0

the question:

Prove by contraposition, if $n$ is a positive integer such that $n(\mod 3)=2$ then $n$ is not a perfect square.

I've started by negating the statement, "Not q then not P": Suppose if n is a perfect square, then n is not $2(mod3)$

Then: if k(mod3) = 2, then k=3q, for some integer q. then $n=k^2, = 9k^2 = 3(3q^2)$ there $n(mod3) = 0$

Is what im doing correct? After this ill attempt $3n+1$ then $3n+2$

I hate math
  • 123

2 Answers2

1

$8\mod 3=2$ but $8$ is not a...

1

The statement "if $n$ is $2$ modulo $3$, then $n$ is not a perfect square" has the contrapositive "if $n$ is a perfect square, then $n$ is not $2$ modulo $3$." What you need to do is verify that this is true: for every perfect square...

I suggest looking at the squares of $3n$, $3n+1$, and $3n+2$, and verifying that they all satisfy this proposition (every number has one of these three forms).

Andrew Dudzik
  • 30,074