Prove equivalent fractions with $\pm$ sign

Question

If $$ \lambda=\frac{n_{1}}{n_{2}}=\frac{m_{1}}{m_{2}} $$ prove that $$ \lambda=\frac{n_{1}\pm m_{1}}{n_{2}\pm m_{2}} $$

I know this is true if I add numbers to it $$ \frac{1}{2}=\frac{2}{4} $$ $$ \frac{1+2}{2+4}=\frac{3}{6}=\frac{1}{2} $$ $$ \frac{1-2}{2-4}=\frac{-1}{-2}=\frac{1}{2} $$

Answer 1

Compute the difference: From $\frac{n_1}{n_2}=\frac{m_1}{m_2}$, we know tha tthe numerator of $$ \frac{n_1}{n_2}-\frac{m_1}{m_2}=\frac{n_1m_2-n_2m_1}{n_2m_2}$$ is zero. Hence $$\begin{align}\frac{n_1\pm m_1}{n_2\pm m_2}-\frac{n_1}{n_2}&=\frac{(n_1\pm m_1)n_2-n_1(n_2\pm m_2)}{(n_2\pm m_2)m_1}\\&= \frac{n_1n_2\pm m_1n_2-n_1n_2\mp n_1m_2}{(n_2\pm m_2)m_1}\\&=\frac{\pm(n_1m_2-n_2m_1)}{(n_2\pm m_2)m_1}\\&=0\end{align}$$

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Remark: Essentially, we use $\frac ab=\frac cd\iff ad=bc$.

Answer 2

$$λ=\frac{n_{1}}{n_{2}}=\frac{m_{1}}{m_{2}} \\\to \\n_1=\lambda.n_2 \\m_1=\lambda .m_2$$so $$\dfrac{n_1+ m_1}{n_2 +m_2}=\dfrac{(\lambda.n_2)+ (\lambda.m_2)}{n_2+ m_2}=\\ \lambda\dfrac{(n_2)+ (m_2)}{n_2+ m_2}=\lambda.1=\lambda$$also for $-$ $$\dfrac{n_1- m_1}{n_2 -m_2}=\dfrac{(\lambda.n_2)- (\lambda.m_2)}{n_2- m_2}=\\ \lambda\dfrac{(n_2)- (m_2)}{n_2- m_2}=\lambda.1=\lambda$$