text: # 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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