Hard A-Level Disproof by counterexample Questions

Challenging, exam-style A-Level Disproof by counterexample questions with worked solutions. Stretch yourself on the hardest prime-generating polynomials, counterexample, number theory, Euler problems.

prime-generating polynomialscounterexamplenumber theoryEulerMersenne numbersprimes
A-Level34 questionsStep-by-step solutions
Question 1
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?
Show worked solution

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)
Question 2
5 markschallenging
A student claims: \"For every integer n>1n > 1, the triangular number Tn=n(n+1)2T_n = \tfrac{n(n+1)}{2} is never a perfect square.\" State the smallest integer n>1n > 1 that is a counterexample.
Show worked solution

Worked solution

  1. Understand the claim

    Tn=n(n+1)2 never a square for n>1?T_n = \tfrac{n(n+1)}{2}\ \text{never a square for }n>1?

    We compute triangular numbers and look for the first one (beyond n = 1) that is a perfect square.

  2. Recall the formula

    Tn=n(n+1)2T_n = \tfrac{n(n+1)}{2}

    The n-th triangular number is n(n+1)/2, the sum 1 + 2 + ... + n.

  3. Compute T_2

    T2=2⋅32=3T_2 = \tfrac{2\cdot 3}{2} = 3

    For n = 2 we get 3, which is not a perfect square.

  4. Compute T_3

    T3=3⋅42=6T_3 = \tfrac{3\cdot 4}{2} = 6

    For n = 3 we get 6, not a perfect square.

  5. Compute T_4

    T4=4⋅52=10T_4 = \tfrac{4\cdot 5}{2} = 10

    For n = 4 we get 10, not a perfect square.

  6. Compute T_5

    T5=5⋅62=15T_5 = \tfrac{5\cdot 6}{2} = 15

    For n = 5 we get 15, not a perfect square.

  7. Compute T_6

    T6=6⋅72=21T_6 = \tfrac{6\cdot 7}{2} = 21

    For n = 6 we get 21, not a perfect square.

  8. Compute T_7

    T7=7⋅82=28T_7 = \tfrac{7\cdot 8}{2} = 28

    For n = 7 we get 28, not a perfect square.

  9. Compute T_8

    T8=8⋅92=36T_8 = \tfrac{8\cdot 9}{2} = 36

    For n = 8 we get 8 times 9 over 2 = 36.

  10. Recognise the square

    36=6236 = 6^2

    But 36 is 6 squared, a perfect square.

  11. Confirm the conclusion fails

    T8=36=62T_8 = 36 = 6^2

    So T_8 is a perfect square, breaking the claim.

  12. Check nothing smaller worked

    T2,…,T7=3,6,10,15,21,28T_2,\dots,T_7 = 3,6,10,15,21,28

    None of the triangular numbers for n = 2 to 7 is a perfect square, so n = 8 is the first.

  13. Double-check the arithmetic

    8×92=722=36\tfrac{8\times 9}{2} = \tfrac{72}{2} = 36

    Rechecking, 8 times 9 is 72, and half of that is 36.

  14. Note the next such value

    T49=1225=352T_{49} = 1225 = 35^2

    Interestingly the next square triangular number is much later, at n = 49.

  15. State the smallest counterexample

    n=8n = 8

    So the smallest counterexample with n > 1 is n = 8.

Answer
n=8 (T8=36=62)n=8\ (T_8 = 36 = 6^2)
Question 3
5 markschallenging
A student claims: \"If a function ff satisfies f(a)=f(b)f(a) = f(b) (with a<ba<b), then there must be a point cc in (a,b)(a,b) where f′(c)=0f'(c) = 0.\" Which counterexample disproves this?
Show worked solution

Worked solution

  1. Understand the claim

    f(a)=f(b)⇒∃c∈(a,b): f′(c)=0?f(a)=f(b) \Rightarrow \exists c\in(a,b):\ f'(c)=0?

    This is Rolle's theorem, but the student has dropped the crucial condition that f must be differentiable on (a,b).

  2. State Rolle's theorem properly

    continuous on [a,b], differentiable on (a,b), f(a)=f(b)\text{continuous on }[a,b],\ \text{differentiable on }(a,b),\ f(a)=f(b)

    Rolle requires continuity on the closed interval AND differentiability on the open interval, plus equal endpoints.

  3. Choose a function with a corner

    f(x)=∣x∣f(x) = |x|

    Take f(x) = |x|, which has a sharp corner at x = 0.

  4. Pick a symmetric interval

    [a,b]=[−1,1][a,b] = [-1,1]

    Use the interval from -1 to 1, symmetric about the corner.

  5. Check equal endpoint values

    f(−1)=1,f(1)=1f(-1) = 1,\quad f(1) = 1

    Here f(-1) = |-1| = 1 and f(1) = |1| = 1, so f(a) = f(b).

