# What number is rational and irrational

1. Aug 30, 2004

### Tina

When is x rational and irrational?

Also When is r positive and negative?

2. Aug 30, 2004

### Tide

By definition an irrational number is one that is not rational. Therefore, it is not possible for a number to both rational and irrational. If you're thinking of the number zero (0) then you're mistaken since 0 is clearly rational - it can be expressed as a ratio of whole numbers.

Also, a positive number is one that is greater than zero and a negative number is one that is less than 0. The only alternative to those two conditions is a number that is neither greater than zero nor less than zero - which is zero itself and it is neither positive nor negative.

3. Aug 31, 2004

### HallsofIvy

The definition of "rational number" is that it can be written as a fraction.
(1/2, 3/5, 2/3, etc.)
It is also true that rational numbers are those that, in decimal notation, are either terminating or repeating. (1/2= 0.5 is terminating, 3/5= 0.6 is terminating, 2/3= 0.66666... is repeating)

An irrational number is any number that is NOT rational. $\sqrt{2}$ for example, as well as $\pi$. Irrational numbers tend to be written in "funny" ways exactly because they are not as simple to write as rational numbers.

Are you really asking what a negative number is? That seems very elementary for someone who is also working with rational and irrational numbers. About the best answer is that positive numbers are greater than 0 and negative numbers are less that 0. That negative sign in front of a number is a pretty good clue!

Since you specifically said "when is r positive or negative", I wonder if you are not thinking of polar coordinates. If that is the case the answer is easy: r is never negative, it can only be 0 or positive.

I just realized that the question was NOT "what numbers are rational or irrational" or "what numbers are positive or negative" as I must have thought when I wrote this.

Hurkyl is completely correct- there are NO numbers that are both rational and irrational and there are NO numbers that are both positive and negative.

Last edited by a moderator: Mar 7, 2010
4. Aug 31, 2004

### Alkatran

A = not A?

5. Aug 31, 2004

### arildno

Tina (welcome to PF, BTW!),
It would be best if you give the context of your questions:

Is it for example that you wonder:
What can I assume in some calculation/proof, and what must I prove?
Or just: how do I KNOW this?

6. Aug 31, 2004

### gravenewworld

Aren't all irrational numbers just sequences of rational numbers?

7. Aug 31, 2004

### Alkatran

A sequence of rational numbers would be rational.

1 then 2 -> 1.2

8. Aug 31, 2004

### gravenewworld

Not all the time. The real numbers have the property that they are dense, i.e. for any real number a there is a sequence of rational numbers {r_n} so r_n--->a. So say for the irrational number pi

r_1=3.0
r_2=3.1
r_3=3.14
r_4=3.141

and so on you get the idea. If a is irrational you can just choose r_n to be the rational numbers of the first n terms of the decimal expansion of a followed by zeroes. If r has its decimal expansion that agrees with the expansion of a to the mth place then the number differs from a less than 10^-m. So obviously the sequence of rationals {r_n} converges to a.

9. Aug 31, 2004

### Hurkyl

Staff Emeritus
Anyways, you're missing a subtlety.

Any irrational number is equal to the limit of some sequence of rational numbers.

Except for a few particular models of the real numbers, it is incorrect to say that an irrational number is a sequence of rational numbers.

Also a correction:

You meant to say

The rational numbers are a dense subset of the real numbers.

Or more compactly,

The rationals are dense in the reals.

Last edited: Aug 31, 2004
10. Aug 31, 2004

### Ethereal

Is there a proof of this somewhere?

11. Aug 31, 2004

### Hurkyl

Staff Emeritus
Yes. Gravenworld's last post (#8) is a good demonstration of this fact.

A rigorous proof would require use of the completeness axiom, though.

Last edited: Aug 31, 2004
12. Aug 31, 2004

### gravenewworld

Ah yes thanks for catching my mistake. The rational numbers are dense in the real numbers.

13. Sep 1, 2004

### HallsofIvy

Ooops! I will edit that. Thanks.

14. Sep 1, 2004

### HallsofIvy

Hmm, couldn't edit- the post is too old. Of course, what I meant was that "an irrational number is any number that is not RATIONAL".

By the way, the fact that, for any real number there exist a sequence of rationals that converge to it can be used as a definition of "real number".

