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        riddles >> putnam exam (pure math) >> cos(1 deg) = irrational
        
(Message started by: william wu on Aug 21^st, 2003, 2:22pm)

Title: cos(1 deg) = irrational
Post by william wu on Aug 21^st, 2003, 2:22pm

Prove that cos 1° is irrational.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Aug 22^nd, 2003, 5:43pm

Nice problem...
::[hide]
Let c=cos(x) and s=sin(x).

So, e^xi=cos(x)+i sin(x)=c+si

Therefore e^5xi=cos(5x)+i sin(5x)=(c+si)⁵.

(c+si)⁵=c⁵+5c⁴si–10c³s²–10c²s³i+5cs⁴+ s⁵.

Taking real parts, cos(5x)=c⁵–10c³s²+5cs⁴.

Using s²=1–c² throughout, cos(5x)=16c⁵–20c³+5c.

We know that cos(45–30)=cos45cos30+sin45sin30=([sqrt]3+1)/(2[sqrt]2), and this is clearly irrational.

Therefore, cos15=16cos3–20cos3+5cos3=([sqrt]3+1)/(2[sqrt]2).

If cos3 was rational, 16cos3–20cos3+5cos3 would be rational, which is a contradiction, so we deduce that cos3 is irrational.

Finally, using cos(3x)=4cos³x–3cos(x), cos3=4cos³1–3cos1. And by the same reaosning as above, we conclude that cos1 is irrational. quod erat demonstrandum.
[/hide]::

[e]Edited to add in new radical symbol. :)[/e]

Title: Re: cos(1 deg) = irrational
Post by Icarus on Aug 23^rd, 2003, 7:31pm

A somewhat simpler demonstration - though along the same lines:
[hide]
cos nx = 2cos (n-1)x cos x - cos (n-2)x

Repeated application of this formula allows you to express cos n^o = P(cos 1^o) for some polynomial P with integer coefficients.

Since the value of such a polynomial is rational if the argument is rational, if cos 1^o were rational, then so would be all values of cosine for integer degrees. Since we know that several integer degrees for which the cosine is irrational (for instance 45^o), cos 1^o must be irrational as well.[/hide]

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Aug 24^th, 2003, 6:03am

I'm really interested to know how you got this...

cos nx = 2cos (n-1)x cos x - cos (n-2)x

Could you please explain?

However, I don't think that your method works, Icarus.

Quote:

Repeated application of this formula allows you to express cos n^o = P(cos 1^o) for some polynomial P with integer coefficients.

After each iterative step, the new formula will contain cos1 raised to increasing powers.

For example, after the first stage,
cosn = 2cos(n–1)cos1–cos(n–2) = 2[2cos(n–2)cos1–cos(n–3)]cos1–cos(n–2) = 4cos(n–2)cos²1–2cos(n–3)cos1–cos(n–2).

Title: Re: cos(1 deg) = irrational
Post by towr on Aug 24^th, 2003, 8:01am

on 08/24/03 at 06:03:55, Sir Col wrote:

After each iterative step, the new formula will contain cos1 raised to increasing powers.

If cos(1) were rational, cosⁿ(1) would still be rational. So the sum of any powers of cos(1) would still be rational.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Aug 24^th, 2003, 8:21am

Good point, towr.

Do you know how he derived the iterative formula?

Title: Re: cos(1 deg) = irrational
Post by towr on Aug 24^th, 2003, 9:50am

I wouldn't claim he did..
http://mathworld.wolfram.com/Multiple-AngleFormulas.html has it (equation 37).. So while he might have derived it himself I would sooner think he learned it from somewhere..

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Aug 24^th, 2003, 10:49am

I really need to start making more use of that website. You won't believe how long it took me to get an expansion in terms of cos(x), for cos(5x). Before I thought of using Euler's formula – of which I was quite proud ;D – I was battling with trying to simplify cos(2x+3x),but to no avail. I couldn't get is exclusively in terms of cosine.

