THE MATHS TAILOR

Logic & Proof | The Maths Tailor — UK Methods

Logic & Proof

10 methods · 48 questions

A universal statement ("for all", "every", "always", "must") is false the moment one case satisfies its hypothesis yet fails its conclusion. So do not try to prove it: hunt for one offender. Read off the antecedent, list the candidates, and discard any candidate that already fails the antecedent before you even test the conclusion.

Trigger: "for all", "every", "must be true", "which provides a counterexample".

Instances:
(i) for a function-based claim, test the candidates first against the hypothesis, then check the conclusion only on survivors;
(ii) one valid counterexample is a complete disproof, you never need to list all of them;
(iii) a candidate that fails the hypothesis (for example the antecedent is already false) can never be a counterexample.

Linked questions (7)

TMUA 2017 Paper 2 — Q16

Consider the following statement:

(*)

f (x)

is an integer for every integer

x

, then

f^{'} (x)

is an integer for every integer

x

. Which one of the following is a counterexample to

(*)

?

(A)

f (x) = \frac{x ^{3} + x + 1}{4}

(B)

f (x) = \frac{x ^{4} + x ^{2} + x}{2}

(C)

f (x) = \frac{x ^{4} + x ^{3} + x ^{2} + x}{2}

(D)

f (x) = \frac{x ^{4} + 2 x ^{3} + x ^{2}}{4}

Voir la correction

A counterexample to

(*)

must be a function for which

f (x)

is an integer for every integer

x

, but

f^{'} (x)

is not an integer for some integer

x

. Check option A at

x = 0

f (0) = \frac{1}{4}

, not an integer, so A fails the hypothesis and is not a counterexample. Check B at

x = 1

f (1) = \frac{1 + 1 + 1}{2} = \frac{3}{2}

, not an integer, so B is not a counterexample. For C:

f (x) = \frac{x ^{4} + x ^{3} + x ^{2} + x}{2} = \frac{x ( x ^{3} + x ^{2} + x + 1 )}{2} = \frac{x ( x + 1 ) ( x ^{2} + 1 )}{2}

. Since

x (x + 1)

is always even,

f (x)

is an integer for every integer

x

. Now

f^{'} (x) = \frac{4 x ^{3} + 3 x ^{2} + 2 x + 1}{2}

, and

f^{'} (0) = \frac{1}{2}

, which is not an integer. So C is a counterexample. For D:

f (x) = \frac{x ^{2} ( x + 1 ) ^{2}}{4}

; since

x (x + 1)

is always even,

x^{2} (x + 1)^{2}

is always divisible by 4, so

f (x)

is an integer for every integer

x

. Also

f^{'} (x) = \frac{x ( x + 1 ) ( 2 x + 1 )}{2}

and the numerator is always even, so

f^{'} (x)

is an integer for every integer

x

, meaning D is not a counterexample. Hence the answer is C.

TMUA 2018 Paper 2 — Q6

Which one of the following functions provides a counterexample to the statement:

[[B]] i f [[/ B]] f^{'} (x) > 0 for all real x, [[B]] t h e n [[/ B]] f (x) > 0 for all real x .

(A)

f (x) = x^{2} + 1

(B)

f (x) = x^{2} - 1

(C)

f (x) = x^{3} + x + 1

(D)

f (x) = 1 - x

(E)

f (x) = 2^{x}

Voir la correction

A counterexample requires

f^{'} (x) > 0

for all

x

(antecedent true) but

f (x) > 0

for all

x

is false (consequent false).

A:

f^{'} (x) = 2 x

, which is not positive for all

x

. Not a counterexample. B:

f^{'} (x) = 2 x

, same issue. Not a counterexample. C:

f^{'} (x) = 3 x^{2} + 1 \geq 1 > 0

for all

x

, so the antecedent holds. But

f (x) = x^{3} + x + 1

is a cubic and takes all real values, including negative ones (e.g. for very negative

x

). So the consequent is false. This is a counterexample. D:

f^{'} (x) = - 1 < 0

for all

x

. Antecedent false. Not a counterexample. E:

f^{'} (x) > 0

for all

x

(exponential is increasing), and

2^{x} > 0

for all

x

, so both antecedent and consequent are true. Not a counterexample.

Hence the answer is C.

TMUA 2020 Paper 2 — Q4

Consider the following statement: every positive integer

N

that is greater than 6 can be written as the sum of two non-prime integers that are greater than 1. Which of the following is/are counterexample(s) to this statement?

I N = 5 II N = 7 III N = 9

(A) none of them
(B) I only
(C) II only
(D) III only
(E) I and II only
(F) I and III only
(G) II and III only
(H) I, II and III

Voir la correction

A counterexample must be a positive integer

N > 6

that cannot be written as the sum of two non-prime integers each greater than 1. The non-prime integers greater than 1 are: 4, 6, 8, 9, 10, ... (composites). I:

N = 5

is not greater than 6, so it does not satisfy the hypothesis of the statement and is not a counterexample. II:

N = 7 > 6

. We need

7 = a + b

with

a, b > 1

both non-prime. The only ways to write 7 as a sum of two integers both

> 1

are

2 + 5

3 + 4

. Both 2 and 5 are prime; 3 is prime. So 7 cannot be written in the required form: it is a counterexample. III:

N = 9 > 6

. Ways:

2 + 7

3 + 6

4 + 5

. In

3 + 6

: 3 is prime; in

4 + 5

: 5 is prime; in

2 + 7

: both prime. So 9 also cannot be written as required: it is a counterexample. Hence II and III are counterexamples, and the answer is G.

