Paxton Hoffman

2022-07-21

Prove equivalent fractions with $±$ sign
If
$\lambda =\frac{{n}_{1}}{{n}_{2}}=\frac{{m}_{1}}{{m}_{2}}$
prove that
$\lambda =\frac{{n}_{1}±{m}_{1}}{{n}_{2}±{m}_{2}}$
I know this is true if I add numbers to it
$\frac{1}{2}=\frac{2}{4}$
$\frac{1+2}{2+4}=\frac{3}{6}=\frac{1}{2}$
$\frac{1-2}{2-4}=\frac{-1}{-2}=\frac{1}{2}$

eishale2n

Compute the difference: From $\frac{{n}_{1}}{{n}_{2}}=\frac{{m}_{1}}{{m}_{2}}$, we know tha tthe numerator of
$\frac{{n}_{1}}{{n}_{2}}-\frac{{m}_{1}}{{m}_{2}}=\frac{{n}_{1}{m}_{2}-{n}_{2}{m}_{1}}{{n}_{2}{m}_{2}}$
is zero. Hence
$\begin{array}{rl}\frac{{n}_{1}±{m}_{1}}{{n}_{2}±{m}_{2}}-\frac{{n}_{1}}{{n}_{2}}& =\frac{\left({n}_{1}±{m}_{1}\right){n}_{2}-{n}_{1}\left({n}_{2}±{m}_{2}\right)}{\left({n}_{2}±{m}_{2}\right){m}_{1}}\\ & =\frac{{n}_{1}{n}_{2}±{m}_{1}{n}_{2}-{n}_{1}{n}_{2}\mp {n}_{1}{m}_{2}}{\left({n}_{2}±{m}_{2}\right){m}_{1}}\\ & =\frac{±\left({n}_{1}{m}_{2}-{n}_{2}{m}_{1}\right)}{\left({n}_{2}±{m}_{2}\right){m}_{1}}\\ & =0\end{array}$

$\lambda =\frac{{n}_{1}}{{n}_{2}}=\frac{{m}_{1}}{{m}_{2}}\phantom{\rule{0ex}{0ex}}\to \phantom{\rule{0ex}{0ex}}{n}_{1}=\lambda .{n}_{2}\phantom{\rule{0ex}{0ex}}{m}_{1}=\lambda .{m}_{2}$
so
$\frac{{n}_{1}+{m}_{1}}{{n}_{2}+{m}_{2}}=\frac{\left(\lambda .{n}_{2}\right)+\left(\lambda .{m}_{2}\right)}{{n}_{2}+{m}_{2}}=\phantom{\rule{0ex}{0ex}}\lambda \frac{\left({n}_{2}\right)+\left({m}_{2}\right)}{{n}_{2}+{m}_{2}}=\lambda .1=\lambda$
also for
$\frac{{n}_{1}-{m}_{1}}{{n}_{2}-{m}_{2}}=\frac{\left(\lambda .{n}_{2}\right)-\left(\lambda .{m}_{2}\right)}{{n}_{2}-{m}_{2}}=\phantom{\rule{0ex}{0ex}}\lambda \frac{\left({n}_{2}\right)-\left({m}_{2}\right)}{{n}_{2}-{m}_{2}}=\lambda .1=\lambda$

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