By the way, thanks very much for pointing out that formula, towr. When I get some time I may try to derive it for myself.

Title: Re: cos(1 deg) = irrational
Post by Icarus on Aug 24^th, 2003, 1:09pm

Towr has it right. I looked it up in my copy of CRC Standard Mathematical Tables - something we pre-internet types had to depend on for looking up all those formulas that you don't want to memorize (at least I'm young enough to have never used a slide rule! :P). But I once worked it out myself as well. You can prove it and the corresponding formula for sine,

sin nx = 2sin (n-1)x cos x - sin (n-2)x,
together by induction.

However, the page that towr linked has a nicer proof of the thing that I needed in my proof above: That cos nx is a polynomial with rational coefficients of cos x.

Title: Re: cos(1 deg) = irrational
Post by SWF on Aug 24^th, 2003, 2:21pm

This can be done without induction. Use the identity for cos(x+y) to find cos((m+1)x)=cos(mx+x) and cos((m-1)x)=cos(mx-x). Adding cancels out the sin() terms and leaves 2*cos(mx)*cos(x)=cos((m+1)x)+cos((m-1)x). Let m=n-1.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Aug 24^th, 2003, 2:32pm

Wow! Nice proof; thanks, SWF.

Title: Re: cos(1 deg) = irrational
Post by Icarus on Aug 24^th, 2003, 6:20pm

Related question: Show that cos x and sin x are algebraic if x is a rational multiple of [pi], and are transcendental if x is rational.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Aug 25^th, 2003, 4:55am

Great puzzle, Icarus, but are you sure about the last part? cos(1/2) is transcendental? :P

I'll assume that you meant algebraic for that too...

It is sufficient to show that if cos(x) is algebraic, sin(x)=[sqrt](1–cos²(x)) will also be algebraic.

We have established that a polynomial for cos(qx), where q is integer, exists in terms of cos(x).
Therefore a polynomial for cos(x) must exist in terms of cos(x/q).
E.g. cos(3x)=4cos³(x)–3cos(x) [smiley=bigto.gif] cos(x)=4cos³(x/3)–3cos(x/3).

So if cos(x) is algebraic, it follows that cos(x/q) is algebraic.

If x=p[pi], cos(p[pi])=[smiley=pm.gif]1, which is algebraic, hence cos(p[pi]/q) will also be algebraic.

If x=1, cos(1) is algebraic (see proofs above), therefore cos(p) is algebraic, and so too will be cos(p/q).

Title: Re: cos(1 deg) = irrational
Post by Icarus on Aug 25^th, 2003, 5:35pm

Sorry, but you have mis-interpreted something. cos(1) [ne] cos(1^o).

cos(1^o) = cos([pi]/180), which is algebraic.

When i refered to cos x. I meant exactly that, not "cosine of x degrees".

However, your proof that cos r[pi] is algebraic [forall] r [in] [bbq] is good.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Aug 25^th, 2003, 5:44pm

Why, thank you. :-[

At the time I thought to myself, "Icarus just doesn't make mistakes, except for flying too close to the sun, that is." But I couldn't think where I was misunderstanding the problem. I should have realised that cos(1^o)=cos([pi]/180). Duh!

Anyway, I must get some shut-eye now and work on it in the morning – well later this morning, as it's 1:45am. I just wish I hadn't peeked at the thread before going to bed; I'll spend most of the night awake now thinking about it. :)

Wonders if he can do something with the power series expansion of cos(x)... zzz

Title: Re: cos(1 deg) = irrational
Post by Icarus on Aug 25^th, 2003, 6:02pm

on 08/25/03 at 17:44:08, Sir Col wrote:

"Icarus just doesn't make mistakes, except for flying too close to the sun, that is."

Don't I wish! (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1027806383;start=75#78)

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Aug 26^th, 2003, 6:02am

I don't know if there's a more elementary proof for this, but I think the following method works; although it does make use of a theorem I researched and don't really know a whole lot about:

Given x is some rational value, suppose that cos(x)=y. Clearly y [ne] 0, as x = [pi](4k[pm]1)/2 and we have already stated that x is rational.

