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