Cardinality of infinity (3)

[kingwinner]

Homework Statement

1) Find the cardnality of the set of constructible angles whose cosine is
a) irrational
b) rational


2. The attempt at a solution
For the other questions I have posted my attempt, but for this one I really have no clue...
Could someone please give me some general hints?

Your help is greatly appreciated!

 

[morphism]

Isn't this problem simply equivalent to the problem of determining the cardinality of the sets of constructible rational and irrational numbers, respectively? (Since an angle is constructible iff its cosine is.)

 

[kingwinner]

Isn't this problem simply equivalent to the problem of determining the cardinality of the sets of constructible rational and irrational numbers, respectively? (Since an angle is constructible iff its cosine is.)

Hi,

I don't think they are equivalent.
cos(x) has to be between -1 and 1 inclusive, so it can't be e.g. 2, 3 , -4, etc. while the set of constructible numbers have these as elements.

 

[HallsofIvy]

That is not relevant. What he said was that an angle is constructible if and only if its cosine is. Yes, 2, 3, etc. are constructible and so are cos(2), cos(3), etc. And a number is constructible if and only if it is algebraic of order a power of 2 which includes all rational numbers. Now- there are at most a countable number of possible polynomials, with integer coefficients, of degree n, each of which has at most n zeros so the set of all numbers algebraic of order n is countable. What does that tell you about the cardinality of the set of all algebraic numbers?

 

[kingwinner]

That is not relevant. What he said was that an angle is constructible if and only if its cosine is. Yes, 2, 3, etc. are constructible and so are cos(2), cos(3), etc. And a number is constructible if and only if it is algebraic of order a power of 2 which includes all rational numbers. Now- there are at most a countable number of possible polynomials, with integer coefficients, of degree n, each of which has at most n zeros so the set of all numbers algebraic of order n is countable. What does that tell you about the cardinality of the set of all algebraic numbers?

"And a number is constructible if and only if it is algebraic of order a power of 2 which includes all rational numbers."
I haven't learned this in my class, so I can't take this for granted...there is so little tool I can use :(

 

[kingwinner]

Isn't this problem simply equivalent to the problem of determining the cardinality of the sets of constructible rational and irrational numbers, respectively? (Since an angle is constructible iff its cosine is.)

So I think the problem is equivalent to the problem of finding the cardnality of the sets of consturctible rational and irrational numbers between -1 and 1 inclusive. To justify this, we need to first prove the equivalence.
e.g. Is there a 1-1 and onto map from {consturctible irrational numbers between -1 and 1} and {consturctible angles whose cosine is irrational}? I can't think of one...for example cos360^o=cos720^o, how can it be one-to-one, then?

I know that {all consturctible numbers} is countably infinite, but how about {consturctible irrational numbers between -1 and 1} and {consturctible rational numbers between -1 and 1 inclusive}? How to find their cardinalities?

Thanks!

 

[kingwinner]

An angle is constructible iff its cosine is, but the trouble is that the correspondence is not one-to-one. There are infinitely many angles whose cosine is the same value.

 

[morphism]

That doesn't matter. x is a constructible angle iff cos(x) is constructible. So if x is a constructible angle whose cosine is rational, that means cos(x) is a constructible rational number, etc. etc.

 

[kingwinner]

But if cos(theta) and theta does not have one-to-one correspondence, then we CANNOT say that:
|{constructible angles whose cosine is irrational}| = |{constructble irrational numbers between -1 and 1}|
which seems to create a lot of trouble...

Does anyone have any idea how to fix this problem? Please help!

 

[HallsofIvy]

It seems to me that a large part of this problem is know what numbers are constructible. Since you say that you do not know that "a number is constructible if and only if it is algebraic of order a power of 2" what do you know about constructible numbers? Since you are asking about the "cardinality of the set of constructible angles whose cosine is rational", are you aware that all rational numbers are constructible?

 

[kingwinner]

It seems to me that a large part of this problem is know what numbers are constructible. Since you say that you do not know that "a number is constructible if and only if it is algebraic of order a power of 2" what do you know about constructible numbers? Since you are asking about the "cardinality of the set of constructible angles whose cosine is rational", are you aware that all rational numbers are constructible?

Yes, I am aware that all rationals are constructible.

 

[HallsofIvy]

Good! So you know the answer to part b! Now since you know that cos(x) is irrational if and only if x is, and, between 0 and $2\pi$ there at most two angles that have the same cardinality, you know that the answer to part a is just the cardinality of the constructible numbers.

 

[kingwinner]

Good! So you know the answer to part b! Now since you know that cos(x) is irrational if and only if x is, and, between 0 and $2\pi$ there at most two angles that have the same cardinality, you know that the answer to part a is just the cardinality of the constructible numbers.

"cos(x) is irrational if and only if x is"
Um...I am not too sure about this, why is this true?


"between 0 and $2\pi$ there at most two angles that have the same cardinality"
Why only between 0 and 2pi? Constructible angles can be e.g.720 degrees, 1440 degrees, etc.

 

[kingwinner]

Still struggling to understand...

 

[kingwinner]

If I restrict theta to be between 0 and 180 degrees, then cos(theta) and theta would have one-to-one corrspondence. Thus |{constructible angles between 0 and 180 degrees whose cosine is irrational}| = |{constructble irrational numbers between -1 and 1}|

{constructble numbers} is countably infinite, so {constructble irrational numbers between -1 and 1} as a subset of it must be countable. But how can we justify that {constructble irrational numbers between -1 and 1} is an infinite set?[/color]

Now, assuming {constructble irrational numbers between -1 and 1} is countably infinte, then {constructible angles between 0 and 180 degrees whose cosine is irrational} is countably infinite since they have the same cardinality

{constructible angles whose cosine is irrational} is the union of {constructible angles between 180k and 180k+180 degrees whose cosine is irrational} where k is from 0 to infinity
This is a countable union, and a countable union of countable sets is countable, so {constructible angles whose cosine is irrational} is countable.

Is this a valid proof?[/color]

 

[HallsofIvy]

Yes, once you have cleared up the question of whether the set of all constructible numbers between -1 and 1 is infinite.

 

[kingwinner]

Yes, once you have cleared up the question of whether the set of all constructible numbers between -1 and 1 is infinite.

How can I prove this? Any hints?

 

[kingwinner]

The irrationals are dense in the reals, so they must be infinitely many irrationals between -1 and 1. But here they must be constructbile, too! How can I prove that this set is also infinte?