# Can irrational numbers exist on the numberline?

1. Jul 7, 2010

### Mu naught

This may be an elementary question, but I've been thinking about it a little bit and wondering what other people thought.

First, let me say that I'm talking about a number line not as a set but in the more literal sense, like a partitioned line that might exist as the axis of a graph.

So, if you want to represent a number as a point on this line, you can do so by starting at zero, and moving a certain distance to a point which corresponds to n units, the direction depending on if it's positive or negative. So what about an irrational number? It seems to me that you'd have to approach some point on the line, but continuously move toward it at a slower and slower rate, moving a thousandth of a unit, then a millionth, then a billionth and so on, constantly moving but constantly slowing down and never actually reaching any fixed point on the line. Is this a sound conception?

This also raises questions about real life objects having lengths which we can calculate to be irrational, but that's probably another thread.

2. Jul 7, 2010

### LCKurtz

The "real line" is an abstract idea. We draw a line on a sheet of paper to help us visualize its properties. But in the sense of a geometry construction, you can certainly plot a point that represents an irrational number. Just draw an x axis and a y axis perpendicular to it. Construct the 45 degree line y = x and mark 1 unit from the origin on it in the first quadrant. Drop a perpendicular from that point to the x axis. It will hit the x axis at sqrt(2) / 2, which is irrational. But, of course, your pencil representation won't be exact.

3. Jul 7, 2010

### D H

Staff Emeritus
A much better question is can rational numbers exist on the numberline? After all, almost all of the numbers on the number line are irrational. A point picked at random on that number is almost surely irrational. There are lots of (an infinite number of) irrationals that are constructible on that numberline -- and yet the set constructible numbers is still of measure zero. Just because you cannot construct a number doesn't mean the number doesn't exist.

4. Jul 8, 2010

### Char. Limit

So rational numbers don't exist on a numberline?

5. Jul 8, 2010

### disregardthat

Irrational numbers are often defined in terms of rational cauchy-sequences. Any real number is identified with an equivalence class of rational cauchy-sequences. The equivalence is as such: two rational cauchy-sequences are identical if their difference converges to 0. In a way an irrational number can be seen as an algorithm of rational numbers growing closer together.

So you could actually identify sqrt(2) with an algorithm which produces a rational sequence q_n such that |sqrt(2)-q_n| --> 0 as n --> infinity. The 'infinity' of information is not more a problem. Here, sqrt(2) have been identified in the (arguably) finitistic terms of a sequence, which can be considered a recursive rule.

However, the debate of whether irrational numbers exists more or less than rational numbers is actually irrelevant when it comes to the number line. The number line is merely an abstraction from an ordered set. A set is ordered if; given any two elements (a,b), then either a=b, a>b or b>a. If we identify two points a and b on the number line such that a < b, then the line segment in between them represent the set { x | a<x<b }. The number line serves as nothing but a consistent picture of an ordered field such as the real numbers.

6. Jul 8, 2010

### Mu naught

I still don't see how an irrational number can be represented by a definite point on a number line. Any point you choose, you can always move to the right by n x 10-j for whatever decimal you want to bring the number out to. This is what leads me to understand that an irrational number can only be represented by a point which is moving infinitesimally slowly towards some value but cant ever reach it, and if that is a valid way of thinking of things, then I don't believe you can ever represent an irrational number as a point on a number line.

Is this directed at me? If so, please explain how anything I've said would imply that.

7. Jul 8, 2010

### Landau

I don't quite understand your problem. Why do you think that an irrational number cannot be represented by a definite point on a number line, while a(ny) rational number can? If you worry about constructibility, LCKurtz has given you a prescription to construct sqrt(2)/2.

8. Jul 8, 2010

### Mu naught

I think I explained why I think this pretty clearly:

Any point you choose, you can always move to the right by n x 10-j for whatever decimal you want to bring the number out to. This is what leads me to understand that an irrational number can only be represented by a point which is moving infinitesimally slowly towards some value but cant ever reach it

9. Jul 8, 2010

### Office_Shredder

Staff Emeritus
I don't understand. How can I move to the right of a point that I'm not actually at? And if you move by some power of ten to the side you're on another irrational number, which also doesn't exist apparently

10. Jul 8, 2010

### Hurkyl

Staff Emeritus
Let's get away from the idea of a "moving point" and an "infinitessimal" and just say the following: you are thinking about the rational number line, and you are representing a number by a Cauchy sequence. (e.g. by a sequence of decimal approximations)

We can go further. We can define the sum, difference, or product of two sequences. Then, we might define two Cauchy sequences to be equivalent if their difference converges to zero. We can then define an ordering, and we might even be inclined to make a "Cauchy sequence line". We can even work out how to do calculus with these things.

