Where are the irrational numbers?

[agentredlum]

There exists a bijection between $\mathbb{R}$ and any arbitrary small interval. So the reals can be "fitted" in any arbitrary small interval.

And what is the length of the real numbers?

 

[micromass]

And what is the length of the real numbers?

You mean the Lebesgue measure? It's infinite. I don't see what this has to do with anything.

 

[agentredlum]

You mean the Lebesgue measure? It's infinite. I don't see what this has to do with anything.

Yes, the Lebesgue measure.
disregardthat talked about length and made it seem that the length of the reals is the same as the length of the rationals or i misunderstood him.

There is a 1-1 correspondence between the reals and any arbitrarily small interval of reals because in between any 2 real numbers there are as many real numbers as there are real numbers from -infinity to +infinity

 

[disregardthat]

How would you fit the reals?

mathwonk fits the rationals by making a list and using a 1-1 correspondence between every member of his list and a mathwonk 'cut'

Cantor proved the reals cannot be listed.

How would you fit the reals?

Mathwonk made a 1-1 correspondence between the rationals and intervals of increasingly smaller length.

The fact that there is a bijection between [0,a] and the reals for any a means that the reals "fit" into an interval of any length. Though this is not completely analogous to Mathwonk's example, it does mean that one can't conclude much from this kind of measuring the "length" of a set. The reals does not have to be listed to be put in a 1-1 correspondence with an interval.

The lebesgue measure can't either "measure" the length or size of a set, as it is completely dependent of its definition according to the predefined sigma algebra. We could equally well have a measure of the reals for which each measurable set is 0.

 

[agentredlum]

Mathwonk made a 1-1 correspondence between the rationals and intervals of increasingly smaller length.

The fact that there is a bijection between [0,a] and the reals for any a means that the reals "fit" into an interval of any length. Though this is not completely analogous to Mathwonk's example, it does mean that one can't conclude much from this kind of measuring the "length" of a set. The reals does not have to be listed to be put in a 1-1 correspondence with an interval.

The lebesgue measure can't either "measure" the length or size of a set, as it is completely dependent of its definition according to the predefined sigma algebra. We could equally well have a measure of the reals for which each measurable set is 0.

The fact is that complicated explanations don't help anyone who doesn't know the answer already. All you and micromass are doing is confusing me.

mathwonk made his point in a clear and concise way and you guys are showing off your knowledge of technical terms. You are both smart but can you explain what you know in a way that anyone with some mathematical knowledge can understand?

 

[agentredlum]

mathwonk...heeeeelp!

 

[ArcanaNoir]

I rather of appreciate the exposure to stuff I haven't learned yet. I get excited when it pops up in class instead of being wary of it. Of course it's best if the complicated stuff is only a bonus after a less complicated explanation.

 

[agentredlum]

I don't think you guys understood what mathwonk was trying to say. He simply gave a clever illustration of why rational numbers are insignificant in measure compared to the measure of the real numbers.

You guys are saying the measure of the real numbers is insignificant compared to the measure of the real numbers. That's very interesting but it does not diminish the worth of HIS argument. All of you are saying interesting things, personally i like mathwonk explanation because it is fascinating to me

[EDIT] If you can show me this bijection instead of saying 'there exists' maybe i'll find it fascinating.

 

[micromass]

The fact is that complicated explanations don't help anyone who doesn't know the answer already. All you and micromass are doing is confusing me.

mathwonk made his point in a clear and concise way and you guys are showing off your knowledge of technical terms. You are both smart but can you explain what you know in a way that anyone with some mathematical knowledge can understand?

We use technical terms because mathematics is technical. Mathematics uses very precise statements, and I feel that I am lying if I do not use these statements.
You can always ask for more explanations if you don't understand something, but eventually it will be up to you to learn these precise statements.
One cannot do mathematics while using handwaving arguments.

I don't think you guys understood what mathwonk was trying to say. He simply gave a clever illustration of why rational numbers are insignificant in measure compared to the measure of the real numbers.

You guys are saying the measure of the real numbers is insignificant compared to the measure of the real numbers. That's very interesting but it does not diminish the worth of HIS argument. All of you are saying interesting things, personally i like mathwonk explanation because it is fascinating to me

[EDIT] If you can show me this bijection instead of saying 'there exists' maybe i'll find it fascinating.

Fair enough,

$$\mathbb{R}\rightarrow ]-a,a[:x\rightarrow \frac{2a}{\pi} atan (x)$$

is a bijection between the reals and an interval ]-a,a[. So any open interval can be put in one-to-one correspondance with the reals in this manner.

