Mathematics · May / June 2023 · Paper 41 · Question 7
YMA01
As printed in the paper
The question as text
7. Use proof by contradiction to prove that 7 is irrational. (You may assume that if k is an integer and k 2 is a multiple of 7 then k is a multiple of 7) (4)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
a
a
2
Assume that
7 is rational so that 7
=
7
=
b
b
2
(where a and b have no factors in common)
a
2
7
=
7
b
2
=
a
2
b
2
So a2 is a multiple of 7 which means a is a multiple of 7
2
2
2
2
a
=
7
k
7
b
=
49
k
b
=
7
k
So b2 is a multiple of 7 which means b is a multiple of 7. As a and b are both
multiples of 7, this contradicts the fact that a and b have no factors in common.
Hence √7 must be irrational.
Any variables may be used
M1: Starts the proof by contradicting the given statement and squares both sides.
Condone the omission of the statement about common factors here.
a
p
a
2
As a minimum allow ''assume 7
'' or similar ''let 7
''
=
7
=
=
p
2
=
7
q
2
b
b
2
q
A1: Reaches for example
7b
2
=
a
2
and deduces that a2 is a multiple of 7 which means a is a multiple of 7
Condone these being the ''wrong way around'' as long as they state both a and a2 are multiples of 7
Look for
7b
2
=
a
2
a
=
7
k
for example)
M1: Sets
7
and proceeds to
2
7
2
a
=
k
b
=
k
A1: Fully Correct proof. This must include
statements such as a
b is a fraction in simplest form or coprime, relatively prime, mutually
prime
E.g. a2 is a multiple of 7 which means a is a multiple of 7 (not the other way around)
and b2 is a multiple of 7 which means b is a multiple of 7 (not the other way around)
Be tolerant of slight issues with terminology/language used as this may not be the students first language.
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