(Prove by contrapositive) If 3 is a factor of n^2, then 3 is a factor of n?

Answer 1

To prove this by contrapositive, we have to prove the following:

"If 3 is not a factor of n, then 3 is not a factor of n^2"

Assume 3 is not a factor of n. Then we have two cases:

1) n = 1 + 3k, for some integer k.
2) n = 2 + 3k, for some integer k.

Claim: In both of these cases, 3 is not a factor of n^2.

Case 1: n = 1 + 3k.

Then n^2 = (1 + 3k)^2 = 1 + 6k + 9k^2, so
n^2 = 1 + 3(2k + 3k^2), which implies 3 is not a factor of n^2.

Case 2: n = 2 + 3k.

Then n^2 = (2 + 3k)^2 = 4 + 12k + 9k^2.

Splitting up 4 as 1 + 3 we get

n^2 = 1 + 3 + 12k + 9k^2. Factoring a 3 out of the last 3 terms,
n^2 = 1 + 3(1 + 4k + 3k^2).

This implies 3 is not a factor of n^2.

Therefore, by proof by contrapositive, 3 is not a factor of n^2.

Answer 2

Contra → If 3 is not a factor of n, then 3 is not a factor of n².

Assuming 3 is not a factor of n, we have
n = 3m + r, where m, r are integers and r=1 or r=2.
→ n² = 9m² + 6rm + r²

The first two terms are multiples of 3, and r²(=1 or r²=4) is not. Therefore the sum, which equals n², is not divisible by 3.

QED.

Answer 3

n and n^2 have the same prime factors only that the exponents in n^2 is double of those in n. So if n^2 contains the prime factor 3 ,n mus t contain it because if not,as n and n^2 have the same prime factors n^2 would not contain it