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Is a Whole Number part of A Rational Number or a whole number??

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    $\begingroup$ What do you already know about these things? What are the definitions of the words and some examples you can think of? What thoughts do you have on the problem? $\endgroup$
    – Zach L.
    Jul 14, 2013 at 4:14
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    $\begingroup$ Can an administrator pick an answer? This question is clearly answered and pretty old. $\endgroup$
    – YoTengoUnLCD
    Sep 10, 2015 at 23:24

4 Answers 4

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Every whole number is a rational number: for example, $3 = \dfrac 31$. So it is rational.

Every whole number $n$ can be written as a fraction of integers: $n =\dfrac n1$. We aren't required to write it that way; we just need to know that it is possible to express every whole number as a fraction of integers, and hence it is rational.

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  • $\begingroup$ You're welcome, Alfred. $\endgroup$
    – amWhy
    Jul 14, 2013 at 3:04
  • $\begingroup$ I am downvoting because I think this is stated too forcefully. In the definitions used in many texts, a rational number is an ordered pair of integers, or an equivalence class of such ordered pairs, and it is impossible for any object to simultaneously be a whole number and a rational number, $\endgroup$
    – Carl Mummert
    Jul 18, 2013 at 18:21
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    $\begingroup$ @CarlMummert: Thanks for commenting. I understand what you're saying. This answer was a "judgment call" based on the OPs earlier questions. I chose not to "muddy the waters". That very well be an error on my part. $\endgroup$
    – amWhy
    Jul 18, 2013 at 18:24
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It depends on how you set things up.

First, some notation. Write

  • $\mathbb{N}$ for the natural numbers. (or 'whole' numbers, if you prefer).
  • $\mathbb{Q}$ for the rational numbers.

Okay. So usually, we set things up such that if $x \in \mathbb{N}$, then $x \in \mathbb{Q}$. More tersely: $$\mathbb{N} \subseteq \mathbb{Q}.$$

However, we can also set things up so that natural numbers and rational numbers are different things that have a special relationship.

Specifically, they are different in that whenever $x \in \mathbb{N}$, it follows that $x \notin \mathbb{Q}$. (And vice versa). More tersely:

$$\mathbb{N} \cap \mathbb{Q} = \emptyset.$$

Whichever way you set it up, there's a function that maps each natural number $n$ to its corresponding rational number $\frac n 1$.

There is always a special natural number called $1_\mathbb{N}$, and a special rational number called $1_\mathbb{Q}$. This gives one (of possibly many ways) of defining this function (called the canonical embedding) of the natural numbers into the rationals. In particular, we can define this embedding as the unique $$f\colon \mathbb{N} \to \mathbb{Q}$$ such that $$f(1_\mathbb{N} + \cdots + 1_\mathbb{N}) = 1_\mathbb{Q} + \cdots + 1_\mathbb{Q},$$ where the number of terms we're summing on both sides are the same.

Okay, what's so special about this function $f$? Well, it preserves equality, order, addition, multiplication, exponentiation, etc. If this last comment is unclear, please say so and I will clarify.

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  • $\begingroup$ This is the point I was getting at, but you expressed it better. However, you did not express it at a level the OP is likely to understand. $\endgroup$
    – dfeuer
    Jul 14, 2013 at 3:29
  • $\begingroup$ @dfeuer, yes I agree, but how to simplify it? You're welcome to edit the post if you wish it. $\endgroup$
    – goblin GONE
    Jul 14, 2013 at 4:10
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The real answer, as usual, is "it depends". As the other answers have indicated, it is possible to identify whole numbers with certain rational numbers. On the other hand, it's also possible to identify rational numbers with certain ordered pairs of integers. So it really depends on your perspective/purpose. If you're doing something like number theory, you'll be thinking in terms of a whole number being a rational number. If you're thinking in terms of "mathematical foundations", you'll most likely be looking at it from the other direction.

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    $\begingroup$ I'm sorry, but your post is simply confusing the issue, and the OP, I suspect. A whole number is a rational number that also happens to be an integer. Every integer is a rational number. $\endgroup$
    – amWhy
    Jul 14, 2013 at 3:02
  • $\begingroup$ @amWhy: I figured that the question of whether a whole number is "part of" a rational number related to the usual notation $\frac m n$. In that sense, the integers $m$ and $n$ are "part of" the rational number $\frac m n$. And that's a perfectly sensible way to get to some of the usual approaches to constructing the rationals, and then the reals, from the integers. $\endgroup$
    – dfeuer
    Jul 14, 2013 at 3:04
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    $\begingroup$ I like this answer. It did not deserve a down-vote. $\endgroup$
    – goblin GONE
    Jul 14, 2013 at 3:21
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    $\begingroup$ @amWhy "Every integer is a rational number"... No. Or rather, it depends (for more explanations see dfeuer's post). $\endgroup$
    – Did
    Jul 14, 2013 at 8:26
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A whole number is a rational number. Write a whole number, $n$, as $\dfrac{n}{1}$.

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