 

[agentredlum]

We use technical terms because mathematics is technical. Mathematics uses very precise statements, and I feel that I am lying if I do not use these statements.
You can always ask for more explanations if you don't understand something, but eventually it will be up to you to learn these precise statements.
One cannot do mathematics while using handwaving arguments.

Fair enough,

$$\mathbb{R}\rightarrow ]-a,a[:x\rightarrow \frac{2a}{\pi} atan (x)$$

is a bijection between the reals and an interval ]-a,a[. So any open interval can be put in one-to-one correspondance with the reals in this manner.

Can you post a picture of the bijection, my browser does not decode TeX

 

[SteveL27]

Can you post a picture of the bijection, my browser does not decode TeX

It's just the arctan function. The way I like to think of this bijection is by imagining a horizontal line in the plane through the origin. It makes an angle of zero with the x-axis and its slope is zero.

As you rotate the line counterclockwise, as the line goes from horizontal to vertical the angle goes from 0 to pi/2; and the slope goes from 0 to +infinity.

Likewise as you rotate a horizontal line clockwise, the angle goes from from 0 to -pi/2 (= 3*pi/2), and the slope goes from 0 to -infinity.

What we've just described is a continuous bijection between the open interval (-pi/2, pi/2) and the entire real line (-infinity, +infinity). That's the way to visualize the tangent, which is the slope of a given angle; and the arctangent, which is the angle given the slope.

So topologically, the entire real line is exactly the same as the open interval (-pi/2, pi/2). You can in fact do the same trick with any open interval (a,b) by mapping the interval (a,b) to (-pi/2, pi/2) via the equation of the straight line between that passes between the two points (a,b) and (-pi/2, pi/2).

[Hmmm, now I see why people like to use ]a,b[ to denote an open interval. In the previous paragraph I overloaded the notation (a,b) to mean both a point and an interval. I hope the meaning's clear.]

Anyway the point is that you can mentally rotate a line through the origin to visualize a continuous bijection between an interval and the entire real line.

 

[spamiam]

Another visualization that may be helpful:

Take an open interval of whatever length and bend it into a semicircle. Now, project each point on this semicircle onto the real line by drawing a straight line from the middle of the semicircle, through a point on the semicircle. Where this line crosses the real line is the image of the corresponding point on the semicircle. Since the interval is open, the "endpoints" map to "infinity." If that's not clear, take a look at the attached picture.

 

[agentredlum]

It's just the arctan function. The way I like to think of this bijection is by imagining a horizontal line in the plane through the origin. It makes an angle of zero with the x-axis and its slope is zero.

As you rotate the line counterclockwise, as the line goes from horizontal to vertical the angle goes from 0 to pi/2; and the slope goes from 0 to +infinity.

Likewise as you rotate a horizontal line clockwise, the angle goes from from 0 to -pi/2 (= 3*pi/2), and the slope goes from 0 to -infinity.

What we've just described is a continuous bijection between the open interval (-pi/2, pi/2) and the entire real line (-infinity, +infinity). That's the way to visualize the tangent, which is the slope of a given angle; and the arctangent, which is the angle given the slope.

So topologically, the entire real line is exactly the same as the open interval (-pi/2, pi/2). You can in fact do the same trick with any open interval (a,b) by mapping the interval (a,b) to (-pi/2, pi/2) via the equation of the straight line between that passes between the two points (a,b) and (-pi/2, pi/2).

[Hmmm, now I see why people like to use ]a,b[ to denote an open interval. In the previous paragraph I overloaded the notation (a,b) to mean both a point and an interval. I hope the meaning's clear.]

Anyway the point is that you can mentally rotate a line through the origin to visualize a continuous bijection between an interval and the entire real line.

Thanks steve, I get it now. So you are creating a bijection between angles of the line and slopes of the line. What is the role of arctanx in this bijection?

I am a bit uncomfortable using your analogy because the line intersects arctanx twice for any given angle and zero angle gives plus or minus infinity depending on direction of rotation.

[EDIT] Also the arc length of arctan (pun not intended,lol) is infinite. At least spamiam semicircle has finite arc length but i have a problem with that too.

I am not attacking your arguments, like i say i can see it your way. I am merely making what i believe are interesting observations. [END EDIT]

However having said that, i can still see it your way.

 

[agentredlum]

Another visualization that may be helpful:

Take an open interval of whatever length and bend it into a semicircle. Now, project each point on this semicircle onto the real line by drawing a straight line from the middle of the semicircle, through a point on the semicircle. Where this line crosses the real line is the image of the corresponding point on the semicircle. Since the interval is open, the "endpoints" map to "infinity." If that's not clear, take a look at the attached picture.

Thank you for the picture, it helps a lot. I have seen this picture before but as i understood it, this analogy was used to prove that the cardinality of points on the open semicircle is equal to the cardinality of the real numbers.

