Disproof by counterexample Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Disproof by counterexample questions. See exactly how to solve problems on counterexample, universal statement, integers, squares.

counterexampleuniversal statementintegerssquaresreal numbersprimes
A-Level70 questionsStep-by-step solutions
Question 1
2 markseasy
A student claims: \"For every positive integer nn, n2>nn^2 > n.\" Which of the following is a valid counterexample that disproves this claim?

Worked solution

  1. Understand what must be shown

    ∀n∈Z+:  n2>n\forall n\in\mathbb{Z}^+:\; n^2>n

    The word "every" means the claim is a universal statement. To disprove it we only need ONE positive integer for which the statement is false.

  2. Test the smallest positive integer

    n=1:n2=12=1n=1:\quad n^2 = 1^2 = 1

    Always try the smallest allowed values first. Here the smallest positive integer is 1, so we work out 1 squared.

  3. Compare the two sides

    1≯1(since 1=1)1 \not> 1\quad(\text{since }1=1)

    We need n^2 to be bigger than n, but 1 is not bigger than 1, it is equal. So the statement fails.

  4. State the counterexample

    n=1 disproves the claimn=1\ \text{disproves the claim}

    Because we found one value that breaks the rule, the whole claim is false. n = 1 is our counterexample.

Answer
n=1, since 12=1≯1n=1,\text{ since }1^2=1\not>1
Question 2
2 markseasy
Consider the statement: \"For all real numbers xx, x2>0x^2 > 0.\" Which value of xx shows this statement is false?

Worked solution

  1. Read the claim carefully

    ∀x∈R:  x2>0\forall x\in\mathbb{R}:\; x^2>0

    The claim says every real number squared is strictly greater than zero. To disprove it we hunt for one real number whose square is NOT greater than 0.

  2. Try the special value zero

    x=0:x2=02=0x=0:\quad x^2 = 0^2 = 0

    Zero is the value that most often breaks 'greater than 0' claims, so test it first.

  3. Compare with the claim

    0≯00 \not> 0

    The claim needs the square to be greater than 0, but 0 is not greater than 0. So x = 0 breaks the rule.

  4. Conclude

    x=0 is a counterexamplex=0\ \text{is a counterexample}

    One counterexample is enough to disprove a 'for all' statement, so the claim is false.

Answer
x=0, since 02=0≯0x=0,\text{ since }0^2=0\not>0
Question 3
2 markseasy
A pupil states: \"All prime numbers are odd.\" Give a counterexample.

Worked solution

  1. Recall the definition of prime

    p prime  ⟺  p>1, factors are only 1 and pp\text{ prime}\iff p>1,\ \text{factors are only }1\text{ and }p

    A prime number has exactly two factors: 1 and itself. Remember this from earlier number work.

  2. Look for an even prime

    2: factors are 1,22:\ \text{factors are }1,2

    We want a prime that is even. The number 2 has only the factors 1 and 2, so it is prime.

  3. Check it breaks the claim

    2 is prime and 2 is even2\text{ is prime and }2\text{ is even}

    The claim says all primes are odd, but 2 is prime and even. That single case is enough.

Answer
2 is an even prime number2\text{ is an even prime number}
Question 4
2 markseasy
Consider: \"For all real numbers xx, x2=x\sqrt{x^2} = x.\" Which counterexample disproves it?

Worked solution

  1. State what we test

    x2=x for all x?\sqrt{x^2}=x\ \text{for all }x?

    The square root symbol always gives the non-negative root. So we suspect negative numbers will cause trouble.

  2. Try a negative value

    x=−3:(−3)2=9=3x=-3:\quad \sqrt{(-3)^2} = \sqrt{9} = 3

    Choose x = -3. Square it to get 9, then take the positive square root to get 3.

  3. Compare with x

    3≠−33 \neq -3

    The rule claims the answer should equal x = -3, but we got 3. They are not equal.

  4. Conclude

    x=−3 is a counterexample; x2=∣x∣x=-3\ \text{is a counterexample};\ \sqrt{x^2}=|x|

    So the correct statement is that the square root of x^2 is the modulus |x|, not x itself.

Answer
x=−3, since (−3)2=3≠−3x=-3,\text{ since }\sqrt{(-3)^2}=3\neq-3
Question 5
2 markseasy
A student writes: \"For all real numbers aa and bb, (a+b)2=a2+b2(a+b)^2 = a^2 + b^2.\" Which counterexample disproves this?

Worked solution

  1. Recall the correct expansion

    (a+b)2=a2+2ab+b2(a+b)^2 = a^2 + 2ab + b^2

    Expanding the bracket properly gives an extra 2ab term. The claim has forgotten it, so we expect it to fail whenever ab is not zero.

  2. Pick simple non-zero numbers

    a=1, b=1a=1,\ b=1

    Choose easy values where a and b are both non-zero, so the missing 2ab term matters.

  3. Work out both sides

    (1+1)2=4,12+12=2(1+1)^2 = 4,\qquad 1^2+1^2 = 2

    The left side is 2 squared = 4. The right side is 1 + 1 = 2.

  4. Compare

    4≠24 \neq 2

    The two sides are different, so the claim is false. a = b = 1 is a counterexample.

Answer
a=1, b=1, since (1+1)2=4≠2=12+12a=1,\ b=1,\text{ since }(1+1)^2=4\neq 2=1^2+1^2

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