Answer:
Explanation:
Let n be an integer
Given that [tex]n^3+5[/tex] is odd
To prove that n is even
a) Proof by contraposition
Let [tex]n^3+5[/tex] be non odd
Then this would be a multiple of 2 being even
[tex]n^3+5 = 2m\\n^3=2m-5\\n^3=2(m-3)+1[/tex]
i.e. we get cube of n is odd since gives remainder 1 when divided by 2
It follows that n is odd.
Thus proved by contraposition
b) contradiction method:
If possible let [tex]n^3+5[/tex] is odd for n odd.
Then we get
since n is odd,
[tex]n^3[/tex] is odd being the product of three odd numbers
When we add 5, we get
[tex]n^3+5[/tex] is even being the sum of two odd numbers
A contradiction
Hence our assumption was wrong
if n is an integer and n^3 + 5 is odd, then n is even
