A-Level Disproof by counterexample Practice Questions

Free A-Level Disproof by counterexample practice questions with full step-by-step worked solutions. Covers counterexample, universal statement, integers, squares. Practise exam-style problems and check your method.

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?
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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: \"For all real numbers xx, if x>1x > 1 then 1x>1\tfrac{1}{x} > 1.\" Which counterexample disproves it?
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Worked solution

  1. Think about reciprocals

    x>1  ⟹  0<1x<1x>1 \implies 0<\tfrac1x<1

    When x is bigger than 1, its reciprocal is actually a fraction less than 1. So the claim looks wrong.

  2. Test x = 2

    x=2:1x=12x=2:\quad \tfrac1x = \tfrac12

    Choose x = 2, which satisfies x > 1. Its reciprocal is 1/2.

  3. Compare

    12≯1\tfrac12 \not> 1

    The claim needs 1/x to be greater than 1, but 1/2 is less than 1. So it fails.

  4. Conclude

    x=2 is a counterexamplex=2\text{ is a counterexample}

    So x = 2 disproves the claim.

Answer
x=2, since 12≯1x=2,\text{ since }\tfrac12\not>1
Question 3
3 marksintermediate
A student claims: \"For all integers aa, bb and cc, if a∣bca \mid bc then a∣ba \mid b or a∣ca \mid c.\" Which counterexample disproves this?
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Worked solution

  1. Understand the claim

    a∣bc⇒a∣b or a∣c?a \mid bc \Rightarrow a\mid b\ \text{or}\ a\mid c?

    This 'prime-like' property only holds for primes. A composite divisor should break it, so try a = 4.

  2. Choose a = 4 and split its factors

    a=4, b=2, c=6a=4,\ b=2,\ c=6

    Take a = 4 and write bc so that 4's factors are shared between b and c, e.g. b = 2, c = 6.

  3. Check the hypothesis

    bc=12,4∣12bc = 12,\quad 4 \mid 12

    Here bc = 12, and 4 divides 12, so the 'if' part holds.

  4. Check divisibility of b

    4∤24 \nmid 2

    But 4 does not divide 2.

  5. Check divisibility of c

    4∤64 \nmid 6

    And 4 does not divide 6 either.

  6. Conclude

    a=4, b=2, c=6 is a counterexamplea=4,\ b=2,\ c=6\text{ is a counterexample}

    So 4 divides bc without dividing b or c, disproving the claim (it needs a to be prime).

Answer
a=4, b=2, c=6: 4∣12 but 4∤2 and 4∤6a=4,\ b=2,\ c=6:\ 4\mid 12\text{ but }4\nmid 2\text{ and }4\nmid 6
Question 4
4 markshard
A student claims: \"For all differentiable functions ff and gg (with g≠0g \neq 0), ddx ⁣(fg)=f′g′\dfrac{d}{dx}\!\left(\dfrac{f}{g}\right) = \dfrac{f'}{g'}.\" Which counterexample disproves this?
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Worked solution

  1. Understand the claim

    (fg)′=f′g′?\left(\tfrac{f}{g}\right)' = \tfrac{f'}{g'}?

    The genuine quotient rule is (f/g)' = (f'g - fg')/g^2. The 'divide the derivatives' shortcut should fail on simple powers.

  2. Choose simple functions

    f(x)=x2,g(x)=xf(x) = x^2,\quad g(x) = x

    Take f(x) = x^2 and g(x) = x, so their quotient is easy to simplify.

  3. Simplify the quotient

    fg=x2x=x\tfrac{f}{g} = \tfrac{x^2}{x} = x

    The quotient x^2 / x simplifies to x (for x not 0).

  4. Differentiate the quotient

    ddx(x)=1\tfrac{d}{dx}(x) = 1

    The derivative of x is 1. This is the true left-hand side.

  5. Differentiate each part

    f′(x)=2x,g′(x)=1f'(x) = 2x,\quad g'(x) = 1

    Separately, f'(x) = 2x and g'(x) = 1.

  6. Apply the claimed rule

    f′g′=2x1=2x\tfrac{f'}{g'} = \tfrac{2x}{1} = 2x

    The claimed rule would give 2x / 1 = 2x.

  7. Compare

    1≠2x (e.g. at x=1, 1≠2)1 \neq 2x\ \text{(e.g. at }x=1,\ 1\neq 2)

    But 1 is not equal to 2x in general; at x = 1 we get 1 versus 2.

