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Can irrational numbers exist on the numberline?

  1. Jul 7, 2010 #1
    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.
     
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  3. Jul 7, 2010 #2

    LCKurtz

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    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. :frown:
     
  4. Jul 7, 2010 #3

    D H

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    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.
     
  5. Jul 8, 2010 #4

    Char. Limit

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    So rational numbers don't exist on a numberline?
     
  6. Jul 8, 2010 #5

    disregardthat

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    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.
     
  7. Jul 8, 2010 #6
    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.
     
  8. Jul 8, 2010 #7

    Landau

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    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.
     
  9. Jul 8, 2010 #8
    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
     
  10. Jul 8, 2010 #9

    Office_Shredder

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    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
     
  11. Jul 8, 2010 #10

    Hurkyl

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    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? :confused:
     
  12. Jul 8, 2010 #11

    LCKurtz

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

    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.
     
  13. Jul 9, 2010 #12
    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.
     
  14. Jul 9, 2010 #13

    Hurkyl

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    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)
     
  15. Jul 10, 2010 #14
    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
  16. May 27, 2012 #15
    (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?
     
  17. May 27, 2012 #16
    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.
     
  18. May 27, 2012 #17
    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.
     
  19. May 29, 2012 #18
    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.
     
  20. May 29, 2012 #19
    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.
     
  21. May 29, 2012 #20
    Of course!
     
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