agentredlum
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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.
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.

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
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