If you take any finite straight line interval this trick does not work, so something special happens when you bend it into a semicircle. IMHO an open semicircle is not the same 'type' of interval as a finite straight line interval.

I guess you can say that the real numbers fit in any interval of real numbers because they have the same cardinality as the interval but this is a very non intuitive idea of what 'fit' means.

I can see it your way too and i love your illustration because its fascinating.

You see, where i am fascinated is, what is so special about bending the interval into a semicircle? One can use half a rectangle with the endpoints missing and achieve the same result, or many other 2 dimensional geometric figures. The semicircle itself can be thought of as a polygon whose number of sides goes to infinity.

As a thought experiment, one can use a very thin rectangle of infinite height, this will cause some problems cause it will be very difficult to hit the first right angle, however if you overcome that difficulty by defining 'you hit it when the line is exactly 90 degrees', it seems to work.

So the technique in your illustration appears to give the same result for finite intervals (open semicircle) and infinite intervals (very thin rectangle of infinite height).

IMHO i believe the open semicircle is special because it is a 2 dimensional geometric figure, the straight line interval is only 1 dimensional.

 

[SteveL27]

Thanks steve, I get it now. So you are creating a bijection between angles of the line and slopes of the line. What is the role of arctanx in this bijection?

The arctan function is the bijection. The arctan function maps the reals bijectively to a bounded open interval.

Any non-vertical line through the origin has slope y/x, where (x,y) is any point on the line. In particular if you choose a point on the unit circle, then the line intersects the unit circle at the point (cos(t), sin(t)) where t is the angle the line makes with the positive x-axis.

What's the slope of the line passing through the origin and the point (cos(t), sin(t))? It's sin(t)/cos(t) = tan(t).

We are interested in the restriction of the tangent function to the open interval ]-pi/2, pi/2[. That restriction maps an angle in the open interval ]-pi/2, pi/2[ to a slope in the reals. And the map is bijective.

Since the (restricted) tan is bijective, it has an inverse. What's its inverse? It's the arctan. So the arctan function maps all the reals to the interval ]-pi/2, pi/2[.

It's helpful to look at the graphs of the tan and arctan to see how we're selecting one of the many connected components of the graph of the tan; and using that as a bijection.

I am a bit uncomfortable using your analogy because the line intersects arctanx twice for any given angle

Not sure exactly what you mean. The arctan is the function that maps the real numbers to the angles between -pi/2 and pi/2. Nothing "intersects arctan." And the line only goes halfway around the circle, if that's your concern. We don't care about angles you get when you go past the y-axis. Was that your concern? That's the restriction idea above.

and zero angle gives plus or minus infinity depending on direction of rotation.

No, that's not true. The tangent function is not defined at +/- pi/2. We are only concerned about tan on the open interval ]-pi/2, pi/2[. It's not correct to say that it's "plus or minus infinity."

There are some situations in general where it's useful to define the values of a function in the extended real numbers; but this is not one of those situations! If we restrict our attention to the open interval where tan does not blow up, we avoid exactly the problem you mentioned.

[EDIT] Also the arc length of arctan (pun not intended,lol) is infinite. At least spamiam semicircle has finite arc length but i have a problem with that too.

Not sure what the concern is. These are just visualizations to show that a bounded line segment is bijectively equivalent to an unbounded one. In fact they're topologically equivalent: you can choose a bijection that's continuous in both directions. This example shows that a continuous function can transform a bounded set into an unbounded one and vice versa.

However having said that, i can still see it your way.

Credit where credit's due. Micromass already gave the function that maps the reals to the open interval ]-a, a[ using the arctan function. Earlier you mentioned you can't see the TeX, here's the ASCII:

R -> ]-a, a[ : x -> (2a/pi) * arctan(x)

This entire discussion is already implicit in that symbology. I'm just providing the visualization.

 

[agentredlum]

The arctan function is the bijection. The arctan function maps the reals bijectively to a bounded open interval.

Any non-vertical line through the origin has slope y/x, where (x,y) is any point on the line. In particular if you choose a point on the unit circle, then the line intersects the unit circle at the point (cos(t), sin(t)) where t is the angle the line makes with the positive x-axis.

What's the slope of the line passing through the origin and the point (cos(t), sin(t))? It's sin(t)/cos(t) = tan(t).

We are interested in the restriction of the tangent function to the open interval ]-pi/2, pi/2[. That restriction maps an angle in the open interval ]-pi/2, pi/2[ to a slope in the reals. And the map is bijective.

Since the (restricted) tan is bijective, it has an inverse. What's its inverse? It's the arctan. So the arctan function maps all the reals to the interval ]-pi/2, pi/2[.

