I was thinking about fractions with prime denominators and wanted to confirm something. If I have two fractions less than 1 with different prime denominators, is it certain that these two fractions can never be equal. I will try to state it more formally:
Given two fractions with prime denominators
$\frac{a}{p_1}$ and $\frac{b}{p_2}$
Where $a,b \in \Bbb Z$ ; $a<p_1$ ; $b < p_2$ ; $p_1 \not = p_2$
Show that there are no $a$ and $b$ for which $\frac{a}{p_1} = \frac{b}{p_2}$
$\frac a{p_1} = \frac b{p_2} \iff$
$ap_2 = bp_1$.
This means $p_2|bp_1$ and as $p_2$ is prime $p_2|b$ or $p_2|p_1$ (Euclid's Lemma). But $p_1$ is a prime not equal to $p_2$ so $p_2|p_1$ is impossible so $p_2|b$.
But this means $\frac b{p_2}$ is an integer.
If we let $k:=\frac b{p_2}$ this would mean $a = p_1*k$ and $b=p_2*k$ and $\frac a{p_1} = \frac b{p_2} = k$. That's simple enough...
...but you said $b < p_2$ (and $a<p_1$ and that's impossible of $p_2|b$. So we have a contradiction.
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As a side note you should be aware that every rational number when written as a fraction $\frac ab$ so that $b\in \mathbb N$ and $a \in \mathbb Z$ will have only one unique expression in lowest terms.
So if $\frac a{p_1} = \frac b{p_2}$ but $p_1 \ne p_2$ and $a \ne b$ then it is not possible that these are both in lowest terms.
But if $\frac a{p_1}$ is not in lowest terms then there must be $k > 1$ so that $a= N*k$ and $p_1 = q*k$. But $p_1$ is prime so that would mean $k=p_1$ and $q=1$ so $\frac a{p_1} = N$. And $a=Np_1$. That would mean $\frac b{p_2}=N$ so $b=N{p_2}$ and we $a = Np_1$ and $b=Np_2$.
We'd have gotten the exact same results if we tried figuring $\frac b{p_2}$ wasn't in lowest terms first.
As you said $\frac a{b_2} = \frac b{p_2} < 1$ we know this case the above is impossible.