  8. Apply the correct quotient rule

    f′g−fg′g2=2x⋅x−x2⋅1x2=1\tfrac{f'g - fg'}{g^2} = \tfrac{2x\cdot x - x^2\cdot 1}{x^2} = 1

    The real quotient rule gives (2x·x - x^2·1)/x^2 = x^2/x^2 = 1, matching the direct answer.

  9. Interpret

    the shortcut ignores the cross terms\text{the shortcut ignores the cross terms}

    The claimed shortcut drops the required terms and is therefore wrong.

  10. Conclude

    f(x)=x2, g(x)=x is a counterexamplef(x)=x^2,\ g(x)=x\text{ is a counterexample}

    So the derivative of a quotient is not the quotient of the derivatives; this pair disproves the claim.

Answer
f(x)=x2, g(x)=x: (x2x)′=1≠2x=f′g′f(x)=x^2,\ g(x)=x:\ \left(\tfrac{x^2}{x}\right)'=1\neq 2x=\tfrac{f'}{g'}
Question 5
5 markschallenging
A student claims: \"If ff is continuous with f(0)<0f(0) < 0 and f(2)>0f(2) > 0, then ff has exactly one root in (0,2)(0,2).\" Which counterexample disproves this?
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Worked solution

  1. Understand the claim

    f(0)<0, f(2)>0⇒exactly one root in (0,2)?f(0)<0,\ f(2)>0 \Rightarrow \text{exactly one root in }(0,2)?

    The Intermediate Value Theorem guarantees AT LEAST one root, but a wiggly function can cross the axis several times.

  2. Recall the IVT

    sign change⇒at least one root\text{sign change} \Rightarrow \text{at least one root}

    A sign change between the endpoints forces at least one root, but says nothing about 'exactly one'.

  3. Choose a wiggly cubic

    f(x)=(x−12)(x−1)(x−32)f(x) = (x-\tfrac12)(x-1)(x-\tfrac32)

    Take a cubic with three roots inside the interval, at x = 1/2, 1 and 3/2.

  4. Check the endpoint at 0

    f(0)=(−12)(−1)(−32)=−34f(0) = (-\tfrac12)(-1)(-\tfrac32) = -\tfrac34

    At x = 0 the product is (-1/2)(-1)(-3/2) = -3/4, which is negative.

  5. Confirm f(0) < 0

    f(0)=−34<0f(0) = -\tfrac34 < 0

    So f(0) is negative, matching the hypothesis.

  6. Check the endpoint at 2

    f(2)=(32)(1)(12)=34f(2) = (\tfrac32)(1)(\tfrac12) = \tfrac34

    At x = 2 the product is (3/2)(1)(1/2) = 3/4, which is positive.

  7. Confirm f(2) > 0

    f(2)=34>0f(2) = \tfrac34 > 0

    So f(2) is positive, again matching the hypothesis.

  8. Confirm continuity

    polynomials are continuous\text{polynomials are continuous}

    As a polynomial, f is continuous everywhere, so the 'continuous' condition holds.

  9. Count the roots in (0,2)

    x=12, 1, 32x = \tfrac12,\ 1,\ \tfrac32

    The factorised form shows three roots, all lying in the interval (0,2).

  10. Compare with the claim

    3 roots≠1 root3\ \text{roots} \neq 1\ \text{root}

    The claim insists on exactly one root, but there are three.

  11. Explain the extra crossings

    the curve wiggles across the axis\text{the curve wiggles across the axis}

    Between the endpoints the curve dips and rises, crossing the x-axis three times.

  12. State the correct conclusion

    IVT gives at least one root\text{IVT gives at least one root}

    The IVT only promises at least one root, never exactly one.

  13. Give a simpler idea

    any odd number of crossings is possible\text{any odd number of crossings is possible}

    With opposite end signs, the number of crossings is odd (1, 3, 5, ...), not necessarily 1.

  14. Interpret

    sign change⇏uniqueness\text{sign change} \not\Rightarrow \text{uniqueness}

    A single sign change cannot rule out additional roots.

  15. Conclude

    f(x)=(x−12)(x−1)(x−32) is a counterexamplef(x)=(x-\tfrac12)(x-1)(x-\tfrac32)\text{ is a counterexample}

    So the claim is false; this cubic has three roots in (0,2), disproving it.

Answer
f(x)=(x−12)(x−1)(x−32): f(0)<0, f(2)>0, but there are three roots in (0,2)f(x)=(x-\tfrac12)(x-1)(x-\tfrac32):\ f(0)<0,\ f(2)>0,\text{ but there are three roots in }(0,2)

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