It's helpful to look at the graphs of the tan and arctan to see how we're selecting one of the many connected components of the graph of the tan; and using that as a bijection.
Not sure exactly what you mean. The arctan is the function that maps the real numbers to the angles between -pi/2 and pi/2. Nothing "intersects arctan." And the line only goes halfway around the circle, if that's your concern. We don't care about angles you get when you go past the y-axis. Was that your concern? That's the restriction idea above.
No, that's not true. The tangent function is not defined at +/- pi/2. We are only concerned about tan on the open interval ]-pi/2, pi/2[. It's not correct to say that it's "plus or minus infinity."

There are some situations in general where it's useful to define the values of a function in the extended real numbers; but this is not one of those situations! If we restrict our attention to the open interval where tan does not blow up, we avoid exactly the problem you mentioned.

Not sure what the concern is. These are just visualizations to show that a bounded line segment is bijectively equivalent to an unbounded one. In fact they're topologically equivalent: you can choose a bijection that's continuous in both directions. This example shows that a continuous function can transform a bounded set into an unbounded one and vice versa.

Credit where credit's due. Micromass already gave the function that maps the reals to the open interval ]-a, a[ using the arctan function. Earlier you mentioned you can't see the TeX, here's the ASCII:

R -> ]-a, a[ : x -> (2a/pi) * arctan(x)

This entire discussion is already implicit in that symbology. I'm just providing the visualization.

Oh i get it now, is x any real number? The domain of arctanx is -infinity, +infinity the range is -pi/2, pi/2 this shows a fit of all real numbers in that interval ]-pi/2,pi/2[ why couldn't you guys say so to begin with?

Concerning my comment about hitting arctanx twice...if you rotate a line on the x-axis counterclockwise using origin as pivot then the left part of the line hits arctanx as well as the part on the right. You can fix this if you use half a line not the whole x-axis. but that does not mean using half a line won't cause other difficulties, i can think of a few.

You talked about rotating a line sitting on the x-axis this will hit arctanx twice, once on the right once on the left except when the line makes angle 90 degrees, then it hits arctanx only once. Have i misunderstood your original post?

about my use of infinity, didn't you use it first?

Steve, if you approach zero angle from above on the x-axis the right part of your line aproaches x=+infinity in arctanx and y approaches pi/2. However the left part of your line aproaches x=-infinity in arctanx and y approaches -pi/2 so your observation that tan(pi/2) is undefined is a bit misleading

 

[SteveL27]

Oh i get it now, is x any real number? The domain of arctanx is -infinity, +infinity the range is -pi/2, pi/2 this shows a fit of all real numbers in that interval ]-pi/2,pi/2[ why couldn't you guys say so to begin with?

The domain is the open interval ]-infinity, +infinity[. That notation is a shorthand for "the domain is all of the real numbers." That's a legitimate use of infinity. The open brackets mean that +/- infinity are NOT part of the domain; nor are they in the range of the tangent function. Using infinity that way is just a shorthand. And it's essential to understand that +/- infinity are not elements of the domain of the arctan.

Concerning my comment about hitting arctanx twice...if you rotate a line on the x-axis counterclockwise using origin as pivot then the left part of the line hits arctanx as well as the part on the right. You can fix this if you use half a line not the whole x-axis. but that does not mean using half a line won't cause other difficulties, i can think of a few.

If you prefer to think of the directed ray emanating from the origin, that's fine. But you don't actually need to.

Consider the line y = 2x. It passes through the point (1,2) so its slope is 2. But if we instead take the point in the third quadrant (-1, -2), the slope is still -2/-1 = 2. The tangent is the slope, period. And the angle is the angle made with the positive x-axis in the counterclockwise direction. That's the standard convention.

Why you keep saying it "hits arctan" is a complete mystery to me. It shows that you are misunderstanding something. The tangent is the slope as a function of the angle. The arctangent is the angle as a function of the slope.

You talked about rotating a line sitting on the x-axis this will hit arctanx twice, once on the right once on the left except when the line makes angle 90 degrees, then it hits arctanx only once. Have i misunderstood your original post?

In your latest post you seem to have some misunderstandings. I never said any such thing as "hitting arctan." You keep saying that, and I keep trying to correct that misunderstanding.

The slope of a vertical line is undefined.

about my use of infinity, didn't you use it first?

I used the notation ]-infinity, +infinity[ as a shorthand for "all the real numbers. That's a legitimate usage. The slope of a vertical line is undefined. The tangent of pi/2 is undefined.

