# If I simplify an expression containing 1/x and 1/x goes away in the simplification, is the expression defined in $\Bbb{R}$ for x=0??

I'm having trouble understanding what happens when an algebraic expression in $\Bbb{R}$ that has a 1/x can be simplified into a form not containing 1/x.

For example, suppose I have:

$\frac{x^2-\frac{1}{x}}{x+\frac{1}{x}+1}$

If I simplify it in this way:

$\frac{\frac{x^3-1}{x}}{\frac{x^2+1+x}{x}}$ (Step 1)

$\frac{x^3-1}{x^2+1+x}$ (Step 2)

$\frac{(x-1)(x^2+1+x)}{x^2+1+x}$ (Step 3)

I finally get:

$x-1$ ,which is defined at $x=0$.

When I plug the expression $\frac{x^2-\frac{1}{x}}{x+\frac{1}{x}+1}$ in a graphing program it plots the line x-1 and WolframAlpha also simplifies this as x-1. But to do Step 1, are we not assuming that $\exists1/x\in\Bbb{R}$, which implies that $x\neq 0$?

Another example:

$x+\frac{1}{x+\frac{( 1+x)}{x}}$

$x+\frac{1}{\frac{x^2+ 1+x}{x}}$ (Step 1)

$\frac{x(\frac{x^2+ 1+x}{x})+1}{\frac{x^2+ 1+x}{x}}$ (Step 2)

$\frac{x^2+1+x+1}{\frac{x^2+ 1+x}{x}}$ (Step 3)

$\frac{x^2+x+2}{\frac{x^2+ 1+x}{x}}$ (Step 4)

And I finally get:

$\frac{x^3+x^2+2x}{x^2+x+1}$, which is defined at $x=0$. And again WolframAlpha gives this as a simplification and plotting software shows a curve that goes through the point $(0;0)$. But can Step 1 be done without assuming that $x\neq 0$??

• The new expression may be defined at 0, but it's not equivalent to the original. Commented Aug 20, 2017 at 21:13
• The fancy name for this is a removable singularly. Defining a value at $x=0$ is what removes it. Commented Aug 20, 2017 at 22:20

This is sort of meta, but: These are examples of your proving that a given function defined on $\mathbb R{\setminus}\{0\}$ can be extended to a continuous function on $\mathbb R$. Formally, you are defining a new, different, function with a different domain, which happens to agree with the original function at all places where the original one is defined, and at $0$ as well, in such a way as to be continuous. The original expression assumes $x\ne 0$, and so do all the steps of your derivation, until the last step, where you say, "oh look, this formula makes sense for $x=0$ as well, and even looks better than the formula we started with." Yes, but this is an act of mathematical creation on your part: the end formula is a different one from the one you started with.