TMUA 2020 Paper 2 — Q8

A student is asked to prove whether the following statement

(*)

is true or false:

(*)

For all real numbers

a

and

b

∣ a + b ∣ < ∣ a ∣ + ∣ b ∣

. The student's proof is as follows: Statement

(*)

is false. A counterexample is

a = 3

b = 4

, as

∣3 + 4∣ = 7

and

∣3∣ + ∣4∣ = 7

, but

7 < 7

is false. Which of the following best describes the student's proof?

(A) The statement

(*)

is true, and the student's proof is not correct.
(B) The statement

(*)

is false, but the student's proof is not correct: the counterexample is not valid.
(C) The statement

(*)

is false, but the student's proof is not correct: the student needs to give all the values of

a

and

b

where

∣ a + b ∣ < ∣ a ∣ + ∣ b ∣

is false.
(D) The statement

(*)

is false, but the student's proof is not correct: the student should have instead stated that for all real numbers

a

and

b

∣ a + b ∣ \leq ∣ a ∣ + ∣ b ∣

.
(E) The statement

(*)

is false, and the student's proof is fully correct.

Voir la correction

The statement

(*)

claims

∣ a + b ∣ < ∣ a ∣ + ∣ b ∣

for ALL real

a, b

(strict inequality). The student provides

a = 3, b = 4

: then

∣3 + 4∣ = 7

and

∣3∣ + ∣4∣ = 7

, so

∣ a + b ∣ = ∣ a ∣ + ∣ b ∣

, meaning the strict inequality fails. This is a valid counterexample to the universal statement. To disprove a 'for all' statement, one valid counterexample is sufficient. The counterexample is arithmetically correct and properly negates the statement. Hence the student's proof is fully correct, and the answer is E.

TMUA 2021 Paper 2 — Q10

The first seven terms of a sequence of positive integers are:

u_{1} = 15, u_{2} = 21, u_{3} = 30, u_{4} = 37, u_{5} = 44, u_{6} = 51, u_{7} = 59.

Consider the following statement about this sequence:

(*)

n

is a prime number, then

u_{n}

is a multiple of 3 or

u_{n}

is a multiple of 5. What is the smallest value of

n

that provides a counterexample to

(*)

?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
(F) 6
(G) 7

Voir la correction

A counterexample requires

n

to be prime but

u_{n}

to be neither a multiple of 3 nor a multiple of 5. Check the prime values of

n

n = 2

u_{2} = 21 = 3 \times 7

, multiple of 3 (no counterexample).

n = 3

u_{3} = 30 = 2 \times 3 \times 5

, multiple of both 3 and 5 (no counterexample).

n = 5

u_{5} = 44

. Is 44 a multiple of 3?

44 = 3 \times 14 + 2

, no. Is 44 a multiple of 5? No. So

n = 5

is prime and

u_{5} = 44

is neither a multiple of 3 nor 5: this is a counterexample. Check

n = 7

u_{7} = 59

, prime itself, neither multiple of 3 nor 5, also a counterexample, but

n = 5

is smaller. Hence the answer is E.

TMUA 2021 Paper 2 — Q4

Consider the following claim about positive integers

a

b

and

c

: if

a

is a factor of

b c

, then

a

is a factor of

b

a

is a factor of

c

. Which of the following provide(s) a counterexample to this claim?

I.

a = 5, b = 10, c = 20

II.

a = 8, b = 4, c = 4

III.

a = 6, b = 7, c = 12

(A) none of them
(B) I only
(C) II only
(D) III only
(E) I and II only
(F) I and III only
(G) II and III only
(H) I, II and III

Voir la correction

A counterexample requires

a ∣ b c

but

a ∤ b

and

a ∤ c

. Statement I:

b c = 200

5 ∣ 200

(true),

5 ∣ 10

(true), so the conclusion holds; not a counterexample. Statement II:

b c = 16

8 ∣ 16

(true),

8 ∤ 4

and

8 ∤ 4

(true), so the conclusion fails; this is a counterexample. Statement III:

b c = 84

6 ∣ 84

(true),

6 ∤ 7

but

6 ∣ 12

(true), so the conclusion holds; not a counterexample. Hence the answer is C.

TMUA Specimen Paper 2 — Q4

A set of five cards each have a letter printed on their front and a number printed on their back, as follows: Card A has front A and back 3; Card B has front B and back 4; Card C has front C and back 1; Card D has front D and back 7; Card E has front E and back 8. Which one of the five cards (A, B, C, D or E) provides a counterexample to the following statement? "Every card that has a vowel on its front has an even number on its back."

(A) Card A
(B) Card B
(C) Card C
(D) Card D
(E) Card E

A table showing five cards. Row 'Fronts': boxes labelled A, B, C, D, E. Row 'Backs': boxes containing the numbers 3, 4, 1, 7, 8 respectively. The cards are labelled Card A through Card E.

Voir la correction

A counterexample to "every card with a vowel on its front has an even number on its back" is a card that has a vowel on its front AND an odd number on its back. The vowel cards are A (back 3) and E (back 8). Card E has an even back (8), so it satisfies the statement. Card A has a vowel (A) on its front and an odd number (3) on its back, which directly contradicts the statement. Hence the answer is A.