Steve, if you approach zero angle from above on the x-axis the right part of your line aproaches x=+infinity in arctanx and y approaches pi/2. However the left part of your line aproaches x=-infinity in arctanx and y approaches -pi/2 so your observation that tan(pi/2) is undefined is a bit misleading

Do you understand the slope of a line? What is the slope of the line y = 2x? Does it matter whether you compute the slope using a point in the first quadrant or in the third quadrant?

The angle a line makes with the positive x-axis in the counterclockwise direction is unambiguous.

 

[agentredlum]

The domain is the open interval ]-infinity, +infinity[. That notation is a shorthand for "the domain is all of the real numbers." That's a legitimate use of infinity. The open brackets mean that +/- infinity are NOT part of the domain; nor are they in the range of the tangent function. Using infinity that way is just a shorthand. And it's essential to understand that +/- infinity are not elements of the domain of the arctan. If you prefer to think of the directed ray emanating from the origin, that's fine. But you don't actually need to.

Consider the line y = 2x. It passes through the point (1,2) so its slope is 2. But if we instead take the point in the third quadrant (-1, -2), the slope is still -2/-1 = 2. The tangent is the slope, period. And the angle is the angle made with the positive x-axis in the counterclockwise direction. That's the standard convention.

Why you keep saying it "hits arctan" is a complete mystery to me. It shows that you are misunderstanding something. The tangent is the slope as a function of the angle. The arctangent is the angle as a function of the slope.

In your latest post you seem to have some misunderstandings. I never said any such thing as "hitting arctan." You keep saying that, and I keep trying to correct that misunderstanding.

The slope of a vertical line is undefined. I used the notation ]-infinity, +infinity[ as a shorthand for "all the real numbers. That's a legitimate usage. The slope of a vertical line is undefined. The tangent of pi/2 is undefined.

Do you understand the slope of a line? What is the slope of the line y = 2x? Does it matter whether you compute the slope using a point in the first quadrant or in the third quadrant?

The angle a line makes with the positive x-axis in the counterclockwise direction is unambiguous.

Are you not using a line and arctanx to establish a one-to-one correspondence between points on the line and points on the graph of arctanx?

Saying tan(pi/2) is undefined only helps up to the level of precalculus, it does not help after that when limits are explored. The tan(pi/2) depends on which way you approach pi/2 on the x-axis, if you approach pi/2 from the left, with positive dx, tan(pi/2-dx) increases without bound, if you approach pi/2 from the right, with positive dx, tan(pi/2+dx) decreases without bound so these answers are not meaningless because they explain the behavior of tanx. To say tan(pi/2) is undefined doesn't help anyone beyond precalculus.

Like i said i can see it your way, but asking me what the slope of y=2x is hurts my feelings a little bit.


agentredlum at rest.

 

[SteveL27]

Are you not using a line and arctanx to establish a one-to-one correspondence between points on the line and points on the graph of arctanx?

Not in the slightest. I can't imagine where you got that idea.

Not only aren't we doing that; but it wouldn't even be interesting to try! If you look at the graph of arctan you see it's a curvy line in very obvious 1-1 correspondence with the points on the x-axis. Each vertical line in the plane passes through exactly one point of the x-axis and one corresponding point on the graph of arctan(x). This is the least interesting thing anyone could say about the arctan.

Saying tan(pi/2) is undefined only helps up to the level of precalculus, it does not help after that when limits are explored.

That remark is irrelevant to the discussion. What on Earth do limits have to do with this discussion?

The tan(pi/2) depends on which way you approach pi/2 on the x-axis, if you approach pi/2 from the left, with positive dx, tan(pi/2-dx) increases without bound, if you approach pi/2 from the right, with positive dx, tan(pi/2+dx) decreases without bound so these answers are not meaningless because they explain the behavior of tanx. To say tan(pi/2) is undefined doesn't help anyone beyond precalculus.

It would help you to understand what micromass was saying when he gave the arctan as a specific function that maps the reals to a bounded open interval.

I honestly cannot tell if you are just a little confused, or deliberately obfuscating the discussion.

Like i said i can see it your way, but asking me what the slope of y=2x is hurts my feelings a little bit.

If you understood that slope of a line through the origin is the same as the tangent of the angle the line makes with the positive x-axis, there would be no more confusion. I'm not trying to hurt your feelings, I'm just trying to explain the arctan function.

In any event, it's often the case that we may have studied math to a particular level, yet be totally confused about much more elementary things. We are using the high-school math idea of slope to visualize a bijection between the reals and a bounded open interval. So we're taking a more sophisticated look at something elementary here, and there's no harm in trying to review the basics.In any event ... micromass already gave a bijection between the reals and a bounded open interval. I mentioned a visualization that helps me to understand that bijection. However if it's not helpful to you, it's not worth further flogging this deceased equine.

