# Taking the same thing of both sides of an equation

I have questions regarding the validity of taking:

1. The same function (inverse or otherwise) on both sides of an equation. For example, is it possible to conclude $f(x)=f(y)$ from $x=y$ or $f(x)=x$ from $f^{-1}(f(x))=f^{-1}(f^2(x))$ from $f(x)=f^2(x)$? If it is valid, what are some other rules that you can conclude using this? For example, consider the functional equation $f(x)+f(\frac{x-1}{x})=x+1$; is it possible to conclude $x+\frac{x-1}{x}=f^{-1}(x+1)$?

2. Logs on both sides of an equation. I am more accustomed to this operation, but I don't understand why it is valid.

Please tell me if the above operations are legal and why.

• 1. $x=y \Rightarrow f(x) = f(y)$ is true for any well-defined function, the converse is note necessarily true. As to your functional equation, $f^{-1}(a+b) \neq f^{-1}(a) + f^{-1}(b)$ in geral as you assume. Also, you're assuming the inverse exists.  2. $f(x) = \log x$, then it's just applying a function to both sides of an equation.  Functions have to map one element of the domain to exactly one element in the image, it can map different elements to the same thing if it wishes, but it is not possible to map one element to two different things, it's no longer a function then. – Zain Patel Jul 31 '16 at 19:37

## 1 Answer

You can usually apply a function to both sides of an equation as long as you are careful. Let's look at some of the examples you mentioned.

1. $X=Y$ implies $f(X)=f(Y)$. This is true! If $X$ and $Y$ are equal then applying any function to both should give the same thing. This is what people mean when they say that a function is 'well-defined'.

2. $f(X)=f^2(X)$ implies $f^{-1}(f(X))=f^{-1}(f^2(X))$ implies $X=f(X)$. This is true so long as $f^{-1}$ indeed exists as a function. For instance, the function $f(X)=X^2$ does not have a well defined inverse since $f(X)=f(-X)$ so $f^{-1}(f(X))$ could be either $X$ or $-X$.

3. $X+ \frac{X-1}{X}= f^{-1}(X+1)$ implies $f(X)+f(\frac{X-1}{X})=X+1$. Careful! this is not true. The problem is that when you apply $f$ to both sides, the left becomes $f(X+\frac{X-1}{X})$ and this need not be the same as $f(X)+f(\frac{X-1}{X})$.

4. You can take logs of both sides so long as you keep in mind that $\mathrm{log}(X)$ only makes sense for $X>0$.