THE LATEX BRAGGER IS WILSON!!! THERES A NEW ONE

PLEASE READ THIS BEFORE YOU OKAY JUST READ THAT ALRIGHT

Since the whole topic was about functions, the solution probably has stuff to do with functions...

NOPE

I used a proof by contradiction. Very easy to understand. Thought it was clever. All algebra.

:D


https://latex.artofproblemsolving.com/miscpdf/wnvoimxv.pdf?t=1563576156993



We can explain why $f$ and $g$ are not the same function through a proof by contradiction. Let's pretend that $\sqrt{\frac{x+1}{x-1}}=\frac{\sqrt{x+1}}{\sqrt{x-1}}$, and prove that that can not be the case.


We can begin by multiplying both sides by $\sqrt{\frac{x+1}{x-1}}$

$$\sqrt{\frac{x+1}{x-1}}=\frac{\sqrt{x+1}}{\sqrt{x-1}}$$
$$\left(\sqrt{\frac{x+1}{x-1}}\right)^2=\frac{\sqrt{x+1}}{\sqrt{x-1}}\cdot \sqrt{\frac{x+1}{x-1}}$$
$$\frac{x+1}{x-1}=\frac{\sqrt{x+1}\cdot \sqrt{\frac{x+1}{x-1}}}{\sqrt{x-1}}$$
$$\frac{x+1}{x-1}=\frac{\sqrt{(x+1)\cdot \frac{x+1}{x-1}}}{\sqrt{x-1}}$$
$$\frac{x+1}{x-1}=\frac{\sqrt{\frac{(x+1)^2}{x-1}}}{\sqrt{x-1}}$$


We can go back to our original equation of $\sqrt{\frac{x+1}{x-1}}=\frac{\sqrt{x+1}}{\sqrt{x-1}}$. Letting $x+1=a$ and $x-1=b$, we come up with a property in this pretended case:
$$\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$$


We can apply this property to the numerator of the right hand side.

$$\frac{x+1}{x-1}=\frac{\sqrt{\frac{(x+1)^2}{x-1}}}{\sqrt{x-1}}$$
$$\frac{x+1}{x-1}=\frac{\frac{\sqrt{(x+1)^2}}{\sqrt{x-1}}}{\sqrt{x-1}}$$
$$\frac{x+1}{x-1}=\frac{\frac{(x+1)}{\sqrt{x-1}}}{\sqrt{x-1}}$$
$$\frac{x+1}{x-1}=(x+1)\cdot \frac{\sqrt{x-1}}{\sqrt{x-1}}$$
$$\frac{x+1}{x-1}=x+1$$


Obviously, this is not true, meaning that $\boxed{\boxed{\sqrt{\frac{x+1}{x-1}}\neq \frac{\sqrt{x+1}}{\sqrt{x-1}}}}$



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