Mathematics · Oct / Nov 2021 · Paper 41 · Question 10
YMA01
As printed in the paper
Show the answer and mark scheme
10(a)[1]
+
+
10(b)[5]
Assumption:
p
(where p
E.g. States that there exists integers p and q such that 3 2
=
q is
q
p
3
in its simplest form) and then cubes to get
2
=
q
3
p
3
2
=
⇒
p
3
=
2
q
3
and concludes that
p is even so therefore p is even
3
q
3
so (
)
If p is even then it can be written
p
=
2
m
2
m
3
=
2
q
3
States that
q
3
=
4
m
3
and concludes that
q is even so therefore q is even
3
This contradicts our initial statement, as if they both have a factor of 2 it means
that p
q is not in its simplest form, so 3 2 is irrational *
(a)
B1: See scheme. Requires correct reason/algebra and a statement of the expression being odd.
(
)
Allow even + even + even +1 = odd. Allow
2
p
4
p
2
+
6
p
+
3
+ = odd
1
(b)
M1: Sets up the contradiction AND cubes. Condone the omission of the fact that p
q is in its simplest form for this
3
p
p
mark. Condone as a minimum 3 2
2
=
followed by
=
o.e.
q
3
q
A1: States that
p
3
=
2
q
3
and concludes both that
p is even so therefore p is even.
3
Accept other equivalent statements to even such as ''multiple of 2''
3
Condone poor explanations so long as they state that both
p and p are even
so (
)
M1: Writes
p
=
2
m
2
m
3
=
2
q
3
and then attempts to find q3 = ...
A1: States that
q
3
=
4
m
3
and concludes that both
q is even so therefore q is even
3
Accept other equivalent statements to even such as ''multiple of 2''
q and q are even
3
Condone poor explanations so long as they state that both
A1*: Completely correct proof and conclusion with no missing statements.
To score this final mark the statements now need to be the correct way around .
q is even so therefore q is even
3
E.g.
It requires p
q to be in simplest form (or equivalent such as no common factor) in the initial assumption.
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