I think of the number zero as a whole number.

It can certainly be a ratio = $\frac{0}{x}, x \neq 0.$

Therefore it is rational.

But any ratio equaling zero involves zero, or is irrational, e.g.$\frac{x}{\infty}, x \neq 0$ is not a ratio of integers.

Can a rational number that is rational only when the number itself is involved still be rational?

I realize that it "should" be rational but it doesn't seem to fit into the same category with other rational numbers. (I'm glad mse has a soft question category for questions like this...).

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    $\begingroup$ 0 is a rational number, as you have shown. $\endgroup$ – tylerc0816 Oct 11 '13 at 20:36
  • $\begingroup$ "Can a rational number that is rational only when the number itself is involved still be rational" Yes...? Why wouldn't it be? The definition of rationality is that it can be expressed as a ratio. Any ratio. There are no other requirements on what that ratio should be like. $\endgroup$ – Jack M Oct 11 '13 at 20:44
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    $\begingroup$ Your question is not about 0 being rational or not, but that it is rational with a unique property that it has infinitely many representations that are can be simplified to the same thing like $\frac{1}{2} = \frac{2}{4}$ $\endgroup$ – Arjang Oct 11 '13 at 21:11
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    $\begingroup$ @JackM: $\pi/1$ does not count :) $\endgroup$ – Ben Millwood Oct 11 '13 at 21:26
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    $\begingroup$ @Arjang And $\frac02$ reduces to $\frac01,$ the unique representation of $0$ in the form $\frac mn$ where $m,n$ are integers, $n\gt0,$ and $\gcd(m,n)=1.$ Note that $\frac0n$ is not in lowest terms when $n\gt1,$ because $\gcd(0,n)=n.$ $\endgroup$ – bof Nov 2 '15 at 19:50

One perspective on this is to regard $0$ as a symbol that actually refers to several different mathematical objects:

  • $0$-the-natural-number, usually the first natural number defined. Let's write $0_n$ for this specifically.
  • $0$-the-integer; an integer is a natural number with a sign, and so we can write $0_i = +0_n$. In fact it's also $-0_n$, but that doesn't really matter.
  • $0$-the-rational-number: a rational number is an integer divided by another, so $0_r = 0_i/1_i$

From this point of view, there is nothing circular about $0 = 0/1$, because we're actually just using the same symbol to refer to two very similar objects, one an integer and one a rational number.


A real number $x$ is rational if and only if it can be written as a fraction $a/b$ with $a$ and $b$ integers and $b\neq 0$. In particular, integers are rational, and $0$ is an integer, so it is rational.

  • $\begingroup$ Yes, certainly. I should have made more clear that I do understand it is rational, but I have a 'soft question' (or maybe speculative, reflecting that 0 is different and I don't completely understand why). Some of the answers given so far have given me some insight into it and if nothing else it has certainly sparked conversation. $\endgroup$ – Circulwyrd Oct 13 '13 at 19:15

That is a good observation, the question is not really about 0 being rational or not but that wether 0 is the only rational with infinitely many representation that can not be simplified.

Use the alternative definition of rationals :Rational can be represented by finite or repeating decimal representation.

Using that definition there is no ambiguity trying to find two numbers that ratio will be 0. But according to that definition there are infinite numbers that their ratio is 0, where we only needed 1. So it is a rational with the property that can be expressed with infinity many different ratios (all other rational have only one ratio in its simplest terms ).

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    $\begingroup$ How do you define simplest terms? Why is $0/3$ in simplest terms? $\endgroup$ – Ben Millwood Oct 11 '13 at 21:29
  • $\begingroup$ @BenMillwood : sorry that would have meant to be lowest terms, for example $0.5=\frac{1}{2}=\frac{2}{4}=\cdots$ but we disregard everything but $\frac{1}{2}$, $\frac{0}{3}$ is in it's lowest terms as 3 can not be any further divided by any other number (although 0 can be simplified further). $\endgroup$ – Arjang Oct 11 '13 at 22:18
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    $\begingroup$ $3$ can be divided by $3$... $\endgroup$ – Ben Millwood Oct 11 '13 at 23:34
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    $\begingroup$ In particular, the usual definition of reduced applies here: $\gcd(0,3)=3 \neq 1$, so $0/3$ is not reduced. $\endgroup$ – Hurkyl Oct 12 '13 at 0:45

As you stated $0$ is whole number, and we know that whole numbers are subset of rational numbers. So what does that says?

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    $\begingroup$ You want to be careful with $\frac{x}{\infty}$. $\infty$ is not a natural number or an integer. $\endgroup$ – Ben Millwood Oct 11 '13 at 21:28
  • $\begingroup$ My bad, but anyway $\frac {x}{\infty} = 0$, or? $\endgroup$ – Stefan4024 Oct 11 '13 at 21:35
  • $\begingroup$ @Stefan4024 it's definition varies with the branch of mathematics you're using; it would be undefined in algebra, but in calculus (where we have limits) it represent a limit of zero. $\endgroup$ – KutuluMike Oct 12 '13 at 0:31
  • $\begingroup$ I wasn't clear enough that I do understand it is rational according to the definitions. See comment on Keenan Kidwell's post above. $\endgroup$ – Circulwyrd Oct 13 '13 at 19:41

The division of two irrational quantities can obviously have rational results. For instance, both π and 2π are irrational; but the result of their respective division is either 2 or $\tfrac12$ , both of which are of course rational. And not just division, but other operations as well. Let's try a simple addition, for instance: both π + 1 and 1 - π are irrational; yet their sum, 2, is not...

In case you weren't aware, there is a reason why there is no symbol specifically for irrationals, as in the case of natural numbers $(\mathbb{N})$, integers $(\mathbb{Z})$, rationals $(\mathbb{Q})$, reals $(\mathbb{R})$, complex $(\mathbb{C})$, algebraics $(\mathbb{A})$, etc. This has to do with the fact that -unlike these other numerical sets- the set of irrationals is not self-contained when mixed with even the most basic arithmetic operations, like addition or multiplication, for instance. E.g., despite the fact that both $\sqrt2$ and $\sqrt2$ are irrational, their product, 2, isn't. The same observation holds for $\sqrt3$ and $^1/_\sqrt3$ , etc.


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