P(x)” — and an existence statement is settled by producing one witness (Lecture 9).

Definition 1. A counterexample to the statement “for every $x \in S$x∈S, $P(x)$P(x)” is a single element $x_0 \in S$x0​∈S for which $P(x_0)$P(x0​) is false. One counterexample disproves the statement completely.

Example 1. Claim: “Every prime number is odd.” Counterexample: $2$2 is prime and even. One number ends the discussion — the claim is false. (Note $2$2 is the only counterexample; it takes just one.)

Example 2. Claim: “For every real number $x$x, $\ x^2 \ge x$ x2≥x.” This looks plausible: it holds for $x = 2$x=2 ($4 \ge 2$4≥2), for $x = 5$x=5, for every $x \ge 1$x≥1, and for every $x \le 0$x≤0. But take

$x_0 = \tfrac12: \qquad x_0^2 = \tfrac14 \ \not\ge\ \tfrac12 = x_0.$x0​=21​:x02​=41​ ≥ 21​=x0​.

So $x_0 = \tfrac12$x0​=21​ is a counterexample and the claim is false. The inequality fails exactly on the open interval $(0,1)$(0,1), where the parabola dips below the line (Figure [fig:x2]).

On $(0,1)$(0,1) the parabola $y=x^2$y=x2 lies below the line $y=x$y=x, so $x^2 \ge x$x2≥x fails there. The witness $x_0=\tfrac12$x0​=21​ suffices to disprove the universal claim.

A verified pattern is not a proof. The danger is subtler when a claim survives many test cases. Checking a hundred — or a million — instances establishes nothing about the infinitely many you did not check. The next result is the classic warning.

Proposition 1. The statement “for every $n \in \mathbb{N}$n∈N with $n \ge 1$n≥1, the number $\,n^2 + n + 41\,$n2+n+41 is prime” is false.

Proof. The expression $n^2+n+41$n2+n+41 is genuinely prime for $n = 1, 2, 3, \dots, 39$n=1,2,3,…,39 (a striking run worth checking for a few small $n$n). Nevertheless it is not always prime. Take $n_0 = 40$n0​=40:

$40^2 + 40 + 41 = 1600 + 40 + 41 = 1681 = 41 \times 41 = 41^2,$402+40+41=1600+40+41=1681=41×41=412,

which is composite. Thus $n_0 = 40$n0​=40 is a counterexample, and the universal claim is false.   ◻

Remark 1. Notice why the counterexample was inevitable: $n^2+n+41$n2+n+41 at $n=41$n=41 is $41(41+1+1)$41(41+1+1), plainly divisible by $41$41, and $n=40$n=40 fails for the same structural reason. When a pattern begs to be believed, look for the hidden algebraic reason it must eventually break — do not trust the streak.

Proving uniqueness (§5.4)

Many theorems assert not just that an object with a certain property exists, but that there is exactly one. “Exactly one” splits into two claims:

$\underbrace{\text{at least one}}_{\text{existence (Lecture 9)}} \qquad\text{and}\qquad \underbrace{\text{at most one}}_{\text{uniqueness}}.$existence (Lecture 9)at least one​​anduniquenessat most one​​.

Existence we prove by exhibiting a witness. Uniqueness has its own standard template.

Definition 2 (Uniqueness argument). To prove there is at most one object with property $P$P: assume $x$x and $y$y both have property $P$P, and deduce $x = y$x=y. Combined with an existence proof, this establishes “there exists a unique object with property $P$P,” written $\exists!\,x,\ P(x)$∃!x, P(x).

Example 3. Claim: there is a unique real number $x$x with $x + 3 = 7$x+3=7. Existence: $x = 4$x=4 works. Uniqueness: if $x + 3 = 7$x+3=7 and $y + 3 = 7$y+3=7, then $x + 3 = y + 3$x+3=y+3, and subtracting $3$3 gives $x = y$x=y. So the solution is unique.

The next theorem is the linear-equation version, and its uniqueness half is the model argument to imitate.

Theorem 2. Let $a, b \in \mathbb{R}$a,b∈R with $a \neq 0$a=0. Then there is a unique real number $x$x with $ax = b$ax=b.

Proof. Existence. Since $a \neq 0$a=0, the number $x = b/a$x=b/a is defined, and $a \cdot (b/a) = b$a⋅(b/a)=b. So at least one solution exists.

Uniqueness. Suppose $x_1$x1​ and $x_2$x2​ both satisfy the equation, so $a x_1 = b$ax1​=b and $a x_2 = b$ax2​=b. Then $a x_1 = a x_2$ax1​=ax2​, hence $a(x_1 - x_2) = 0$a(x1​−x2​)=0. Because $a \neq 0$a=0, we may divide by $a$a to get $x_1 - x_2 = 0$x1​−x2​=0, i.e. $x_1 = x_2$x1​=x2​. Thus at most one solution exists.

Existence and uniqueness together give exactly one solution, namely $x = b/a$x=b/a. ◻

Geometry of uniqueness: the line $y=ax$y=ax (slope $a \neq 0$a=0) meets the horizontal line $y=b$y=b in exactly one point. Two solutions would force the same line to cross $y=b$y=b twice — impossible for a non-horizontal line.

A uniqueness argument shines when there is no formula to point at — when the object is defined only by its property. The additive identity of $\mathbb{R}$R is the cleanest instance.

Proposition 3. The real numbers have a unique additive identity: there is exactly one $e \in \mathbb{R}$e∈R with $x + e = x$x+e=x for every $x \in \mathbb{R}$x∈R.

Proof. Existence. $e = 0$e=0 satisfies $x + 0 = x$x+0=x for all $x$x.

Uniqueness. Suppose $e$e and $e'$e′ are both additive identities. Using that $e'$e′ is an identity, $e + e' = e$e+e′=e. Using that $e$e is an identity, $e + e' = e'$e+e′=e′. Reading the same quantity two ways,

$e = e + e' = e'.$e=e+e′=e′.

Hence $e = e'$e=e′, so the identity is unique. ◻

Remark 2. The trick — evaluate one expression, here $e + e'$e+e′, in two ways, once via each hypothesis — is the heart of most uniqueness proofs in algebra. We meet it again when we prove that a bijection has a unique inverse function in Lecture 17.

Homework problems

1. Disprove: “For every $n \in \mathbb{N}$n∈N, $\ 2^n - 1$ 2n−1 is prime.” Give the smallest counterexample and justify it.

2. Disprove: “For all real $x$x and $y$y, $\ \sqrt{x^2 + y^2} = x + y$ x2+y2​=x+y.” State the negation as an existence statement, then produce a witness.

3. True or false, with a proof or a counterexample: “For every $n \in \mathbb{Z}$n∈Z, if $n^2$n2 is even then $n$n is even.”

4. Prove there is a unique real number $x$x with $5x - 2 = 13$5x−2=13: show existence, then run the uniqueness template.

5. Let $a \in \mathbb{R}$a∈R be nonzero. Prove $a$a has a unique multiplicative inverse: exactly one $y \in \mathbb{R}$y∈R with $ay = 1$ay=1. (Mimic the identity proof.)

6. Find a statement “$\forall n \in \mathbb{N},\ P(n)$∀n∈N, P(n)” that is true for $n = 1,\dots,5$n=1,…,5 but false for some larger $n$n, and name the counterexample.

Suggested textbook exercises (Chartrand §5.1, 5.4): 5.20, 5.24, 5.28, 5.32, 5.36, 5.40.