# Thread: a few simple proofs

1. ## a few simple proofs

hey guys
having a problem with a couple of calculus proofs...please help!

(1) prove that $\displaystyle n$ is odd iff $\displaystyle n^2$ is odd
i have done the <--- part of this, im just having a problem proving that $\displaystyle n$ is odd if $\displaystyle n^2$ is odd (the ---> part)

(2) Show that there exists no rational number $\displaystyle x$ such that $\displaystyle x^2$ = 3.
i have attempted to use proof by contradiction but im getting nowhere!

thanks

2. Originally Posted by wik_chick88
hey guys
having a problem with a couple of calculus proofs...please help!

(1) prove that $\displaystyle n$ is odd iff $\displaystyle n^2$ is odd
i have done the <--- part of this, im just having a problem proving that $\displaystyle n$ is odd if $\displaystyle n^2$ is odd (the ---> part)

(2) Show that there exists no rational number $\displaystyle x$ such that $\displaystyle x^2$ = 3.
i have attempted to use proof by contradiction but im getting nowhere!

thanks
1
(a) if n is odd, then n^2 is odd
Let n = 2p + 1.
n^2 = (2p + 1)^2 = 4p^2 + 4p + 1 = 2(2p^2 + 2p) + 1 = 2m + 1 which is odd

Now show if n^2 is odd, then n is odd.

3. i have already proven that, im having problem proving that if n^2 is odd then n is odd!

4. Originally Posted by wik_chick88
i have already proven that, im having problem proving that if n^2 is odd then n is odd!
Contradiction. Assume n is even.

5. but then aren't you just proving that if n is odd, then n^2 is odd? not the other way around?

6. Originally Posted by wik_chick88
(2) Show that there exists no rational number $\displaystyle x$ such that $\displaystyle x^2$ = 3.
i have attempted to use proof by contradiction but im getting nowhere!
Try the rational root theorem. $\displaystyle x^2 = 3 \implies x^2 - 3 = 0$

The only rational roots of this equation are going to be $\displaystyle \pm1,~\pm3$. Since none of these work then the solutions of $\displaystyle x^2 - 3 = 0$ are not rational. Therefore they are irrational.

-Dan

7. Originally Posted by wik_chick88
but then aren't you just proving that if n is odd, then n^2 is odd? not the other way around?
We have already shown that if n is odd, then n^2 is odd. Next, we have to show if n^2 is odd, then n is odd.

Contradiction: n^2 is odd and n is even.

Since n is even, 2|n, and since n^2 is odd, $\displaystyle 2\nmid n^2$

$\displaystyle 2\nmid (n\cdot n)\Rightarrow 2\nmid n \ \text{or} \ 2\nmid n$.

But, we have 2|n. Therefore, we have reached a contradiction, and if n^2 is odd, then n is odd.

8. To show that $\displaystyle \displaystyle n^2\textrm{ is odd} \implies n\textrm{ is odd}$, you can use a proof by exhaustion. You need to consider the cases $\displaystyle \displaystyle n\textrm{ is odd}, n\textrm{ is }0, n\textrm{ is even}$ and show what you get for each when you get $\displaystyle \displaystyle n^2$.