Let X be the set of all increasing, bounded, sequences of rational numbers (convergent or not). We say that two such sequences, {an} and {bn}, are equivalent if and only if the sequence {an- bn} converges to 0. It is easy to show that that is an "equivalence relation" and so partitions the set X into equivalence classes. We identify the real numbers with those equivalence classes (addition, multiplication are defined by using "representative" sequences). That definition makes it easy to prove that any increasing bounded sequence of real numbers converges to a real number and the "completeness axiom" follows from that.

15. Sep 1, 2004

### humanino

... contrary to the rational numbers, the real numbers are dense in themselves :laughing:

But when it goes to :
... the real numbers are not compact in themselves :tongue2:

16. Sep 1, 2004

### Hurkyl

Staff Emeritus
I beg to differ; I think the rationals are also dense in themselves!

17. Sep 1, 2004

### humanino

I was just joking. I thought a discrete set cannot be dense at all, because all the subset of a discrete set are open, thus no subset can be closed, thus no subset can be dense. Now that I think about it seriously, I think you must be right. Is it true that any set is dense in itself ? What's the definite answer ! :uhh:

Just had a thought : the fact that Q is (partially) ordered. It is sufficient to show that for any two rationals, another lies between them. Q is dense in itself. I do not delete the begining, since I would like to know what happens in the case of a set without partial order.

18. Sep 1, 2004

### Hurkyl

Staff Emeritus
Being dense is a topological property...

Of course, any totally ordered space has a natural topology that respects the ordering (let the neighborhoods be open intervals), so it makes sense to speak of being dense in an ordered set.

A set S is dense in a topological space T if the closure of S is all of T. So, clearly, T is dense in T.

The rational numbers is not a discrete space: the property it lacks is being complete.

Also, there's nothing stopping a set from being both open an closed. For any topological space, both the empty set and the whole thing are both open and closed. And, in general, any set that is isolated from the rest of the space (such as any set in a discrete topology) is both open and closed.

As an addendum, the term "dense order" means something else entirely: an ordering is dense if, for any A and B with A < B there is a C so that A < C < B.

edit: changed an erroneous use of the word "set" to "space"

Last edited: Sep 2, 2004
19. Sep 2, 2004

### humanino

Thanks Hurkyl ! Much clearer in my head now... I should never have quit math for physics

20. Mar 7, 2010

### fk378

So any real number is the limit of a sequence of rationals. Is any real number also a limit of a sequence of irrationals? I found this proof of the latter statement...

it seems contradictory to have that every real number is both the limit of a sequences of rationals as well as the limit of a sequence of irrationals.

21. Mar 7, 2010

### snipez90

If you think about the formal definition of a limit, this shouldn't seem that contradictory. For one, there are SO MANY more irrational numbers than rational numbers (the set of irrational numbers is uncountable), so even on an intuitive level, it shouldn't be hard to convince yourself that if you can approximate a real number by a sequence of rationals, then certainly you could approximate the same real number by a sequence of irrationals. The proof given in the yahoo answers link relies fundamentally on the fact that the decimal expansion of a real number is arbitrarily close to approximating rational numbers. Richard Courant gives a pretty intuitive explanation regarding such approximations here:

"[PLAIN [Broken] Real Numbers and Infinite Domains"]http://kr.cs.ait.ac.th/~radok/math/mat6/calc1.htm#1.1.2 [Broken] Real Numbers and Infinite Domains[/URL]

Last edited by a moderator: May 4, 2017
22. Mar 7, 2010

### HallsofIvy

Oops! Thanks. I'll go back and edit so I can pretend I never made that silly error!

23. Mar 7, 2010

### HallsofIvy

Yes.

No, the limit of a sequence of rational numbers does not have to be rational.

One way of defining the real numbers, in terms of rational numbers, is to define them to be equivalence classes of certain kinds of kinds of sequences of rational numbers.

For example, we can define the real numbers to be equivalence classes of Cauchy sequences of rational numbers with two such sequences, $\{a_n\}$ and $\{b_n\}$ defined to be "equivalent" if and only if the sequence $\{a_n- b_n\}$ converges to 0.

Alternatively, we can define the real numbers to be equivalence classes of "increasing sequences of rational numbers having an upper bound" with the same equivalence relation.

For example, the sequence {3, 3.1, 3.14, 3.141, 3.141, 3.1415, 3.14159, 3.141592, ...} is both a Cauchy sequence and an increasing sequence of rational numbers having limit $\pi$, an irrational number. In the senses of the two definitions above, the equivalence class containing that sequence would be the number $\pi$.