# Where are the irrational numbers?

by smolloy
Tags: irrational, numbers
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P: 15,671
 Quote by agentredlum 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.
P: 460
 Quote by micromass 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
P: 1,670
 Quote by agentredlum 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.
P: 460
 Quote by disregardthat 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?
 P: 460 mathwonk...heeeeelp!
 PF Patron P: 753 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.
 P: 460 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.
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 Quote by agentredlum 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.

 Quote by 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.
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.
P: 460
 Quote by micromass 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
P: 791
 Quote by agentredlum 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.
 P: 362 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. Attached Thumbnails
P: 460
 Quote by SteveL27 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.
P: 460
 Quote by 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.
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 acheive 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.
P: 791
 Quote by agentredlum 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.

 Quote by agentredlum 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.

 Quote by agentredlum 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.

 Quote by agentredlum [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.

 Quote by agentredlum 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.
P: 460
 Quote by SteveL27 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
P: 791
 Quote by agentredlum 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.

 Quote by agentredlum 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.

 Quote by agentredlum 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.

 Quote by agentredlum 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.

 Quote by agentredlum 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.
P: 460
 Quote by SteveL27 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.
P: 791
 Quote by agentredlum 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.

 Quote by agentredlum 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?

 Quote by agentredlum 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.

 Quote by agentredlum 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.

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