 

[Robert1986]

Admittedly, I haven't read every single post. But it seems as though agentredlum has gotten (at least) two bijections confused.

Given that most people on this forum are better at explaining stuff that I am, this might be a futile attempt on my part, but here goes:

First, forget about the semi-circle thing for now. This doesn't have anything to do with the arctan bijection. You asked for a bijection and micromass (or someone) gave you one. It is just a bijective function from the entire real line to to the interval (-a,a). (Graph it in WolframAlpha.) However, I am not quite sure how Steve has come up with his visulisation. The slope of arctanx is not 0 at x=0, it is 1. And, as x -> infinity, the slope goes to 0. Now, it is 100% possible that I, too, have misunderstood (or not read) something, but SteveL seems to have gotten arctan confused with tan. Yet, this doesn't really matter. Even in the visualisation that SteveL gave (which I think is of the tanx function on he interval (-a,a)) there is still a bijection between the reals and this interval. As someone (I think Steve) mentioned, graph (I would use wolframalhpa) tanx and arctanx and you will see that there is a bijection.

 

[SteveL27]

However, I am not quite sure how Steve has come up with his visulisation. The slope of arctanx is not 0 at x=0, it is 1. And, as x -> infinity, the slope goes to 0. Now, it is 100% possible that I, too, have misunderstood (or not read) something, but SteveL seems to have gotten arctan confused with tan.

I'm tempted to just let this go. I'd invite you to reread my posts. The slope of arctan, by which I imagine you mean the derivative of the arctan function, has nothing to do with this.

Briefly, as a line through the origin goes between -pi/2 and pi/2, its slope -- the tan of the angle -- goes from -infinity to +infinity. The inverse function is the arctan, mapping the reals to the bounded open interval ]-pi/2, pi/2[. That's all I've ever said. It's how I visualize the arctan.

I'm quite surprised that two people have now read what I've written and decided that I'm trying to say something about the slope of the graph of the arctan function. I'm literally baffled by that interpretation of what I wrote. I can't say anything more in this thread that I haven't already said.

 

[agentredlum]

Not in the slightest. I can't imagine where you got that idea.

From the first paragraph of post #46 which i quote below.

'It's just the arctan function. The way I like to think of this bijection is by imagining a horizontal line in the plane through the origin. It makes an angle of zero with the x-axis and its slope is zero.'

I thought you were trying to create a bijection between a line and arctanx subsequently my objections followed.

 

[I_am_learning]

Although, I haven't followed whole of the thread, I came across the thought experiment, where you randomly place the tip of your pencil on a line marked ----------> 0--------1.
Why do you say that the pencil always lands at irrational number? Because there are just as many rationals as irrationals (both infinite), the chances must be equal.
I know I am wrong (because you appear to be great mathmatician :) ), but I would like to learn. :]

 

[gb7nash]

No. There are more irrational numbers than rational numbers. As I said earlier in the thread, the more technical reason for this is that the rational numbers are countable and the irrational numbers are uncountable.

 

[I_am_learning]

No. There are more irrational numbers than rational numbers. As I said earlier in the thread, the more technical reason for this is that the rational numbers are countable and the irrational numbers are uncountable.

Is there a simple logical explanation for that? (Countable/uncountable don't appear to have enough logic)
Sorry if I am having you to repeat.

 

[HallsofIvy]

Yes, every number, rational or irrational, can be represented as a limit of a sequence of rational numbers. In fact, you make use of that when you represent an irrational number in "decimal form". Saying that $\pi= 3.1415926...$ means precisely that the sequence of rational numbers (any terminating decimal is a rational number) 3, 3.1, 3.14, 3.141, 3.1415, 3.14159, 3.141592, ... has $\pi$ as limit.

One way of defining the real numbers, in terms of the rational numbers, is to use equivalence classes of sequences of rational numbers:

Let S be the set of all increasing, unbounded (equivalently "Cauchy") sequences of rational numbers. (Such sequences do not necessarily converge in the rational numbers.) We say that two such sequences, ${a_n}$ and ${b_n}$ are "equivalent" if and only if the sequence ${a_n- b_n}$ converges to 0.

A "real number" is an equivalence class of such sequences.

 

[HallsofIvy]

Is there a simple logical explanation for that? (Countable/uncountable don't appear to have enough logic)
Sorry if I am having you to repeat.

They do if you understand the technical meaning of "countable" and "uncountable".
An infinite set is said to be "countable" if and only if there is a one to one mapping of the set onto the natural numbers, 1, 2, 3, ... An infinite set is said to be "uncountable" if and only if it is not countable.

