Prove that if $na=mb$ and $(a,b)=1$, then $m=a$ and $n=b$.

Question

I'm sorry but I'm having a massive brain fart. I'm trying to show that if $na=mb$ and $(a,b)=1$, with $a,b,n,m \in \mathbb{N}$ of course, then $m=a$ and $n=b$.

Moving to $\mathbb{Q}_+$, we note that

$$\frac{a}{b}=\frac{m}{n}$$

but $a$ and $b$ are relatively prime, so the fraction on the left-hand side is irreducible. Therefore the right-hand side must be also irreducible (if not, reduce it until it is). Therefore $m=a$ and $n=b$.

This, of course, relies on the assumption that $(a/b)=(m/n)$ and $(a,b)=1$ implies that the numerators and denominators are equal, which is basically the same as the initial problem. I know I'm forgetting a super simple proof for this. Could you please tell me it, or at least hint me towards it?.

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Edit: Oh and $m$ and $n$ are relatively prime.

Answer 1

By the Bezout's identity, there are integers $r,s,p$ and $q$ such that $$ ra+sb=1,\\ pm+qn=1. $$ Multiplying the first equality by $n$ and using $na=mb$, we have $$ n=rna+snb=rmb+snb=(rm+sn)b\implies b|n. $$ Similarly, multiplying $1=pm+qn$ by $b$, we get $$ b=pmb+qnb=pna+qnb=(pa+qb)n\implies n|b. $$ Because $b$ and $n$ are both positive integers, we infer that $b=n$. From this, $a=m$ is immediate.

Answer 2

Your claim is not true. There are infinitely many solutions for $m,n$:

Your assumption is that only the pair (a,b) is relatively prime. It does not stop the other pair from having common factors: Now set $m=ka$ and $n=kb$; they satisfy $na=mb (=kab)$. This is valid for all integers $k$.

Answer 3

I'll have a go at this with my small amount of knowledge from a first year proofs class lol.

Since $n = (mb /a)$, and $n$ is a natural number, $a$ has to either divide $m$ or $b$ evenly. However we stated that $b$ and $a$ are co prime, therefore $a$ has to divide $m$. Dividing both sides by $n$ would result in $1 = (mb/an)$. Since $a$ and $b$ are co prime, $m/n$ would have to be the reciprocal of $b/a$, resulting in $1$. Thus, $m = a$ and $n = b$.

Answer 4

Yes, we can prove such uniqueness via principality of fractions. Applying the Theorem below,

$\quad\rm (m,n)=1\,$ and $\rm\ \dfrac{a}b\, =\, \dfrac{m}n\,\Rightarrow\ \begin{array}{} \rm a\, =\, \color{#c00}k\,m\\ \rm b\, =\, \color{#c00}k\,n\end{array}\ $ so $\rm\ (a,b)=1\,\Rightarrow\, \color{#c00}{k=1}.\ \ $ QED

Theorem $ $ For $\rm\:m,n,x,y\in \Bbb Z,\,$ if $\rm\ (m,n) = 1,\:$ then $\rm\ \dfrac{x}y\, =\, \dfrac{m}n\ \Rightarrow\:\begin{array}{c}x\, =\, k\,m\\ \rm y\, =\, k\,n\end{array}\ \ $ for some $\rm\ k\in \Bbb Z$

Proof $ $ By Euclid's Lemma, $\rm\ (n,m)=1,\ nx = my\,\Rightarrow\,n\mid y,\:$ so $\rm\ \dfrac{x}m = \dfrac{y}n = k,\:$ for some $\rm\:k\in \Bbb Z.$