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Let $a,b,m \in \mathbb{Z} $ prove that $m\gcd(a,b)=\gcd(ma,mb)$ iff $m>0$ and $a,b\neq 0$

def $\gcd(a,b)$

  • $d|a$ and $d|b$
  • $c|a$ and $c|b$ $\Rightarrow$ $c \leq d$

My attempt

$\Rightarrow ]$

If $m>0$ and $a,b \neq 0 $ ( need to show $\Rightarrow m\gcd(a,b)=\gcd(ma,mb)$)

Set $\gcd(a,b)=d$,

Now

  • $md|ma$ and $md|mb$
  • if $c|am$ and $c|bm$

so $m\gcd(a,b)=\gcd(ma,mb)$

$\Leftarrow ]$ if $ \gcd(ma,mb)=m\gcd(a,b)\Rightarrow m>0 $ and $a,b \neq 0$

So,

  • $m\gcd(a,b)|ma$ and $m\gcd(a,b)|mb$
  • $c|ma$ and $c|mb$ $\Rightarrow c \leq m \gcd(a,b)$

$\vdots$ foggy at this point

$\therefore$ $m>0$ and $a,b \neq 0$

I am not sure if my proof is correct. Appreciate any constructive help.

vadim123
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Tiger Blood
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1 Answers1

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Now, this is a bit diff than other problems and the easies way to thread the needle is with .... well ordering axiom and linear combos

$\Rightarrow$

note that $\gcd(0,0)=\gcd(0)$ there is no so such number given the gcd add 1 that is now a gcd contradiction. so $a,b$ must be nonzero both.worst case scenario one to be zero and only one.

if $ m \gcd(a,b)=\gcd(ma,mb)$ Let $d=\gcd(a,b)$ that is the least positive integer of linear combination $$ d=va+ub$$

Now least positive element $md=mvu+ub$ so $$md=mvu+ub$$ making $m \gcd(a,b)=\gcd(ma,mb)$

since d is positive, m is also positive where $a,b\neq 0$ since $vu+ub$ is a positive integer nonzero

$\Leftarrow$

if $m>0$ $a\neq 0 \wedge b\neq 0$ consider $\gcd(a,b)$ it does exists call it d by well ordering axiom there is a least positive element of of linear combo $va+bu $ so the least positive element of $$uvm+bum=md$$ is least combo that is $md=m\gcd(a,b)=\gcd(am,bm)$

Tiger Blood
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