A simple illustration that the set of all rational numbers is countable is given here:
http://www.homeschoolmath.net/teaching/rational-numbers-countable.php

A discussion of Cantor's proof that the set of all real numbers (and hence the set of all irrational numbers) is uncountable is given here:
http://en.wikipedia.org/wiki/Cantor's_diagonal_argument

 

[agentredlum]

Although, I haven't followed whole of the thread, I came across the thought experiment, where you randomly place the tip of your pencil on a line marked ----------> 0--------1.
Why do you say that the pencil always lands at irrational number? Because there are just as many rationals as irrationals (both infinite), the chances must be equal.
I know I am wrong (because you appear to be great mathmatician :) ), but I would like to learn. :]

The thought experiment is designed to be simple enough that any non-expert can understand and to boggle the mind. If it has done this for you then it is a success. It is designed to bring into question the pedestrian understanding of infinity and invite your curiosity to learn more.

Georg Cantor was the first to show that some infinities are equal, some infinities are larger than others. Then he showed that if you consider any infinity, there is always another infinity larger. That is MIND-BOGGLING!

Cantor did not do this by counting numbers. He did it by establishing a one-to-one correspondence between members of sets.

Here is a simplified example of his technique. Suppose there is a classroom with chairs and students outside in the hallway. Someone asks, 'which set is larger, the set of chairs, or the set of students? Somebody else says, 'I know what to do, count the number of chairs and count the number of students, then we will know which set is greater.' 'WAIT!' says Cantor, 'ask the students to sit down, if there are chairs left over then the set of chairs is larger, if there are students left over then the set of students is larger, if all students are seated and all chairs have 1 student then the sets are equal.'

This simplified version shows it is possible to determine which set is larger WITHOUT COUNTING the members of any two sets and without knowledge of the magnitude of each set by using a one-to-one correspondence, to every student is assigned 1 chair, and to every chair is assigned 1 student.

This is very helpfull when considering infinite sets because no one can actually count up to infinity but after Cantor things you can count and things you can't count were forced to have a different meaning. That is where the terms 'countable infinity' and 'uncountable infinity' originated.

 

[Robert1986]

It's just the arctan function. The way I like to think of this bijection is by imagining a horizontal line in the plane through the origin. It makes an angle of zero with the x-axis and its slope is zero.


As you rotate the line counterclockwise, as the line goes from horizontal to vertical the angle goes from 0 to pi/2; and the slope goes from 0 to +infinity.

Likewise as you rotate a horizontal line clockwise, the angle goes from from 0 to -pi/2 (= 3*pi/2), and the slope goes from 0 to -infinity.

Ahhhh. Now I see what you are saying. My apologies; my interpretation of your post was, in fact wrong. When you started off by saying "It's just the arctan function." I thought that what you described next was how you constructed the graph of the arctan function in your mind.



Now that I see what you mean, I see that it is actually a pretty interesting way of looking at it.

My apologies.


EDIT: If this post sounds sarchastic, I don't meant it to be; it is sincere.

 

[I_am_learning]

They do if you understand the technical meaning of "countable" and "uncountable".
An infinite set is said to be "countable" if and only if there is a one to one mapping of the set onto the natural numbers, 1, 2, 3, ... An infinite set is said to be "uncountable" if and only if it is not countable.

A simple illustration that the set of all rational numbers is countable is given here:
http://www.homeschoolmath.net/teaching/rational-numbers-countable.php

A discussion of Cantor's proof that the set of all real numbers (and hence the set of all irrational numbers) is uncountable is given here:
http://en.wikipedia.org/wiki/Cantor's_diagonal_argument

Nice.
So, rational numbers is countable because it can be precisely located in the infinity X infinity (2-dimensional) matrix of that homeschoolmath site.

Natural number is countable because it can be precisely located in 1x inifinity matrix (1-dimensional) (1 2 3 4 5 ... inf)

Irrational numbers are uncountable because no such matrix exists. Infact, we would have to develop a infinite dimensional matrix (infinity X infinity X infinity X infinity ...) so as to make a matrix that would contain all the Irrational numbers. So, that makes it un-countable and infinitely many than the rational number.
Pretty satisfied now.
However, I am sure you folks have much better grasp of the concept than this.
Thanks.

 

[spamiam]

First, forget about the semi-circle thing for now. This doesn't have anything to do with the arctan bijection. You asked for a bijection and micromass (or someone) gave you one. It is just a bijective function from the entire real line to to the interval (-a,a). (Graph it in WolframAlpha.)

