If lcm(a,b) = ab, why is gcd(a,b) = 1?

[Scootertaj]

1. If lcm(a,b)=ab, show that gcd(a,b) = 1

Homework Equations

We can't use the fact that lcm(a,b) = ab / gcd(a,b)

The Attempt at a Solution

I've already shown that gcd(a,b) = 1 → lcm(a,b) = ab but I can't figure out the other direction!

Any hints?

 

[Dustinsfl]

$lcm[a,b]=\prod p_i^{\max{\alpha_i,\beta_i}}$

Where $a=\prod p_i^{\alpha_i}$

and $b=\prod p_i^{\beta_i}$

And the gcd is the min.

Use that fact.

 

[Scootertaj]

We haven't really used that notation. I imagine it deals with divisibility rules just like the other direction.

 

[murmillo]

I think one way to approach it is to prove the contrapositive. If c is an integer not equal to 1 that divides a and divides b, then a = mc and b = nc for positive integers m and n. Then see if you can show that lcm (a,b) is not ab.

 

[Scootertaj]

Awesome, got it, thanks!