Prove that $lcm(a,b)=ab$ iff a,b are relatively prime

Asked Mar 06 '23 at 22:00

Active Mar 07 '23 at 07:52

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Since this is an iff proof, I know it needs to be proved both ways, but I can't figure out the other way. Can someone look over what I have so far and see if it makes sense and help me out going the other way?

Proof

$<==$ If a and b are relatively prime, by definition, then $gcd(a,b)=1$.

We can also. use the. fact that $gcd(a,b)*lcm(a,b)= ab$ Now plugging in $gcd(a,b)=1$ we get:

$$1*lcm(a,b)=ab$$ $$lcm(a,b)=ab$$, just like we wanted.

$==>$ help?

edited Mar 06 '23 at 22:51

Bill Dubuque

272,048

asked Mar 06 '23 at 22:00

Markova

3

If you know that $\gcd(a,b)\times \text {lcm(a,b)}=ab$ then...isn't it clear? If $A\times B=C$ for positive numbers $A,B,C$ then $B=C\iff A=1$. – lulu Mar 06 '23 at 22:04
Simply rewrite your resoning whith equivalences instead of implications. – Anne Bauval Mar 06 '23 at 22:04
Does this answer your question? Show that lcm$(a,b)= ab$ if and only if gcd$(a,b)=1$ - found as the referenced question in Anne's proposed duplicate of $\operatorname{lcm}(a,b)=ab$ if and only if $\gcd(a,b)=1$. – John Omielan Mar 06 '23 at 22:16

Prove that $lcm(a,b)=ab$ iff a,b are relatively prime

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