Simple Rational Function

Proof : $ \displaystyle \frac{{ax+b}}{{cx+d}}$ is one to one correspondence
\$ \displaystyle \begin{array}{l}f(x)=\displaystyle \frac{{ax+b}}{{cx+d}}\\\\f\ \text{is one to one}\ \Rightarrow f(x)=f(y)\Rightarrow x=y\\\\\text{ Assume that}\ f(x)=f(y),\text{then}\\\\\displaystyle \frac{{ax+b}}{{cx+d}}=\frac{{ay+b}}{{cy+d}}\\\\acxy+adx+bcy+bd=acxy+bcx+ady+bd\\\\adx+bcy=bcx+ady\\\\adx-bcx=ady-bcy\\\\(ad-bc)x=(ad-bc)y\\\\\therefore \ x=y\ (ad\ne bc)\\\\\text{Hence}\ f\ \text{is one to one}\text{.}\end{array}$

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