# Calculator's view on the non-permissible values of the expression $\frac{1}{\frac{x}{x-1}+x}$

When I try to find the non-permissible values of the expression $$\cfrac{1}{\frac{x}{x-1}+x}$$ I rewrite it as

$$\cfrac{1}{\cfrac{x^2}{x-1}}$$

and conclude that the non-permissible values are $$x = 0$$ and $$x=1$$ as those values will make the denominators present in the expression undefined.

But this expression can also be written with the $$\div$$ symbol and then, using the "division is multiplying by a reciprocal," we can have $$1 \div \frac{x^2}{x-1} \stackrel{\text{multiply by recip}}{\implies} 1 \cdot \frac{x-1}{x^2}$$ and then one could conclude that $$x=0$$ is the only non-permissible value.

Three different calculators accept this claim:

(1) A TI-89

(2) Desmos graphing calculator,

(3) Google calculator, with $$1 \div (1 \div (1-1) +1)$$:

These three calculators seem to have no problem with the simplification of $$\cfrac{1}{\frac{x}{x-1}+x} = \frac{x-1}{x^2}$$ as they all treated $$x=1$$ as a permissible value (i.e. no errors).

What is happening here? Is my initial conclusion of the non-permissible values being $$x = 0$$ and $$x=1$$ correct?

I have heard that Desmos and other graphers may simplify the original expression to make operations more efficient, but I am not sure how this might apply to the Google calculator.

(Oddly enough, my Windows calculator gives me a divide-by-zero error for $$1 \div (1 \div (1-1) +1)$$.)

• You are right. Both $x=1$ and $x=0$ need to be excluded. I'm not sure what is happening here though – Jakobian Jul 16 '18 at 0:07
• "division by $k$ is multiplying by the reciprocal $1/k$" is only true if $k \neq 0$. – Will Sherwood Jul 16 '18 at 0:08

They are different functions since they have different domains. But, they agree on whatever numbers are common to their domain. By simplifying to $$\frac{x^2}{x - 1}$$ you are changing to a new function. For example, if I say your function is to take a number, $x$, divide $1$ by it, then divide $1$ by that result, you will get: $$x \mapsto \frac{1}{\frac{1}{x}}.$$ I could then define a different function that just repeats the input: $$x \mapsto x.$$ For $x \neq 0$, they agree. But, the first cannot be computed at $x = 0$.