  6. Confirm the hypothesis holds

    f(−1)=f(1)=1f(-1) = f(1) = 1

    So the student's condition f(a) = f(b) is satisfied.

  7. Differentiate away from 0

    f′(x)=−1 (x<0),f′(x)=1 (x>0)f'(x) = -1\ (x<0),\quad f'(x) = 1\ (x>0)

    For negative x the gradient is -1; for positive x it is +1.

  8. Look for a stationary point

    f′(x)=±1≠0 for x≠0f'(x) = \pm 1 \neq 0\ \text{for }x\neq 0

    The derivative is either -1 or +1, never 0, on either side.

  9. Examine x = 0

    f′(0) does not existf'(0)\ \text{does not exist}

    At x = 0 the left and right gradients differ, so f is not differentiable there and f'(0) does not exist.

  10. Conclude no such c exists

    ∄ c∈(−1,1): f′(c)=0\nexists\ c\in(-1,1):\ f'(c)=0

    So there is no point in (-1,1) with zero derivative.

  11. Compare with the claim

    f(a)=f(b) but no c with f′(c)=0f(a)=f(b)\ \text{but no }c\ \text{with }f'(c)=0

    The endpoints match yet the conclusion of Rolle fails, contradicting the student's claim.

  12. Identify the missing hypothesis

    f not differentiable at 0f\ \text{not differentiable at }0

    The claim failed precisely because f is not differentiable at the corner.

  13. Interpret geometrically

    a V-shape has no flat tangent\text{a V-shape has no flat tangent}

    The V-shaped graph never has a horizontal tangent; the bottom is a sharp point, not a smooth minimum.

  14. State the correct theorem

    Rolle needs differentiability on (a,b)\text{Rolle needs differentiability on }(a,b)

    With differentiability restored, Rolle's theorem would guarantee such a c.

  15. Conclude

    f(x)=∣x∣ on [−1,1] is a counterexamplef(x)=|x|\ \text{on }[-1,1]\text{ is a counterexample}

    So equal endpoint values alone do not give a stationary point; |x| disproves the claim.

Answer
f(x)=∣x∣ on [−1,1]: f(−1)=f(1)=1 but f′ is never 0 (corner at 0)f(x)=|x|\text{ on }[-1,1]:\ f(-1)=f(1)=1\text{ but }f'\text{ is never }0\ (\text{corner at }0)
Question 4
5 markschallenging
A student claims: \"For any two real numbers aa and bb with a<ba < b, there is an integer strictly between them.\" Which counterexample disproves this?
Show worked solution

Worked solution

  1. Understand the claim

    a<b⇒∃ n∈Z: a<n<b?a<b \Rightarrow \exists\ n\in\mathbb{Z}:\ a<n<b?

    If two numbers are very close together, there may be no integer squeezed between them. So pick numbers within the same unit gap.

  2. Recall the spacing of integers

    consecutive integers differ by 1\text{consecutive integers differ by }1

    Integers are spaced one apart, so a gap shorter than 1 might contain none.

  3. Choose close numbers

    a=0.1,b=0.2a = 0.1,\quad b = 0.2

    Take a = 0.1 and b = 0.2, both between the integers 0 and 1.

  4. Confirm the hypothesis

    0.1<0.20.1 < 0.2

    Certainly a < b, so the 'if' part holds.

  5. Identify the surrounding integers

    0<0.1 and 0.2<10 < 0.1\ \text{and}\ 0.2 < 1

    Both numbers lie strictly between the integers 0 and 1.

  6. Check integer 0

    0<0.1, so 0≯a0 < 0.1,\ \text{so }0 \not> a

    The integer 0 is less than a = 0.1, so it is not between a and b.

  7. Check integer 1

    1>0.2, so 1≮b1 > 0.2,\ \text{so }1 \not< b

    The integer 1 is greater than b = 0.2, so it is not between them either.

  8. Rule out all other integers

    negative integers<0, integers≥1>0.2\text{negative integers} < 0,\ \text{integers} \ge 1 > 0.2

    Every other integer is either at most 0 or at least 1, so none can lie in (0.1, 0.2).

  9. Conclude no integer fits

    ∄ n∈Z: 0.1<n<0.2\nexists\ n\in\mathbb{Z}:\ 0.1 < n < 0.2

    There is no integer strictly between 0.1 and 0.2.

  10. Compare with the claim

    hypothesis holds, conclusion fails\text{hypothesis holds, conclusion fails}

    So a < b but no integer lies between them, contradicting the claim.

  11. State the correct statement

    b−a>1⇒an integer lies betweenb - a > 1 \Rightarrow \text{an integer lies between}

    The claim only works if the gap exceeds 1 (more precisely, is at least 1).

