1. ## equation

Show that $\displaystyle (m,n) = 1 \Rightarrow (m) \cap (n) = (mn)$. so $\displaystyle mx+ny = 1$. Now $\displaystyle (m) = \{cm: c \in R \}$ and $\displaystyle (n) = \{dm: d \in R \}$. So show that there is some common element which implies that they must be equal?

2. Originally Posted by Sampras
Show that $\displaystyle (m,n) = 1 \Rightarrow (m) \cap (n) = (mn)$. so $\displaystyle mx+ny = 1$. Now $\displaystyle (m) = \{cm: c \in R \}$ and $\displaystyle (n) = \{dm: d \in R \}$. So show that there is some common element which implies that they must be equal?

I think you meant $\displaystyle (n) = \{dn: d \in Z \}$

Let me prove the following for you

if $\displaystyle m|x$ and $\displaystyle n|x$ then $\displaystyle mn|[\gcd(m,n)x]$

we have $\displaystyle k_1m+k_2n = \gcd(m,n)$
= $\displaystyle xk_1m+xk_2n = x\gcd(m,n)$

Now mn|LHS of the equation above. Hence the result.

In this case $\displaystyle gcd(m,n)=1$, so $\displaystyle mn|x$ Hence the result

3. Originally Posted by aman_cc
I think you meant $\displaystyle (n) = \{dn: d \in Z \}$

Let me prove the following for you

if $\displaystyle m|x$ and $\displaystyle n|x$ then $\displaystyle mn|[\gcd(m,n)x]$

we have $\displaystyle k_1m+k_2n = \gcd(m,n)$
= $\displaystyle xk_1m+xk_2n = x\gcd(m,n)$

Now mn|LHS of the equation above. Hence the result.

In this case $\displaystyle gcd(m,n)=1$, so $\displaystyle mn|x$ Hence the result
Can you use fact that either $\displaystyle (m) \cap (n)$ is disjoint or equal? Thus show that if there is some element in $\displaystyle (m) \cap (n)$ then $\displaystyle (m) = (n)$? Thus $\displaystyle (m) \cap (n) = (mn)$?

4. Sorry - but unable to follow you question completely.

5. Originally Posted by aman_cc
I think you meant $\displaystyle (n) = \{dn: d \in Z \}$

Let me prove the following for you

if $\displaystyle m|x$ and $\displaystyle n|x$ then $\displaystyle mn|[\gcd(m,n)x]$

we have $\displaystyle k_1m+k_2n = \gcd(m,n)$
= $\displaystyle xk_1m+xk_2n = x\gcd(m,n)$

Now mn|LHS of the equation above. Hence the result.

In this case $\displaystyle gcd(m,n)=1$, so $\displaystyle mn|x$ Hence the result
How does this show the result?

6. Originally Posted by Sampras
How does this show the result?
Hi - Maybe I am not getting your question, but here is the logic I have used.

We have proved if m|x and n|x then => mn|x

So let
1. x in (m) and x in (n) => x in (mn)
2. x in (mn) => x in (m) and x in (n)

Hence
$\displaystyle (m) \cap (n) = (mn)$

Essentially all I am saying is $\displaystyle (m) \cap (n) = (LCM(mn))$