# Proof

(related to Proposition: Divisibility Laws)

The divisibility laws follow immediately from the definition of divisors for integers:

• Using the definition of divisibilty, $$\frac{n}{1}=n\in\mathbb Z$$ and $$\frac{n}{n}=1\in\mathbb Z$$ for all $$n\in\mathbb Z$$.
• Therefore, $$1\mid n$$ and $$n\mid n$$ holds for all $$n\in\mathbb Z$$.
• Also, for all $n\in\mathbb Z$, we have $0\cdot n=0$, with $0\in\mathbb Z.$ Thus, $n\mid 0.$

• By hypothesis, $$m\mid n \wedge n\mid m$$.
• If $$m\mid n$$, then there exists $$a\in\mathbb Z$$ with $$a=\frac nm$$.
• Correspondingly, if at the same time $$n\mid m$$, then there exists $$b\in\mathbb Z$$ with $$b=\frac mn$$.
• The product of both numbers results in $$ab=\frac nm\cdot\frac mn=1$$.
• Since both, $$a$$ and $$b$$, are natural numbers, their product can only equal 1 if $$a=b=1$$.
• This shows that $$n=m$$.

• By hypothesis, $$m\mid n \wedge n\mid l$$.
• If $$m\mid n$$, then there exists $$a\in\mathbb Z$$ with $$a=\frac nm$$.
• If $$n\mid l$$, then there exists $$b\in\mathbb Z$$ with $$b=\frac ln$$.
• The product of both integers equals $$ab=\frac nm\cdot\frac ln=\frac lm=:c$$.
• Since $$c$$ is an integer ($$a$$ and $$b$$ are, so their product is), it follows that $$m\mid l$$.

• By hypothesis, $$m\mid n$$.
• If $$m\mid n$$, then there exists $$a\in\mathbb Z$$ with $$a=\frac nm$$.
• If we multiply both sides of the equation by any $$r\in\mathbb Z$$, then we get $$ar=\frac {nr}m$$.
• Since $$ar$$ is an integer, it follows that $$m\mid (nr)$$ for all $$r\in\mathbb Z$$.

• By hypothesis, $$m\mid n \wedge m\mid l$$.
• If $$m\mid n$$, then there exists $$a\in\mathbb Z$$ with $$a=\frac nm$$.
• Correspondingly, if at the same time $$m\mid l$$, then there exists $$b\in\mathbb Z$$ with $$b=\frac lm$$.
• If we add both equations, we get $$c:=a+b=\frac{n+l}m$$.
• Since $$c$$ is an integer ($$a$$ and $$b$$ are, so their sum is), it follows that $$m\mid (n+l)$$.
• Note: This holds also for the subtraction of integers $$m\mid (n-l)$$, as a special case of addition.

• By hypothesis, $$m\mid n \wedge m\mid (n+l)$$.
• If $$m\mid n$$, then there exists $$a\in\mathbb Z$$ with $$a=\frac nm$$.
• Correspondingly, if at the same time $$m\mid (n+l)$$, then there exists $$b\in\mathbb Z$$ with $$b=\frac {n+l}m$$.
• The difference $$c=b-a=\frac lm$$ is an integer (since $b$ and $a$ are.)
• This proves that $$m\mid l$$.
• Note: This holds also for the subtraction of integers in the hypothesis $$m\mid n \wedge m\mid (n-l)$$, as a special case of addition.

• By hypothesis, $m\mid n\wedge r\neq 0.$
• Since $$m\neq 0$$, we have $$mr\neq 0$$.
• Furthermore, $$n=qm$$ for some integer $q.$
• It follows $$nr=qmr$$.
• Therefore, $$mr\mid nr$$.

$$"\Rightarrow"$$ * By hypothesis, $m\mid n \wedge m\mid l.$ * From rule 4 above, we have that $$m\mid nx$$ and $$m\mid ly$$ for any integers $x,y\in\mathbb Z.$ * From rule 5, it follows that $$m\mid (nx+ly)$$ for all integers $x,y\in\mathbb Z.$

$$"\Leftarrow"$$ * By hypothesis, $$m\mid (nx+ly)$$ for all integers $x,y\in\mathbb Z.$ * It follows in particular for $$x=0,y=1$$ and $$x=1,y=0$$ that $$m\mid n \wedge m\mid l$$.

Thank you to the contributors under CC BY-SA 4.0!

Github:

### References

#### Bibliography

1. Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927