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I know that you can prove a function is one to one by graphing it and using the horizontal line test. But in my notes it showed another way to prove a function is one to one but I am not sure if I am doing it correctly. The notes say to assume that $f(x_{1})=f(x_{2})$ and show that $x_{1}=x_{2}$.

One of my homework problems asks me to determine whether or not $\dfrac{x+1}{x-1}$ is one to one.

I assume that $f(x_{1})=f(x_{2})$, $\dfrac{x_1+1}{x_1-1}=\dfrac{x_2+1}{x_2-1}$.

I am not sure how to show that $x_1=x_2$ after setting this step up. I have tried cross multiplying but it got messy. Can anyone help me?

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Cross-multiplying works fine: we find $(x_1+1)(x_2-1) = (x_2+1)(x_1-1)$, yielding $x_1x_2+x_2-x_1-1=x_2x_1+x_1-x_2-1$. The $x_1x_2-1$ on both sides disappears. Collecting $x_2$-terms on the left and $x_1$-terms on the right, we get $2x_2 = 2x_1$ or $x_1=x_2$.

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  • $\begingroup$ Could you explain how you got that? $\endgroup$ – Kot Mar 16 '14 at 21:32
  • $\begingroup$ I expanded it; is it more clear now? $\endgroup$ – user133281 Mar 16 '14 at 21:35
  • $\begingroup$ That makes sense now. I never tried distributing it... $\endgroup$ – Kot Mar 16 '14 at 21:36
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Are you sure you actually tried. We have $(x_1 + 1)(x_2 - 1) = (x_2 + 1)(x_1 - 1)$, that is, $2x_2 = 2x_1$, so $x_1 = x_2$, and we are done.

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You can write $\dfrac{x+1}{x-1}$ as $1 + \dfrac{2}{x-1}$. Then $$ 1 + \frac{2}{x_1-1}= 1 + \frac{2}{x_2-1}$$ $$ \implies \frac{2}{x_1-1}=\frac{2}{x_2-1} $$ $$\implies \frac{x_1-1}{2}=\frac{x_2-1}{2}$$ $$\implies x_1-1=x_2-1 $$ $$\implies x_1=x_2$$ as desired.

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