Proving and disproving are not equally hard. To prove "for all $n$, …" you need a general argument that works for every $n$. To disprove it, you need just one value of $n$ where it fails.
Disproof: "Here is one particular $x$ in the set for which $P(x)$ is false."
| To show it is true | To show it is false | |
|---|---|---|
| "For all …" | A general proof (deduction or exhaustion) | One counter-example |
| "There exists …" | One example | A general proof that no case works |
Many false claims are written by someone who only pictured positive whole numbers bigger than $1$. The values below break such claims surprisingly often — try them first.
| Try… | Because… | Breaks the claim… |
|---|---|---|
| $0$ | multiplying by it, or dividing by it, behaves unusually | "$x^2 > 0$ for all real $x$" |
| $1$ | $1^n = 1$, and $1$ is neither prime nor composite | "$x^2 > x$ for all $x > 0$" |
| Negative numbers | squaring changes the order; inequalities reverse | "If $a > b$ then $a^2 > b^2$" |
| Fractions between $0$ and $1$ | powers make them smaller | "$x^3 \geqslant x$ for all $x > 0$" |
| $2$ | the only even prime | "All primes are odd" |
| Squares, multiples of the constant | they make a formula factorise | "$n^2 - n + 11$ is always prime" |
Disprove the statement "$x^2 > x$ for all real numbers $x$."
Disprove "If $p$ is prime, then $p + 1$ is even."
Disprove "For all real $a$ and $b$, if $a > b$ then $a^2 > b^2$."
Disprove "$n^2 - n + 11$ is prime for every positive integer $n$."
Disprove "If $n$ is prime, then $2^n - 1$ is prime."
Disprove "$\sqrt{a + b} = \sqrt{a} + \sqrt{b}$ for all positive $a$ and $b$."
- State the counter-example clearly. "Let $n = 11$."
- Check it meets the conditions. "$11$ is prime."
- Show the calculation that makes the conclusion fail. "$2^{11} - 1 = 2047 = 23 \times 89$."
- Conclude. "So $2^{11} - 1$ is not prime, and the statement is false."
Hence $n = 11$ is a counter-example and the statement is false."
(a) Disprove "$x^3 > x$ for all real $x$." (b) Prove that $x^3 > x$ for all $x > 1$.
Some questions do not tell you whether a statement is true. Test a few cases — especially the awkward ones — before deciding which way to go.
Prove or disprove each statement. (a) The sum of any three consecutive even numbers is a multiple of $6$. (b) The product of any three consecutive positive integers is a multiple of $4$.
Disprove "The sum of two irrational numbers is always irrational."
One is enough
A single counter-example disproves a "for all" statement.
Not a proof
Any number of examples that work prove nothing.
Usual suspects
$0$, $1$, negatives, fractions, $2$.
Conditions
Your example must satisfy every condition of the claim.
Show it
Write the calculation and the factors — don't just say "fails".
Conclude
"Hence … is a counter-example, so the statement is false."
Structure
$n^2 + n + c$ is divisible by $c$ when $n = c$.
Repair
Restrict the set of values to exclude the counter-example, then prove.
Disprove the statement "every odd number is prime."
▶ Show solution
$9$ is odd, but $9 = 3 \times 3$ is not prime.
So $9$ is a counter-example and the statement is false. ($1$ also works, since $1$ is not prime.)
Disprove "$\sqrt{x^2} = x$ for all real $x$."
▶ Show solution
Let $x = -3$. Then $\sqrt{x^2} = \sqrt{9} = 3$, but $x = -3$.
$3 \neq -3$, so $x = -3$ is a counter-example and the statement is false. (In fact $\sqrt{x^2} = |x|$.)
Disprove "$\dfrac{1}{x} < x$ for all real $x \neq 0$."
▶ Show solution
Let $x = \tfrac12$. Then $\dfrac1x = 2$, and $2 > \tfrac12$.
So $\dfrac1x < x$ is false when $x = \tfrac12$, and the statement is false.
Disprove "If $a^2 > b^2$, then $a > b$."
▶ Show solution
Let $a = -3$, $b = 1$. Then $a^2 = 9 > 1 = b^2$, so the condition holds.
But $a = -3 < 1 = b$, so the conclusion fails. The statement is false.
Show that "$n^2 + 3n + 1$ is prime for every positive integer $n$" is false.
▶ Show solution
$n = 1, 2, 3, 4, 5$ give $5, 11, 19, 29, 41$ — all prime.
$n = 6$: $36 + 18 + 1 = 55 = 5 \times 11$, which is not prime.
So $n = 6$ is a counter-example and the statement is false.
Disprove "$2^n + 1$ is prime for every positive integer $n$."
▶ Show solution
$n = 1$: $3$, prime. $n = 2$: $5$, prime.
$n = 3$: $2^3 + 1 = 9 = 3 \times 3$, not prime.
So $n = 3$ is a counter-example and the statement is false.
A student says: "The sum of two prime numbers is never prime, because adding two odd numbers gives an even number." Find a counter-example, and explain the flaw in the student's reasoning.
▶ Show solution
$2 + 3 = 5$, and $5$ is prime. So the statement is false.
The flaw: the student assumed every prime is odd. The prime $2$ is even, so "odd + odd" does not cover every pair of primes.
Disprove "$(a + b)^2 = a^2 + b^2$ for all real $a$ and $b$." Then find all pairs $(a, b)$ for which it is true.
▶ Show solution
Let $a = b = 1$: $(1 + 1)^2 = 4$, but $1^2 + 1^2 = 2$. So the statement is false.
In general $(a+b)^2 = a^2 + 2ab + b^2$, so the equation holds exactly when $2ab = 0$.
It is true precisely when $a = 0$ or $b = 0$.
Prove or disprove each statement. (a) The sum of any three consecutive integers is a multiple of $6$. (b) The product of any two consecutive integers is even.
▶ Show solution
(a) False. $2 + 3 + 4 = 9$, which is not a multiple of $6$. (The sum is always $3(n+1)$, a multiple of $3$, but only a multiple of $6$ when $n + 1$ is even.)
(b) True. Let the integers be $n$ and $n + 1$.
If $n$ is even, $n = 2k$ and $n(n+1) = 2k(n+1)$ is even.
If $n$ is odd, $n + 1 = 2k$ is even, and $n(n+1) = 2kn$ is even.
Every integer is even or odd, so the product is always even.
A student claims that $x^2 \geqslant 4x - 3$ for all real values of $x$.
(a) Find a counter-example to the claim. (b) Find the complete set of values of $x$ for which the claim fails. (c) Prove that $x^2 \geqslant 4x - 4$ for all real $x$, and state when equality holds.
▶ Show solution
(a) Let $x = 2$: $x^2 = 4$ and $4x - 3 = 5$. Since $4 < 5$, the claim fails at $x = 2$.
(b) The claim fails when $x^2 < 4x - 3$, i.e. $x^2 - 4x + 3 < 0$.
$(x - 1)(x - 3) < 0$. The parabola opens upwards with roots $1$ and $3$, so it is negative between them.
The claim fails exactly for $1 < x < 3$, i.e. $\{x : 1 < x < 3\}$.
(c) $x^2 - (4x - 4) = x^2 - 4x + 4 = (x - 2)^2 \geqslant 0$ for all real $x$.
So $x^2 \geqslant 4x - 4$ for all real $x$, with equality only when $x = 2$.