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Proof & Number Theory | The Maths Tailor — UK Methods

Proof & Number Theory

4 methods · 12 questions

A claim of the form 'for all $n$ , P(n)' or 'if $n$ is prime then Q(n)' is refuted by a single case that satisfies the hypothesis yet breaks the conclusion. When the candidates are handed to you (I, II, III) or are small, do not prove anything: substitute and check. Watch the hypothesis first, a case that fails the hypothesis can never be a counterexample.

Trigger: 'for all', 'always', 'is a counterexample', or a list of explicit candidate values I/II/III.

Instances:
(i) plug each listed value into both sides of an identity and keep the one that disagrees;
(ii) for 'if prime then ...', test the small primes $2, 3, 5$ and discard any candidate whose number is not prime;
(iii) for a claim about a family, find the one value where the divisibility or primality pattern breaks.

Linked questions (4)

TMUA 2016 Paper 2 — Q5

Consider the statement:

(*)

A whole number

n

is prime if it is 1 less or 5 less than a multiple of 6.

How many counterexamples to

(*)

are there in the range

0 < n < 50

?

(A) 2
(B) 3
(C) 4
(D) 5
(E) 6

Voir la correction

A counterexample is a whole number

n

that is 1 less or 5 less than a multiple of 6, yet is NOT prime. Numbers 1 less than a multiple of 6 in the range: 5, 11, 17, 23, 29, 35, 41, 47. Numbers 5 less than a multiple of 6 in range: 1, 7, 13, 19, 25, 31, 37, 43, 49. Combined list with primality: 1 (not prime), 5 (prime), 7 (prime), 11 (prime), 13 (prime), 17 (prime), 19 (prime), 23 (prime), 25 (not prime:

= 5^{2}

), 29 (prime), 31 (prime), 35 (not prime:

= 5 \times 7

), 37 (prime), 41 (prime), 43 (prime), 47 (prime), 49 (not prime:

= 7^{2}

). Counterexamples: 1, 25, 35, 49. That is 4 counterexamples. Hence the answer is C.

TMUA 2022 Paper 2 — Q3

Consider the following statement about the positive integer

n

[[B]] i f [[/ B]] n is prime, [[B]] t h e n [[/ B]] n^{2} + 2 is [[B]] n o t [[/ B]] prime

Which of the following is a counterexample to this statement?

I: $n = 2$
II: $n = 3$
III: $n = 4$

(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 have

n

prime but

n^{2} + 2

also prime (violating the conclusion). I:

n = 2

is prime and

n^{2} + 2 = 6

is not prime, so the statement holds and this is not a counterexample. II:

n = 3

is prime and

n^{2} + 2 = 11

is prime, so the conclusion "not prime" fails. This is a counterexample. III:

n = 4

is not prime, so the hypothesis fails and this cannot be a counterexample. Therefore only II provides a counterexample. Hence the answer is C.

TMUA 2023 Paper 2 — Q3

Consider the claim: for all positive real numbers

x

and

y

x^{y} = x^{y} .

Which of the following is/are a counterexample to the claim? I:

x = 1, y = 16

. II:

x = 2, y = 8

. III:

x = 3, y = 4

.

(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

Check each pair. I:

1^{16} = 1 = 1

and

1^{16} = 1^{4} = 1

. Equal, so not a counterexample. II:

2^{8} = 256 = 16

and

2^{8} = 2^{22} \approx 2^{2.83} \approx 7.1

. Not equal, so II is a counterexample. III:

3^{4} = 81 = 9

and

3^{4} = 3^{2} = 9

. Equal, so not a counterexample. Only II is a counterexample. Hence the answer is C.

TMUA Specimen Paper 2 — Q8

Consider the following statement about the positive integer

n

: Statement (*): The sum of the four consecutive integers, the smallest of which is

n

, is a multiple of 6. Which one of the following is true?

(A) Statement (*) is true for all values of

n

.
(B) Statement (*) is true for all values of

n

which are odd, but not for any other values of

n

.
(C) Statement (*) is true for all values of

n

which are multiples of 3, but not for any other values of

n

.
(D) Statement (*) is true for all values of

n

which are multiples of 6, but not for any other values of

n

.
(E) Statement (*) is not true for any value of

n

Voir la correction

The sum of four consecutive integers starting at

n

n + (n + 1) + (n + 2) + (n + 3) = 4 n + 6

. This equals

6

times an integer iff

4 n + 6

is divisible by 6, i.e. iff

4 n

is divisible by 6. Since

g cd (4, 6) = 2

, this requires

2 n

to be divisible by 3, i.e.

n

divisible by 3 (since

g cd (2, 3) = 1

). Check:

n = 0

: sum

= 6

(multiple of 6, tick);

n = 1

: sum

= 10

(not);

n = 3

: sum

= 18

(tick);

n = 6

: sum

= 30

(tick). So (*) is true precisely when

n

is a multiple of 3. Hence the answer is C.