Can irrational numbers exist on the numberline?

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.

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.

shreyakmath

Of course!

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/virtual...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.

Attached Images

irrational_length.bmp (84.6 KB, 10 views)

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.

ECHOSIDE

This thread was very helpful to me.

This was a great explanation.

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

DonAntonio

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

DonAntonio

oay

"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!".

DonAntonio

For you, but for me it is almost "pretty obvious" he's addressing the message immediately before his, so

I think it is safe enough to say that it is, at least, some people can be confused. That's the reason we

have the very nice "quote" option....

DonAntonio

Vorde

All I can add here is that some people were talking about always approaching a point on the number line but never actually getting there because it is an non-terminating decimal.

This seems a lot like Zeno's paradox and how Zeno 'proved' that motion is impossible (fun fact: everyone contemporary with Zeno thought he was crazy, it was only after he died that some people started believing him).

I think the point (as highlighted early on by a couple people) is that the numberline is wholly our own construction, and therefore it seems silly to argue about this thing without a stringent definition of what the numberline is, in which case you can answer this question mathematically.

rajeshmarndi

It is said Zeno's paradox is solved by the help of limit later. But I still wander because in limit, we take the value of the limit to be something that is never reached.

So the paradox to me is still a paradox.


__________________________
I am a beginner in this field.

Number Nine

The premise of Zeno's paradox is that in order to move any distance, a person must first move an infinite number of smaller distances, each requiring a positive amount of time to traverse. Calculus demonstrates conclusively that this infinite number of positive "times" sums to a finite number; therefore, the time taken to move a distance if finite, therefore, there is no paradox. There is nothing that is "never reached". Zeno's paradox asserts that an infinite series diverges when it does not; there is no paradox, only an error.

SteveL27

I'm always puzzled by this argument.

To be sure, an infinite series can converge in mathematics. But how can we be sure that same logic applies to physics? Do you believe in the physical existence of Dedekind cuts? If the real line has a direct analog in the physical world, does that mean that the Continuum hypothesis is subject to physical experiment? Or that it might someday be, given enough technology?

I don't personally believe these things; therefore, I don't believe that the mathematical theory of infinite series resolves Zeno's paradoxes.

After all, Zeno was making a point about physical motion. Does one have to believe that math = physics in order to accept that the theory of infinite series resolves Zeno? Or is there something else I'm misunderstanding?

Number Nine

No, because Zeno makes a mathematical argument. He attaches a positive real number (denoting time) to each infinite distance that must be traversed and argues that motion between two points is thus impossible, because the amount of time required must be infinite. He has explicitly set up an infinite series. There is no need to appeal to physics, because the is already very clear (things do, in fact, move), so all we're left with is the theoretical (mathematical) justification for his paradox.

Studiot

Lot's of posters here presenting arguments about a 'point' or 'points'.

Any care to offer a definition that can be used in the arguments?

A good definition provides the answer to the original question.

mesa

Abstract idea or not, that is a clever solution. You have quite a knack for the cartesian coordiante system LCKurtz.

I liked the question too.

webberfolds

I'm pretty sure I understand what mean and yes the concept that nothing can reach infinity is true. Now if irrational numbers exist is another question and that may depend on the numerical system.