- #36
agentredlum
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micromass said:There exists a bijection between [itex]\mathbb{R}[/itex] 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 said:There exists a bijection between [itex]\mathbb{R}[/itex] and any arbitrary small interval. So the reals can be "fitted" in any arbitrary small interval.
agentredlum said:And what is the length of the real numbers?
micromass said:You mean the Lebesgue measure? It's infinite. I don't see what this has to do with anything.
agentredlum said: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?
disregardthat said: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 said: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 said: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 said: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,
[tex]\mathbb{R}\rightarrow ]-a,a[:x\rightarrow \frac{2a}{\pi} atan (x)[/tex]
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 said:Can you post a picture of the bijection, my browser does not decode TeX
SteveL27 said: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 said: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 said: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?
agentredlum said:I am a bit uncomfortable using your analogy because the line intersects arctanx twice for any given angle
agentredlum said:and zero angle gives plus or minus infinity depending on direction of rotation.
agentredlum said:[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.
agentredlum said:However having said that, i can still see it your way.
SteveL27 said: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.
agentredlum said: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?
agentredlum said: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.
agentredlum said: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?
agentredlum said:about my use of infinity, didn't you use it first?
agentredlum said: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 said: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.
agentredlum said: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?
agentredlum said:Saying tan(pi/2) is undefined only helps up to the level of precalculus, it does not help after that when limits are explored.
agentredlum said: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.
agentredlum said: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.
Robert1986 said: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.
SteveL27 said:Not in the slightest. I can't imagine where you got that idea.
gb7nash said: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.
They do if you understand the technical meaning of "countable" and "uncountable".I_am_learning said: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.
I_am_learning said: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. :]
SteveL27 said: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.
HallsofIvy said: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
Robert1986 said: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.)
spamiam said: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 [itex] g : (-c, c) \to \mathbb{R} [/itex] by
[tex] g(x) = \frac{2c}{\pi}\tan\left(\frac{\pi}{2c}x\right)
[/tex]
and the one given before was [itex]f : \mathbb{R}\rightarrow (-a,a)[/itex] by [tex]f(x) = \frac{2a}{\pi} \arctan (x) \, .[/tex]
But I like describing the bijection I gave pictorially because I think it's much more intuitive.
cant_count said: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 ! )