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Prove that any integer divides zero: $a\in \mathbb Z \implies a\mid0$

How can i prove that any integer divides zero?

i tried using the definition of divisibility, but i dont know if for the formal proof, it can be used as i im using it.

$a\in \mathbb Z \implies a\mid0$

$a\mid0 \implies \exists b\in \mathbb Z$, such that, $0=a\cdot b$

And by knowing that $n\cdot 0=0$, will implie something, but i cant join together the pieces.

PROOF OF $n\cdot 0=0$

We take:

$$0=0$$ by zero property of addition: $$0+0=0$$ by definition of multiplication: $$a\cdot(0+0)=a\cdot0$$ by distributive law: $$a\cdot0 + a\cdot0 = a\cdot0$$ by cancellation law: $$a\cdot0=0$$

The cancellation law isn't under the field axioms and requires a proof for the above to be complete. Here's a proof:

We want to prove that if $a+c=b+c$, then $a=b$.

by the additive inverse property, we have an $c^{-1}$ such that $c^{-1}+c=0$. So by definition of addition:

$$c^{-1}+a+c=c^{-1}+b+c$$

by associativity and commutativity of addition:

$$(c^{-1}+c)+a=(c^{-1}+c)+b$$

by definition of $c^{-1}$:

$$0+a=0+b$$

by zero property of addition:

$$a=b$$

So we have proven the cancellation law.

Jyrki Lahtonen
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Victor F Salazar G
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  • Are you allowed to use that $n \cdot 0 = 0$ ? – DanZimm Jan 09 '14 at 02:48
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    Note that $0=a\times 0$ for any integer $a$. This means $a\mid 0$, by definition. – Pedro Jan 09 '14 at 02:48
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    Too many logical symbols! There exists an $x$ such that $ax=0$, since $a\cdot 0=0$. It follows that $a\mid 0$. – André Nicolas Jan 09 '14 at 02:49
  • @Victor Francisco Salaza : Looks good to me. But I think it is standard, as well as less confusing, to call the additive inverse of $c$ "$-c$", not "$c^{-1}$". – Stefan Smith Jan 09 '14 at 02:56
  • @StefanSmith well $\mathbb{Z}$ isn't a field, what exactly are your multiplicative inverses? Anyhow if you consider a field $F$ then first put $p=0=q \implies p = q, p + 0 = q + 0 \iff p + 0 = q \implies 0 + 0 = 0$ so then $0 = 0 + 0$ and $b (0) = b(0 + 0) = b\cdot 0 + b\cdot 0 = b\cdot 0 \iff b \cdot 0 = 0$ for any $b$. – DanZimm Jan 09 '14 at 03:18
  • @DanZimm : yes, my bad, of course $\mathbb{Z}$ isn't a field. I Googled "ring axioms" and "$n \cdot 0 = 0$" is not one of them. It looks like you and the OP have proven correctly that $n \cdot 0 = 0$ using only the basic ring axioms. – Stefan Smith Jan 09 '14 at 18:51

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Due to we have alredy proved that $a\cdot 0=0$, by the definition of divisibility* it follows that $a\mid 0$. Because $a\mid0 \implies \exists b\in \mathbb Z$, such that, $0=a\cdot b$

*$a\mid b \implies \exists c\in \mathbb Z$, such that, $b=a\cdot c$, whit $a,b\in \mathbb Z$

Victor F Salazar G
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