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guys.

I just started to learn infinitesimal mathematics 1 (I think it's analagous to calculus A - the professor said that it's the most theoretical course on calculus offered in the university (The Technion, in Israel).

So I'm just saying that I'm a noobie.

We weren't really taught things in number theory and things that are related to mathematics, specifically, but the course has a short introduction about $\mathbb{R}$eal numbers.

In the homework I've got, I'm supposed to prove this:

$\forall n\in\mathbb{N}(3|n^2\Rightarrow 3|n)$

(meaning that $3$ divides $n^2\Rightarrow 3$ divides $n$.

What I did so far is this:

$3|n^2 \because given\Rightarrow$

$n^2 = 3k\,|\, k\in\mathbb{N}$

$n^2 = 3k\, \>|:3\Rightarrow$

$\frac{n^2}{3}=k$

I really don't know what to do beyond this.

I need to rely on high school knowledge and intuitive logic and formulate that. The idea of doing this is practicing making formal proofs.

What knowledge am I suppose to rely on?

Thanks for everyone in advance!

Gal Grünfeld
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  • You could have a look here: http://math.stackexchange.com/questions/1908946/confused-by-proof-of-the-irrationality-of-root-2-if-p2-is-divisible-by-2 – Watson Sep 01 '16 at 10:20
  • Oh, okay, thank you! I suppose I should have searched for longer, than. – Gal Grünfeld Sep 01 '16 at 10:21

3 Answers3

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You want to prove the statement $\color\red{3|n^2}\implies\color\green{3|n}$.

Instead, prove the equivalent statement $\neg(\color\green{3|n})\implies\neg(\color\red{3|n^2})$:

$\neg(3|n)\implies$

$3\not|n\implies$

$n\not\equiv0\pmod3\implies$

$[n\equiv1\pmod3]\vee[n\equiv2\pmod3]\implies$

$[n^2\equiv1^2\pmod3]\vee[n^2\equiv2^2\pmod3]\implies$

$[n^2\equiv1\pmod3]\vee[n^2\equiv4\pmod3]\implies$

$[n^2\equiv1\pmod3]\vee[n^2\equiv1\pmod3]\implies$

$n^2\equiv1\pmod3\implies$

$n^2\not\equiv0\pmod3\implies$

$3\not|n^2\implies$

$\neg(3|n^2)$

barak manos
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  • Let me know if you're having a hard-time with the "$\equiv$" notation, I'll change it to something similar to your usage of $k$ in your question... – barak manos Sep 01 '16 at 10:43
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Hint: instead of that prove more generally that if $p$ is a prime number and $p$ divides a product (of integer numbers) $ab$ the $p$ divides $a$ or $p$ divides $b$. Then use the fact that $3$ is prime

b00n heT
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  • Question about division: $\forall n,m,k(n,m,k \in \mathbb{N}, n|k \overset{def}{=} m \cdot n = k)$, right? – Gal Grünfeld Sep 01 '16 at 10:35
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    It should be written $$n\mid k: \Leftrightarrow \exists m \text{ such that } m\cdot n=k$$ – b00n heT Sep 01 '16 at 10:37
  • Alright. Thanks for making it more precise.

    I understand the difference.

    I just have a question about the notation (since I'm a newbie, like I said): about the "$:\Leftrightarrow$" - shouldn't the whole thing be "$n|k \overset{def}{=}n|k\Leftrightarrow\exists m\in Z|m\cdot n =k$"?

     – Gal Grünfeld Sep 01 '16 at 10:45
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Calculate the squares modulo $3$. Any integer is congruent to $0$, $1$ or $-1$, so squares are congruent to $0^2=0$, $1^2=1$ pr $(-1)^2=1$ respectively.

The only case with $n^2\equiv 0\mod 3$ is $n\equiv 0\mod 3$.

Bernard
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