  12. Contrast with rationals

    a rational always lies between a,b\text{a rational always lies between }a,b

    By contrast, a rational (and an irrational) number can always be found between any two distinct reals.

  13. Interpret

    integers are not dense\text{integers are not dense}

    Integers are spread out, so they are not 'dense' in the real line.

  14. Give a variant

    a=2.3, b=2.7 also worksa = 2.3,\ b = 2.7\ \text{also works}

    The same happens for 2.3 and 2.7, which straddle no integer.

  15. Conclude

    a=0.1, b=0.2 is a counterexamplea=0.1,\ b=0.2\text{ is a counterexample}

    So the claim is false; 0.1 and 0.2 disprove it.

Answer
a=0.1, b=0.2: 0.1<0.2 but no integer lies strictly between thema=0.1,\ b=0.2:\ 0.1<0.2\text{ but no integer lies strictly between them}
Question 5
5 markschallenging
A student claims: \"Every function that is continuous on the open interval (0,1)(0,1) is bounded on (0,1)(0,1).\" Which counterexample disproves this?
Show worked solution

Worked solution

  1. Understand the claim

    f continuous on (0,1)⇒f bounded?f\ \text{continuous on }(0,1) \Rightarrow f\ \text{bounded}?

    Boundedness is guaranteed on a CLOSED interval [a,b], but on an OPEN interval a function can blow up near an excluded endpoint.

  2. Recall the boundedness theorem

    continuous on [a,b]⇒bounded\text{continuous on }[a,b] \Rightarrow \text{bounded}

    The Extreme Value Theorem needs a closed, bounded interval; (0,1) is open, so it may not apply.

  3. Choose a candidate

    f(x)=1xf(x) = \tfrac1x

    Consider f(x) = 1/x on (0,1), which grows large as x approaches 0.

  4. Check continuity on (0,1)

    1x is continuous for x≠0\tfrac1x\ \text{is continuous for }x\neq 0

    The function 1/x is continuous everywhere except x = 0, and 0 is not in the open interval, so it is continuous on (0,1).

  5. Examine behaviour near 0

    x→0+⇒1x→+∞x\to 0^+ \Rightarrow \tfrac1x \to +\infty

    As x approaches 0 from the right, 1/x grows without limit.

  6. Test some values

    f(0.1)=10, f(0.01)=100, f(0.001)=1000f(0.1)=10,\ f(0.01)=100,\ f(0.001)=1000

    For x = 0.1, 0.01, 0.001 the outputs are 10, 100, 1000, climbing endlessly.

  7. Show no upper bound exists

    for any M, ∃x: 1x>M\text{for any }M,\ \exists x:\ \tfrac1x > M

    Whatever ceiling M we propose, choosing x < 1/M makes 1/x exceed it.

  8. Deduce f is unbounded

    f is unbounded on (0,1)f\ \text{is unbounded on }(0,1)

    So 1/x has no finite maximum on (0,1); it is unbounded.

  9. Compare with the claim

    continuous yet unbounded\text{continuous yet unbounded}

    The function is continuous on the open interval yet unbounded, contradicting the claim.

  10. Explain the role of the open endpoint

    0∉(0,1)0\notin (0,1)

    The trouble point x = 0 is excluded, so continuity on (0,1) does not tame the growth.

  11. Contrast with a closed interval

    [a,1] (a>0): 1x≤1a[a,1]\ (a>0):\ \tfrac1x \le \tfrac1a

    On any closed subinterval [a,1] with a > 0, the function IS bounded by 1/a.

  12. Emphasise why closedness matters

    EVT needs a closed interval\text{EVT needs a closed interval}

    This shows the closed-interval condition in the boundedness theorem is essential.

  13. Give a related example

    f(x)=11−x blows up near 1f(x) = \tfrac{1}{1-x}\ \text{blows up near }1

    Similarly 1/(1-x) is continuous on (0,1) but unbounded near the other endpoint.

  14. Interpret

    open intervals can let functions escape to infinity\text{open intervals can let functions escape to infinity}

    Removing an endpoint can allow a continuous function to run off to infinity.

  15. Conclude

    f(x)=1x is a counterexamplef(x)=\tfrac1x\text{ is a counterexample}

    So continuity on an open interval does not imply boundedness; 1/x disproves the claim.

Answer
f(x)=1x on (0,1): continuous but unbounded as x→0+f(x)=\tfrac1x\text{ on }(0,1):\ \text{continuous but unbounded as }x\to 0^+

Unlock 29 more Disproof by counterexample questions

Create a free account to work through every A-Level Disproof by counterexample question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Disproof by counterexample practice

Practice QuestionsWorked Solutions

Related Pure Maths topics

Areas and further applicationsMathematical argument and notationArithmetic sequences and seriesBinomial expansion (rational powers)Binomial expansionCircles