Actually, if you write out the function for my semicircle bijection, you'll see it's more or less the inverse to the one given previously. If you work it out, the semicircle method gives $g : (-c, c) \to \mathbb{R}$ by
$$g(x) = \frac{2c}{\pi}\tan\left(\frac{\pi}{2c}x\right)$$
and the one given before was $f : \mathbb{R}\rightarrow (-a,a)$ by $$f(x) = \frac{2a}{\pi} \arctan (x) \, .$$

But I like describing the bijection I gave pictorially because I think it's much more intuitive.

 

[Robert1986]

Actually, if you write out the function for my semicircle bijection, you'll see it's more or less the inverse to the one given previously. If you work it out, the semicircle method gives $g : (-c, c) \to \mathbb{R}$ by
$$g(x) = \frac{2c}{\pi}\tan\left(\frac{\pi}{2c}x\right)$$
and the one given before was $f : \mathbb{R}\rightarrow (-a,a)$ by $$f(x) = \frac{2a}{\pi} \arctan (x) \, .$$

But I like describing the bijection I gave pictorially because I think it's much more intuitive.

Correct. I meant to mention that he should also consider your example.

Your example is certainly more intuitive and should be what he is looking for given he didn't like the more technical stuff. I also like your description, as well, for the same reason that you like it.

 

[cant_count]

Not a mathematician. I've tried to do some reading on this. My intuitive thought is this: between any two rational numbers there's an infinite number of rational numbers because you can invent any fraction (1.1, 1.11, 1.111, ...). I guess everybody would agree on that. However, when you "hit" infinity in your inventing, then you get an irrational number. Physically impossible, but in theory, if there really is such thing as infinity, then you could. Is that a paradox? Does it mean rational numbers tend towards continuity? - Does it mean anything at all? (I think I've blown a few brain-cells.)

It's a bit like: what's the definition of random? If the digits of Pi are random (and I haven't yet understood whether they are or not) - then does that not mean this: that any sequence of digits you can invent will occur somewhere in it ... including Pi "eventually" repeating itself .. ?

I guess it's like saying you're trying to reach infinity.

(This is the trouble with letting anybody into your forums ! )

 

[pessimist]

your intuitive thought is absolutely not the way to go. The notion of infinity in itself is a counter intuitive notion + your thought about hitting infinity doesn't really make much sense.

 

[agentredlum]

Not a mathematician. I've tried to do some reading on this. My intuitive thought is this: between any two rational numbers there's an infinite number of rational numbers because you can invent any fraction (1.1, 1.11, 1.111, ...). I guess everybody would agree on that. However, when you "hit" infinity in your inventing, then you get an irrational number. Physically impossible, but in theory, if there really is such thing as infinity, then you could. Is that a paradox? Does it mean rational numbers tend towards continuity? - Does it mean anything at all? (I think I've blown a few brain-cells.)

It's a bit like: what's the definition of random? If the digits of Pi are random (and I haven't yet understood whether they are or not) - then does that not mean this: that any sequence of digits you can invent will occur somewhere in it ... including Pi "eventually" repeating itself .. ?

I guess it's like saying you're trying to reach infinity.

(This is the trouble with letting anybody into your forums ! )

I like your idea about rational numbers 1/n as n goes to infinity. Yes, i believe some of them tend toward becoming irrational. Never thought of it that way until you mentioned it so thank you.

Heres what i mean, for a prime p, 1/p can have at most p-1 digits after the decimal before it starts to repeat.

Some primes do not exhaust the possibilities, some primes do.

Example for the prime 3, 1/3 can have at most 2 digits after the decimal before it starts to repeat but
1/3 = .3333... repetition begins after 1 digit so 3 does not exhaust the possibilities

Example for the prime 7, 1/7 can have at most 6 digits after the decimal before it starts to repeat
1/7 = .142857142857142857... repetition begins after the 6th decimal position so 7 exhausts the possibilities.

It's interesting to look at primes in this way.

1/11 = .090909... repetition begins after the 2nd position so 11 does not exhaust the possibilities.

1/13 = .076923 076923 076923... repetition begins after the 6th decimal position so 13 does not exhaust the possibilities.

1/17 = .0588235294117647 0588235294117647 0588235294117647 ... repetition begins after the 16th decimal position so 17 exhausts all the possibilities.

Now, we know that the primes are infinite in number, this means they get bigger and bigger. There may be many primes out there larger than 10^1000000 that have somewhere in the neighborhood of 10^1000000 - 1 non repeating digits, and then repetition begins. As far as calculation is concerned this 1/p would be as difficult to calculate as sqrt(2) to the 10^1000000 - 1 decimal position. So in some sense, there are rational numbers that tend to be irrational.

This idea is going to generate a lot of controversy here so pretend that i am joking.

There is a method similar to long division for finding square roots without using a calculator, you can find it toward the bottom of the following page

http://en.wikipedia.org/wiki/Methods_of_computing_square_roots