31

Is the fraction

$$\frac{1}{\frac{1}{0}}$$

undefined?

I know that division by zero is usually prohibited, but since dividing a number by a fraction yields the same result as multiplying the number by the fraction's reciprocal, you could argue that

$$\frac{1}{\frac{1}{0}} = (1)\left(\frac{0}{1}\right) = 0$$

Is that manipulation permissible in this case? Why or why not?

Peter Olson
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7 Answers7

31

Any expression having an undefined term somewhere inside is undefined as a whole. The rule $\frac1{\frac1x}=x$ holds only for $x\ne0$.

Hagen von Eitzen
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  • Why must that be so? Thinking of things in terms of limits, $\frac{1}{x}$ as $x$ goes to $0$ would be $\infty$ or $-\infty$ depending on whether you are looking from the right or left side. The special thing about $\frac{1}{\frac{1}{x}}$ as $x$ goes to $0$ is that regardless of what the denominator approaches, the whole expression still approaches $0$. – Peter Olson Apr 28 '13 at 20:41
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    @PeterOlson, that is definitely true, but simply writing "$\frac{1}{\frac{1}{0}}$" does not, in itself, say that a limit is being taken anywhere. This is why the expression is undefined, just like other similar expressions like $\frac{0}{0}$ and $0^0$. – Antonio Vargas Apr 28 '13 at 21:05
  • @AntonioVargas I still feel like this example is different than $\frac{0}{0}$ and $0^0$. $\frac{0}{0}$, for example, in the limit sense could be whatever number you want, say $1$ or $2$, by changing the expression in the limit: $\lim_{x\to0}\frac{x}{x} = 1$ and $\lim_{x\to0}\frac{2x}{x} = 2$. Is it possible to do something similar with $\frac{1}{\frac{1}{0}}$? – Peter Olson Apr 28 '13 at 21:14
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    @PeterOlson yes, I was just going to mention that those had additional reasons for being undefined. Ultimately if you decide to say that $\frac{1}{\frac{1}{0}} = 0$ then it will only be true because you have adopted that convention based on the behavior of a particular limiting expression. If you find it useful to do so then feel free---anyone is allowed to define a previously undefined expression to suit their needs---but it is still just convention. – Antonio Vargas Apr 28 '13 at 21:17
20

Another way to think about this is order of operations:

$$ \frac{1}{\frac{1}{0}}=1/(1/0) $$

I always compute what's inside the parenthesis first, which gives me undefined, and I have to stop there.

icurays1
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10

The expression is undefined because the value you are trying to divide $1$ by is undefined. Therefore the operation cannot take place. If you had something like $\frac{1}{\frac{1}{\ln 0}}$ then that would also be undefined because you cannot evaluate $\ln 0$. I think your misunderstanding comes from the fact that you treat $\frac{1}{0}$ as wholly separate term and ignore it's value (not even undefined).

Jeel Shah
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  • But there's a big difference in the $\ln0$ example, since $\lim_{x\to0}\frac{1}{\frac{1}{\ln x}}$ does not exist, but $\lim_{x\to0}\frac{1}{\frac{1}{x}} = 0$. – Peter Olson Apr 28 '13 at 20:47
  • @PeterOlson, why do you think it doesn't exist? Can't we get arbitrary close to $-\infty$ by making $x$ close enough to $0$? And while we do that everything stays defined. – Leo Apr 28 '13 at 21:29
  • @mathusiast I suppose it exists, but diverges. That said, in settings where complex numbers are disregarded, the left-hand side limit doesn't exist. – Peter Olson Apr 28 '13 at 21:34
  • @PeterOlson I am not aware of the complex plane and taking limits in that therefore, I was not privy to that information. How can I improve the post so that it is more technically correct? – Jeel Shah Apr 29 '13 at 00:33
5

Yeah, this is undefined. When we divide some number with some N.D. number it results in N.D. number. I think you are making a mistake by considering $1/0$ a fraction. Its not a fraction because denominator is zero, its a number which is not defined. So, here "fraction's reciprocal" doesn't make any sense.

But taking limit of $1/{(1/x)}$ as $x\to 0$ makes sense.

Sara Tancredi
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4

The algebraic identity $\frac{a}{\frac{b}{c}} = a\frac{c}{b}$ only holds when both $\frac{b}{c}$ and $\frac{c}{b}$ are defined. This is similar to why $\sqrt{-2}\sqrt{-3} \neq \sqrt{6}$.

Think of it in terms of "priority of operations:" $\frac{1}{\frac{1}{0}}$ can only be defined if both $1$ and $\frac{1}{0}$ are defined, because that is the ultimate operation in the expression, and both sides of it must be defined first. Otherwise, it is undefined itself also.

John
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3

A fraction $\frac{a}{b}$ is defined as a solution of the equation $bx=a$. Of course, the equation $0x=1$ has no solutions in $\mathbb{R}$. If you very want to solve this equation you could do as follows.

Consider $\mathbb{R}$ as a (multiplicative) semigroup and try to embed it in such a semigroup $S$ that $\exists x\in S: 0x=1$. Of course, you get nothing, since $0=0\cdot 1=0\cdot 0x=0x=1$. Then you can consider more general situation: take as $S$ some magma (see in Wikipedia "Magma (algebra)"). Since the multiplication will be non-associative, the contradiction disappears: you get $0(0x)=0$, i.e. $0\cdot 1=0$. I believe that such a magma exists, but I didn't try to build it.

Boris Novikov
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1

For $\frac{0}{0}$ one can argue that it can take any value, depending on how the $0$s are reached: $\lim _{x→0}{\frac{x}{x}} = 1$ but $\lim _{x→0} \frac{3x}{x} = 3$. So, defining it does not make sense in general.

For $0^{0}$ you can reach $1$ ($\lim _{x→0} x^x$) or e. g. $0$ ($\lim _{x→0} 0^x$), so a general extension of the definition doesn't make sense, too.

I don't find any such "good reason for being undefined" like varying results for the example of $\frac{1}{\frac{1}{0}}$ (all approaches I could think of result in $0$, nothing else). So I just can agree with the other answers here that it being undefined is just because rules like $\frac{1}{\frac{1}{x}} = x$ only hold valid if any sub-term also is defined.

Alfe
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  • For 0^0, you could also say it makes sense to call it 0, since 0 multiplied by itself any number of times is 0. Since there are two contradictory ways of possibly defining it, and no real benefit to defining it, leaving it undefined probably makes the most sense. – Michael Shaw Apr 29 '13 at 14:24
  • Thanks, Michael. I seem to have had a blackout on this. Of course. I took your comment into my answer. – Alfe Apr 29 '13 at 19:24