# Prove that ab divides c.

• Mar 2nd 2010, 10:17 PM
rainyice
Prove that ab divides c.
Let a|c and b|c, and (a,b)=1, prove that ab divides c.

At first, I don't think this is right. For instance, let a=1 and b=2, then (a,b)=1. However, if c=3, then 2 can't divides 3 which implies b can't divides c......
• Mar 2nd 2010, 11:42 PM
Swlabr
Quote:

Originally Posted by rainyice
Let a|c and b|c, and (a,b)=1, prove that ab divides c.

At first, I don't think this is right. For instance, let a=1 and b=2, then (a,b)=1. However, if c=3, then 2 can't divides 3 which implies b can't divides c......

$a|c \Rightarrow ax=c$, then as $\text{gcd}(a, b) = 1$ and $b|c$ we have that $b|x$. Thus, $abx^{\prime} = c \Rightarrow ab|c$.
• Mar 3rd 2010, 06:53 AM
rainyice
Quote:

Originally Posted by Swlabr
$a|c \Rightarrow ax=c$, then as $\text{gcd}(a, b) = 1$ and $b|c$ we have that $b|x$. Thus, $abx^{\prime} = c \Rightarrow ab|c$.

where does that b|x come from???
• Mar 3rd 2010, 09:42 AM
Swlabr
Quote:

Originally Posted by rainyice
where does that b|x come from???

Because the gcd of a and b is 1, so it must divide x.

(The $b | c$ in my previous post was meant to be $b \nmid c$).
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