And our crowning achievement will be to show that we have constructed a complete ordered field!

But wait a moment -- all that says is we have just created a model of the real numbers. So why did we bother going through all of this trouble?

11. Jul 8, 2010

### LCKurtz

Numbers don't move because they don't have legs

When you get a bit farther along in your mathematics training you may run across the idea of Dedekind cuts. That may help you understand the construction of the irrational numbers without thinking about moving approximations. Until then I wouldn't worry too much about it. You are safe in assuming that the pencil and paper representation of the real line is a pretty good physical representation of an abstract idea.

12. Jul 9, 2010

### Max™

A decimal approximation of an irrational number is an arrow pointing towards a limit.

That the limit exists can be proven, that the approximation can be extended to reach the limit can be proven, but you often can't directly calculate the entire string of digits.

Usually you can settle for being able to show that it must be smaller than n, and larger than m.

The rational line is full of holes, within those holes lie the real numbers.

The integers are a nice grassy lawn, big gaps, evenly spaced, the rational numbers are a forest beside it, much smaller gaps, but still spread out nicely. The real numbers are the dense jungle which is visible behind (and indeed all around) the rational forest, and the whole lawn. The complex numbers are a mountain range surrounding the valley of real numbers.

13. Jul 9, 2010

### Hurkyl

Staff Emeritus
Minor correction: a decimal approximation is simply a number. It doesn't have an extent, it doesn't move, it doesn't point.

A sequence of decimal approximations, I suppose, can be viewed in a variety of ways. I wouldn't begrudge thinking of a Cauchy sequence as "pointing" to something.

And just to be clear for the opening poster, the (possibly infinite) decimal expansion of a number is not an approximation. Again, it is simply a number.

(to be overly precise decimals are often used as notations for numbers, rather than literally being numbers themselves. But that's far more pedantic than most people will ever care about, and in some presentations, decimals really are numbers rather than being merely numerals)

14. Jul 10, 2010

### Max™

Well, I simply wanted to give an impression of what a decimal approximation is without having to go into the deep definition of limits and such.

You are right that it doesn't "point" at a spot on the line, but it isn't too far off to say it helps indicate where the real number is located, in a way which could perhaps help deal with the issue the OP had regarding the infinity of the rationals without delving into Cauchy and such.

Like pi for example, the string 3.14159265358979323846264338327950288419716939937510... is only a suggestion of what the actual value is, kinda like saying "were you able to continue this to an infinite number of places, you'd reach the true value of pi", but the string there is not pi exactly, and it could be far from it if I put a 3 or a 9 next instead of a 5. All of those possible variations fan out from the irrational value, which is the only one that follows the sequence exactly.

Last edited: Jul 10, 2010
15. May 27, 2012

### CodyOwen

(Talking about a graph with a number line) If you put your pencil on a whole number for example 3 and move the pencil to 4 then you must have passed through many irrational numbers, therefore you can find an irrational number on a number line. Does that sound credible?

16. May 27, 2012

### SteveL27

Consider a unit square. Put one corner of the square at the origin of your number line. The lower-left corner, for definiteness.

Now rotate the square 45 degrees clockwise, bringing the upper right corner of the square to the line. Now what point on the line is the corner of the square sitting on? Exactly square root of 2, right?

There's your irrational number on the number line.

17. May 27, 2012

### laughingebony

Not really, because that already assumes that irrational numbers exist. Their existence is a consequence of how the system of real numbers is built up from the system of rational numbers (which, as has already been mentioned, is done via Cauchy sequences or Dedekind cuts). Part of the reason we define the real number system like this is aesthetic. To understand this, first assume that we have only rational numbers. It is easy to find (Google it) a proof that there is no rational number whose square is 2. (This is usually phrased, "√2 is irrational," but we are not yet assuming that irrational numbers exist.) Define X to be the set of positive rational numbers x such that x2<2, and define Y to be the set of positive rational numbers y such that y2>2. Then, define two sequences {an} and {bn} as follows:

a1=1, b1=2

an+1=(an+bn)/2, bn+1=bn, if [(an+bn)/2]2<2

an+1=an, bn+1= (an+bn)/2, if [(an+bn)/2]2>2.

{an} and {bn} are increasing and decreasing sequences, respectively, whose difference converges to 0, but neither of these sequences converges to a rational number. In other words, assuming we only have rational numbers, these two sequences are getting arbitrarily close together for large enough indices n, but neither sequence is going anywhere in particular. If you think about this, I hope you'll agree that this is an undesirable property for our number system to have if we are to represent it by a continuous line (the number line). Loosely speaking, it means that there is an infinitely small hole located where √2 should be. Defining the real number system as the set of equivalence classes of Cauchy sequences fixes this. In particular the equivalence class of {an} (or {bn} -- it doesn't matter) has the property that its square is 2.

