Is the Real Number Line Really Infinitely Dense?

[MathAmateur]

Homework Statement

This is a proof (?) that (to quote the author): "any segment of finite length on the real number line is infinitely dense, in that you can squeeze all of the real numbers into it".

I can't follow all the steps and would appreciate an explanation.

Homework Equations

The author drew a straight line y=x from point (0,0) to point (0,1). He then ran a vector a long the line. The vector ran from coordinate (0,1) to the real number line (y= 1-\frac{x}{c}) where c is the number on the number line. The vector traversed from 0 to 1 on the straight line that it crossed (y=x from 0 to 1) and from 0 to infinity along the real number line. He then says that combining the two equations (y= x and y= 1-\frac{x}{c}, you get y = \frac{x}{x+1} and if you plug any positive real value into that equation you always get another number between 0 and 1

The Attempt at a Solution

I didn't see how he got the equation: y = \frac{x}{x+1} from y=x and y= 1-\frac{x}{c}. When I tried to solve for the two equations I got:
y-1 = \frac{x}{c} and then c = \frac{x}{x -1} and then substituting in I get y=x?? I am doing something wrong

 

[JonF]

I’m having difficulty following what you’re trying to explain, for one thing the point (0,1) isn’t on y=x.

But if you want to prove that we can map every number in the Reals uniquely into the segment (0,1) this isn’t so bad. If you’ve had any trig:

Consider f(x) = ½ + arctan(x)/pi

This function will take any real number and pack it into (0,1), and no two numbers go to the same number in (0,1)

 

[MathAmateur]

Sorry I did such a bad job explaining myself above. However I liked your proof.

I looked at the graph of arctan in Wickipedia (http://en.wikipedia.org/wiki/Inverse_trigonometric_functions). Since arctan runs from -1/2 pi to 1/2pi as x goes from -infinity to infinity monotonically, if one divides arctan by pi and adds 1/2 one gets one and only one of each of the reals from -infinity to infinity jammed somehow into the segment 0 to 1.

So the order of infinity of all the reals from 0 to 1 is the same as all the reals from -infinity to infinity? How strange.

 

[Mark44]

Sorry I did such a bad job explaining myself above. However I liked your proof.

I looked at the graph of arctan in Wickipedia (http://en.wikipedia.org/wiki/Inverse_trigonometric_functions). Since arctan runs from -1/2 pi to 1/2pi as x goes from -infinity to infinity monotonically, if one divides arctan by pi and adds 1/2 one gets one and only one of each of the reals from -infinity to infinity jammed somehow into the segment 0 to 1.

You don't really divide "arctan" by anything any more than you would multiply \sqrt by 3. The name of the function is arctan. A value it produces is arctan(x).

The range of the arctan function is (-pi/2, pi/2). Another way to say this is -pi/2 < arctan(x) < pi/2.

If you divide all three members of this inequality, you get
-1/2 < arctan(x)/pi < 1/2
If you add 1/2 to all three members, you get
0 < arctan(x)/pi + 1/2 < 1

The arctan function maps the real line (-infinity, +infinity) to (-pi/2, pi/2). The function arctan(x)/pi maps the real line to (-1/2, 1/2).
The function arctan(x)/pi + 1/2 maps the real line to (0, 1).

So the order of infinity of all the reals from 0 to 1 is the same as all the reals from -infinity to infinity? How strange.

This seems strange because you're used to working with sets of a finite size. When you're working with sets with infinitely many members, a lot of the usual properties no longer apply. The only way to compare two infinite sets is by showing that there is a one-to-one mapping (function) between the two sets. In this problem, the one-to-one mapping is the function y = arctan(x)/pi + 1/2. Since this function is monotonically increasing on its domain (the reals), it is one-to-one. Since we have a one-to-one pairing between the reals and the interval (0, 1), the cardinality of the two sets is equal.

 

[MathAmateur]

Wow! This is really great stuff. Thank-you JonF and Mark44.