1. ## Prove Divisibility

For any integer $\displaystyle a$ and $\displaystyle n, n | a^2$ if and only if $\displaystyle n | a.$

Let P: $\displaystyle n | a^2$
and $\displaystyle Q: n | a$.

I have no trouble proving $\displaystyle Q \Rightarrow P$.

I have trouble with $\displaystyle \sim Q \Rightarrow P$.

I am doing it by showing that if $\displaystyle n \not | a$, then $\displaystyle a \not = ny$ or equivalently,

$\displaystyle a = n y + c$, where $\displaystyle y, c \in \mathbb{Z}$ and 1$\displaystyle \leq c < n$. It doesn't look too good.

How do you prove $\displaystyle P \Rightarrow Q$?

For $\displaystyle P \Rightarrow Q$ is it okay to prove $\displaystyle Q \Rightarrow \sim P$ to get the contradiction?

2. Originally Posted by novice
For any integer $\displaystyle a$ and $\displaystyle n, n | a^2$ if and only if $\displaystyle n | a.$

Let P: $\displaystyle n | a^2$
and $\displaystyle Q: n | a$.

I have no trouble proving $\displaystyle Q \Rightarrow P$.

I have trouble with $\displaystyle \sim Q \Rightarrow P$.

I am doing it by showing that if $\displaystyle n \not | a$, then $\displaystyle a \not = ny$ or equivalently,

$\displaystyle a = n y + c$, where $\displaystyle y, c \in \mathbb{Z}$ and 1$\displaystyle \leq c < n$. It doesn't look too good.

How do you prove $\displaystyle P \Rightarrow Q$?

For $\displaystyle P \Rightarrow Q$ is it okay to prove $\displaystyle Q \Rightarrow \sim P$ to get the contradiction?
$\displaystyle 4\mid 2^2$ but $\displaystyle 4\nmid 2$

3. Originally Posted by Drexel28
$\displaystyle 4\mid 2^2$ but $\displaystyle 4\nmid 2$
How should I device the question so that $\displaystyle n | a$? Never mind. Thanks anyway.

4. Originally Posted by Drexel28
$\displaystyle 4\mid 2^2$ but $\displaystyle 4\nmid 2$
Drexel,
I just realized, you can't use a counter example because postulate is a biconditional logic. It says $\displaystyle n|a^2$ unless $\displaystyle n|a$.

5. Originally Posted by novice
Drexel,
I just realized, you can't use a counter example because postulate is a biconditional logic. It says $\displaystyle n|a^2$ unless $\displaystyle n|a$.
I'm no logician but $\displaystyle n\mid a^2\Longleftrightarrow n\mid a$ is countered by my example.

6. Originally Posted by Drexel28
I'm no logician but $\displaystyle n\mid a^2\Longleftrightarrow n\mid a$ is countered by my example.
You are right; it's logically inconsistent.