writing proof for greatest common divisor and least common multiple of fractions I know that LCM of fractions is equal to (LCM of numerator / GCD of denominator) and GCD of fractions is equal to ( GCD of numerator / LCM of denominator).
However , I wonder  that why these formulas like that. I tried to prove but I could not do it. Moreover these formulas does not make sense for me. Can you give reasonable proof for them.
 A: Let $\frac ab$ and $\frac nm$ be positive fractions.
$\frac ab = [\frac {a}{\gcd(a,n)}\cdot \frac {\operatorname{lcm}(b,m)}b]\times \frac {\gcd(a,n)}{\operatorname{lcm}(b,m)}$
And $\frac nm = [\frac {n}{\gcd(a,n)}\cdot \frac {\operatorname{lcm}(b,m)}m]\times \frac {\gcd(a,n)}{\operatorname{lcm}(b,m)}$
And both $[\frac {a}{\gcd(a,n)}\cdot \frac {\operatorname{lcm}(b,m)}b]$ and $[\frac {n}{\gcd(a,n)}\cdot \frac {\operatorname{lcm}(b,m)}m]$ are integers,
$\frac {\gcd(a,n)}{\operatorname{lcm}(b,m)}$ is a common divisor of $\frac ab$ and $\frac nm$.
Now suppose $\frac \gamma\delta$ (assume in lowest terms) was another common divisor so there are $k,j$ so that $\frac ab = \frac {k\gamma}{\delta}$ and $\frac nm = \frac {j\gamma}{\delta}$.
That means $\frac {a\delta}\gamma = bk$ so $\gamma$ divides $a\delta$ but as $\delta$ and $\gamma$ are relatively prime, that means $\gamma$ divides $a$  similarly $\frac {n\delta}\gamma = mj$ so $\gamma$ divides $n$ so $\gamma$ is a common divisor of $a,n$.
So $\gamma\le \gcd(a,n)$.
Likewise $\frac {a\delta}b = k\gamma$ and $\frac {n\delta}m = j\gamma$. So $a \delta$ is a multiple of $b$ and $n\delta$ is a multiple of $m$. But as $a,b$ and $n,m$ are relatively prime that means $\delta$ is a multiple of $b$ and $\delta$ is a multple of $m$ so $\delta$ is a common multiple of $b$ and $m$.
So $\delta \ge {\operatorname{lcm}(b,m)}m$
So $\frac \gamma \delta \le \frac {\gcd(a,n)}{\operatorname{lcm}(b,m)}$
So $\frac {\gcd(a,n)}{\operatorname{lcm}(b,m)} = \gcd(\frac ab, \frac nm)$.
Proving that $\operatorname{lcm}(\frac ab, \frac nm) = \frac {\operatorname{lcm}(a,n)}{\gcd(b,m)}$ is done similarly.
