Why $x$ cannot be zero in $y=\frac{1}{\frac{1}{x}}$

Question

I have had this question for quite some time now but never tried to look for the answer but now it bugs me.

When simplifying $y=\dfrac1{\frac1x}$ we get $y=x$ where $0$ is defined ,I presume it is because the denominator can not be zero in a fraction ? Even if that is true what is wrong with it being zero as we get zero eventually :

$$y=\frac{1}{\frac{1}{0}}$$

$$y=\frac{1}{\infty}=0$$

Sorry if this is very naive :(

$\frac{1}{0}\ne \infty$, see https://en.wikipedia.org/wiki/Division_by_zero — TheBestMagician, May 30 '21 at 15:29
It is not that we can simplify it as zero. We don't know what $\frac{1}{0}$ is. It doesn't mean that the reciprocal of the reciprocal of zero is also zero. — soupless, May 30 '21 at 15:30
Yes, you are right, $x$ cannot be zero because it is the denominator of a fraction. — Angelo, May 30 '21 at 15:36

score 3 · Accepted Answer · answered May 30 '21 at 15:32

I think we would have a removable discontinuity at $x=0$, and other than that the function would follow $y=x$. This is because, saying $\frac 10=\infty$ is not correct, as $\infty$ is not a number and hence cannot be equated to anything, in this sense. Rather, saying $\frac 10\to \infty$ makes sense. This itself points towards a removable discontinuity, as the limit at $x=0$ would exist, but not the functional value.

score 1 · Answer 2 · answered May 30 '21 at 15:31

Define the map $f$ on the non-zero reals as the reciprocal function $f(x) = \frac 1 x$, and let $g = f \circ f$, the composition of $f$ with itself. The domain and the target of $f$ are $\mathbb R \setminus \{0\}$ by definition, and thus the same must hold for $g$. Diagrammatically: $$\mathbb R\setminus\{0\} \xrightarrow{\ \ f\ \ }\mathbb R\setminus\{0\}\xrightarrow{\ \ f\ \ }\mathbb R\setminus\{0\}.$$ The observation you make is equivalent to noticing that $g$ acts as the identity on its domain; thus, it can be continuously extended to the whole of $\mathbb R$ by prescribing $g(0) = 0$.

score 1 · Answer 3 · answered May 30 '21 at 15:34

1

The quotient $\frac10$ is undefined and $\infty$ is not a number. So, your justification is not appropriate.

It is usual, when we deal with rational fractions, to simplify them in following since: if you have, say, $\require{cancel}\frac{x^2-1}{x^2+x}$, you notice that this is equal to $\frac{\cancel{(x+1)}(x-1)}{\cancel{(x+1)}x}$, and then we simplify this, getting $\frac{x-1}x$. And this expression is defined for every number other than $0$.

In your case, after simplifying $\frac1{1/x}$ as much as possible, what we get is simply $x$. And $x$ is defined everywhere. So, we say that $\frac1{1/x}=x$ and leave it like that, without bothering with the number $0$.

answered May 30 '21 at 15:34

José Carlos Santos

427,504

2

But is it not the same fraction before and after simplification ? Why does the same fraction have different domains ? – Glowingbluejuicebox May 30 '21 at 15:51
Yes, absolutely. And therefore they are not the same function. I am just saying that, in the context of rational functions we usually do this simplification. – José Carlos Santos May 30 '21 at 15:54

score 0 · Answer 4 · answered May 30 '21 at 15:39

0

By definition, rational numbers can't have zero in denominator. $\frac1x$ is rational, so, its denominator can't be zero.

answered May 30 '21 at 15:39

aarbee

8,246

Why $x$ cannot be zero in $y=\frac{1}{\frac{1}{x}}$

4 Answers4