Mathematics · Oct / Nov 2020 · Paper 41 · Question 1
YMA01
As printed in the paper
The question as text
1. Given that n is an integer, use algebra, to prove by contradiction, that if n3 is even then n is even. (4)
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Assume that there exists a number m such that when
m is even, m is odd
(
)
3
If m is odd then
m
2
p
+ (where p is an integer) and
1
m
3
=
2
p
+
1
=
...
=
=
8
p
3
+
12
p
2
+
6
p
+
1
(
)
3n is
2
×
4
p
3
+
6
p
2
+
3
p
+ is odd and hence we have a contradiction so if
1
even, then n is even.
B1: For setting up the contradiction.
3
Eg Assume that there exists a number m such that when
m is even, m is odd
Condone a contrapositive statement here
3
''Assume that there exists a number m such that when
m is even, m is not even''
3
As a minimum accept ''assume if
m is even then m is odd.''
Condone the other way around '' assume if n is odd then n3 is even''
M1: Attempts to cube an odd number. Accept an attempt at (
2
p
1 ,
)
3
(
2
p
1
)
3
+
−
Look for (
)
2
p
+
1
3
=
...
p
3
.....
(
)
A1: (
)
3
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