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If we have 2 fractions and set them equal to each other, can we remove the numerator without changing the equation? So the demonators would just equal eachother, so $4p = 8$ Like this:

$$\frac{1}{4p} = \frac{1}{8}$$ Multiply each side by 4p:

$$\frac{4p*1}{4p} = \frac{4p*1}{8}$$ Then we cross out each 4p on the LHS: $$1 = \frac{4p}{8}$$ Then times each side by 8 to isolate p: $$1*8 = \frac{4p*8}{8}$$ And finnally we cross out each 8 on the LHS and left with: $$8= 4p$$ Can we do this with any number? Assuming of course the number is real.

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    $\begingroup$ Yes, if $1/x=1/y$ then $x=y$ ($x$ and $y$ nonzero reals). $\endgroup$ – Lord Shark the Unknown Nov 19 '18 at 7:05
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Suppose we have $$\frac{a}{b}=\frac{a}{c}$$

where $a \ne 0$.

Multiplying both sides by $\frac{bc}{a}$ gives us $c=b$.

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  • $\begingroup$ Thanks for confirming my thought! $\endgroup$ – Samurai Nov 19 '18 at 7:22
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Assuming $p \neq 0 $ (which is necessary for LHS not being undefined), you can directly multiply by the denominators' Minimum Common Multiple to yield:

$$ \frac{1}{4p}=\frac{1}{8} \quad\bigg|\cdot 32p \iff 8=4p \iff p=2$$

Another point of view:

The function $\,\,f(x):=\frac1x, \quad f: \mathbb{R^*}\longrightarrow \mathbb{R^*}\,\,$ is bijective (or one-to-one), which by definition means that:

$$\textit{If}\quad f(a)=f(b)\quad \textbf{then}\quad a=b$$

Your equation is nothing else than

$$f(4p)=f(8)$$

Considering $f$'s bijective:

$$4p=8 \iff p=2$$

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