Divisibility, LCM and GCD for fractions (rationals)

how do i find lcm of two fractions? For example:

$\frac{2}{3}$ and $\frac{5}{8}$

A notion of gcd and lcm for rationals arises naturally by extending the divisibility relation from integers to rationals, i.e. for rationals $\rm\:r,s,\:$ we define $\rm\:r\:$ divides $\rm\:s,\:$ if $\rm\ s/r\:$ is an integer, $ $ in symbols $\rm\:r\:|\:s\:$ $\!\iff\!$ $\rm\:s/r\in\mathbb Z\: $ [such divisibility relations induced by subrings are discussed further here].

Then, as proved here, we have these dual formulas for $\,\rm\color{#c00}{reduced}\,$ fractions $\rm \,a/b,\ c/d$

$$\rm\ gcd\left(\frac{a}b,\frac{c}d\right) = \frac{gcd(a,c)}{lcm(b,d)}\ \ \ if\ \ \ \color{#c00}{\gcd(a,b) = 1 = \gcd(c,d)}$$

$$\rm\ lcm\left(\frac{a}b,\frac{c}d\right) = \frac{lcm(a,c)}{gcd(b,d)}\ \ \ if\ \ \ \color{#c00}{\gcd(a,b) = 1 = \gcd(c,d)}$$

Some of these ideas date to Euclid, who computed the greatest common measure of line segments, by anthyphairesis (continually subtract the smaller from the larger), i.e. the subtractive form of the Euclidean algorithm. Euclid's method's become clearer when one learns about fractional ideals. The above methods work much more generally since they do not require the existence of a Euclidean (division) algorithm but, rather, only the existence of (certain) gcds.

As others have noted, it's not quite clear what you're asking for. I'll suppose that you want the smallest positive rational number $m$ that is an integer multiple of both $\frac23$ and $\frac58$. Contrary to the comment above, this is well-defined.

To calculate this, we want $m = \frac23a = \frac58 b$ where $a$ and $b$ are integers, and $m$ is as small as possible. We can multiply through by 24 to get $16a = 15b$. The smallest integer $a$ and $b$ satisfying this is clearly $a=15, b=16$, because 16 and 15 are relatively prime. So the least common multiple $m$ of $\frac23$ and $\frac 58$ is $$\frac23\cdot15 = \frac58\cdot16 = 10.$$

In general we can proceed the same way. Say we want the LCM of $\frac ab$ and $\frac{a'}{b'}$. That is, we want to find integers $p$ and $q$ such that $p\frac ab = q\frac{a'}{b'}$. We can remove the denominators by multiplying both sides by $\def\lcm{\operatorname{lcm}}\lcm(b, b')$, which is equal to $bb'\over \gcd(b, b')$ by a theorem you have probably seen already. This gives us $$ pab' = qa'b .$$

This is solved by letting $m = \lcm(ab', a'b)$ and then $p=\frac m{ab'}, q=\frac m{a'b}$. The final answer is then $p\frac ab = q\frac{a'}{b'} = {m\over bb'}$.

In your example above we had $m = \lcm(ab', a'b) = \lcm(2\cdot 8, 5\cdot 3) = \lcm(16, 15) = 240$, so the final answer was ${240\over 3\cdot 8} = 10$, and the general solution is $$ \lcm\left(\frac ab, \frac{a'}{b'}\right) = {\lcm(ab', a'b)\over bb'}$$