# Proof

#### bosman3321

##### New member
How can i prove that for all positive integers of m and n: m and n are multiples of each other if and only if m=n

#### pka

##### Elite Member
How can i prove that for all positive integers of m and n: m and n are multiples of each other if and only if m=n
If $$\displaystyle n$$ is a multiple of $$\displaystyle m$$ then $$\displaystyle \exists k\in\mathbb{Z}^+$$ such that $$\displaystyle km=n$$.
If $$\displaystyle m$$ is a multiple of $$\displaystyle n$$ then $$\displaystyle \exists j\in\mathbb{Z}^+$$ such that $$\displaystyle jn=m$$.
Please finish the proof an post the result.

#### bosman3321

##### New member
If $$\displaystyle n$$ is a multiple of $$\displaystyle m$$ then $$\displaystyle \exists k\in\mathbb{Z}^+$$ such that $$\displaystyle km=n$$.
If $$\displaystyle m$$ is a multiple of $$\displaystyle n$$ then $$\displaystyle \exists j\in\mathbb{Z}^+$$ such that $$\displaystyle jn=m$$.
Please finish the proof an post the result.
I dont know how to.

#### pka

##### Elite Member
I dont know how to.
Suppose that $$\displaystyle m\text{ is a multiple of }n~\&~ n\text{ is a multiple of }m$$.
So $$\displaystyle \exists K\in\mathbb{Z}^+[K\cdot n=m]~\&~\exists J\in\mathbb{Z}^+[J\cdot m=n]$$
$$\displaystyle \therefore,\;$$
So it must be the case that:
\displaystyle \begin{align*}m&=K\cdot n \\&=K\cdot(J\cdot m)\\&=(K\cdot J)\cdot m \\\large1 &= K\cdot J \end{align*}
$$\displaystyle \LARGE\text{Now what?}$$

• Jomo

#### bosman3321

##### New member
If $$\displaystyle n$$ is a multiple of $$\displaystyle m$$ then $$\displaystyle \exists k\in\mathbb{Z}^+$$ such that $$\displaystyle km=n$$.
If $$\displaystyle m$$ is a multiple of $$\displaystyle n$$ then $$\displaystyle \exists j\in\mathbb{Z}^+$$ such that $$\displaystyle jn=m$$.
Please finish the proof an post the result.
Thanks
Got it

#### Jomo

##### Elite Member
Thanks
Got it
You were helped showing that if m and n are multiples of another then m=n. Did you show that if m=n, then m and n are multiples of one another? It is extremely easy but needs to be shown.