18. May 29, 2012

### Vargo

You should get together with Pythagoras and commiserate with him. It is alleged that his school drowned one of its own members for proving that sqrt(2) is not rational.

19. May 29, 2012

### Diffy

Let me ask you a question.

Say you have a number line in front of you. Now put your pen down and draw a line from 0 to 2. The square root of two is somewhere in between 0 and 2. Did your pen cross over the square root of 2?

I think what you are missing is that points on a number line have 0 width. That is how a line or line segment can have an infinite number of points on it.

20. May 29, 2012

### shreyakmath

Of course!

21. Jun 1, 2012

### AcidRainLiTE

Draw two points on the number line as follows: Draw a point 1 unit along the number line. Label it point A. To draw the second point, construct a square which has the line segment extending from the origin to A as one of its sides. Draw the diagonal of this square, and then rotate the diagonal down to the number line (see attached image). Label this point B.

Now create two line segments: one from the origin of the number line to point A and one from the origin to point B. Call these segments SA and SB respectively.

SB corresponds to an irrational number. Here is why:

• First, I need one definition: Two line segments S1, S2, are called commensurate if there exists a third line segment, E, that can be successively lined up with itself N times and fit perfectly across S1 and be successively lined up with itself M times and fit perfectly across S2. (N and M are integers).
• SA and SB are not commensurate. For a proof of this (and you should thoroughly look at this proof because this is the most important step of my argument here), see the section "Geometric proof of the irrationality of the sqrt(2)" of the following paper: http://www.bsu.edu/libraries/virtualpress/mathexchange/04-01/Coleman.pdf.
• Now, suppose the length of SB was rational. Then Length(SB) = n/m for some integers n and m. But consider the segment, E, that is formed by chopping SA up into m equal parts. E can be successively lined up with itself m times and fit perfectly across SA. Also, E can be successively lined up with itself n times and fit perfectly across SB (because Length(SB) = n/m = n*(1/m)). Thus, SA and SB are commensurate.
• From this, we can conclude that if the length of SB is rational, then SA and SB are commensurate. However, the link I included proved that SA and SB are not commensurate. Hence, the length of SB is not rational.

So, B is a point that corresponds to an irrational number.

That there are irrationals on the number line basically means that, if you chop your unit length up into pieces of equal length, there will always be points on the number line that you cannot get to by successively lining up an integer number of these pieces, no matter how small you try to make the pieces. There are points on the number line that simply 'do not submit' to that type of process.

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22. Jun 1, 2012

### AcidRainLiTE

I think this is a good question. If you will allow me to rephrase what you are saying:
(1) We claim that the numbers (both rational and irrational) correspond to points on the number line.
(2) The way to find out what point a given number corresponds to is to by looking at its decimal expansion. If x=0.d1d2d3d4... then, to find the point it corresponds to, you start at the origin and move to the right d1 10ths, then d2 100ths, then d3 1000ths, and so on.
(3) But for numbers whose decimal expansion does not terminate (which, by the way, does not include only irrationals but also many rationals such as 1/3), we will find ourselves stuck in an infinite process when trying to apply (2) and we will never happen upon any final point.

So, how can we say that an irrational number corresponds to a point if, when we attempt to figure out what point it corresponds to, we end up eternally jumping from point to point, never landing on a final one. Might it not be the case that every point on the number line corresponds to a rational number and that the irrationals just specify an infinite sequence of points just as the sequence {1,2,3,4,5,...} specifies a infinite sequence of points but does not correspond to a point on the number line?

But, what if you ask the question a different way? Given the points on the number line, do we find any whose decimal expansion happens to be non-terminating. The answer is yes. An example is the point B constructed in my previous post.

You may be interested to hear that the question you raise was an open question back in early Greek mathematics. It was not settled whether or not every length could be expressed rationally.

23. Aug 9, 2012

### ECHOSIDE

This was a great explanation.

This is the statement that allowed me to lock it into my memory. Well put, folks.

24. Aug 9, 2012

### DonAntonio

If you don't quote someone else's answer/question, that "of course!" is almost impossible to guess what its meaning is.

DonAntonio

25. Aug 9, 2012

### skiller

"Almost impossible"? Seems pretty obvious to me. Generally speaking, if you aren't quoting a previous post, then you are, by default, replying to the OP, which was a question to which the answer is "Of course!".