Using the identity, y=cos(x)=(e^ix+e^–ix)/2, we get 2y=e^ix+e^–ix.

Rearranging this and writing 2y=2y*e⁰,

e^ix+e^–ix–2y*e⁰=0 (1)

However, Linderman's Theorem states that A*e^a+B*e^b+C*e^c+... [ne] 0 if A,B,C,... and a,b,c,... are algebraic, each of A,B,C,... presented in the sum is non-zero, and a,b,c,... are distinct terms.

As y is not zero and both x and y are algebraic numbers, equation (1) has no solution. Hence we prove that cos(x) is transcendental for all rational values of x. In fact, the proof is much stronger and holds for all algebraic values of x.

I must say that I really enjoyed solving that problem. Thanks for asking the question, Icarus.

Now we've proved that cos(x) is transcendental for all algebraic values of x and cos(x) is algebraic when x is a rational multiples of [pi]. I wonder was can be said, if anything, about algebraic multiples of [pi]... ?

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Aug 26^th, 2003, 7:05am

I've just been thinking about Linderman's Theorem.

If we say that, A*e^a+B*e^b+C*e^c+...+ [alpha]*e⁰ [ne] 0; then A,B,C,...,[alpha] are non-zero algebraic numbers and a,b,c,... are also non-zero algebraic numbers – the non-zero clause is necessary now as the Theorem states that the exponents must be distinct and we have made the last term's exponent zero.

However, we can write, A*e^a+B*e^b+C*e^c+... [ne] -[alpha]*e⁰ = -[alpha].

In other words, a slightly different, and perhaps more useful form, would be: "The sum of exponential terms with non-zero algebaric coefficients and non-zero algebraic exponents cannot be algebraic and must be transcendental."

Title: Re: cos(1 deg) = irrational
Post by Icarus on Aug 26^th, 2003, 3:08pm

Nice! I was thinking of the Gelfond-Schneider Theorem, which William quoted recently:

Gelfond-Schneider Theorem: If [alpha] and [beta] are algebraic numbers with [alpha][ne]0, [alpha][ne]1, and [beta][notin][bbq], then [alpha][smiley=supbeta.gif] is transcendental.

I believe you can get the result I actually stated by a clever choice of exponent. However, here Lindemann has trumped Gelfond.

Well done. 8)

Title: Re: cos(1 deg) = irrational
Post by ultrafilter on Aug 31^st, 2003, 9:12pm

I'm a little rusty on this sort of thing, but I think I've got a simple proof.

As we all know, every rational number is constructible. Since cos(n deg) can be expressed as a polynomial in cos(1 deg), it follows that, if cos (1 deg) is rational, the cosine of every angle is rational. But cos(20 deg) isn't constructible, which implies that it's not rational.

QED?

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Sep 1^st, 2003, 9:59am

Greetings, and welcome to the forum, ultrafiler!

I'm afraid that your proof doesn't quite work, in that the assumption at the heart of your proof is, in fact, what we are trying to prove.

To summarise your proof:
There exists a polynomial for cos(n^o) in terms of cos(1^o). If cos(1^o) was rational, then cos(n^o) would be rational for all n. As cos(20^o) is irrational, cos(1^o) cannot be rational.

What you have stated is the perfect introduction to the proof. However, my objection is that it is incomplete. That is, it does not demonstrate/prove that, (i) there exists a polynomial for cos(n^o) in terms of cos(1^o), and, (ii) cos(20^o) is irrational.

Of course (ii) could quickly be resolved by picking cos(30^o) instead. However, (i) is neither trivial nor simple, and as I've already mentioned, is pretty much equivalent to the fact that cos(1^o) is irrational in the first place.

Your post, however, does beg an interesting challenge...

Prove that cos(20^o) is irrational.

Title: Re: cos(1 deg) = irrational
Post by ultrafilter on Sep 1^st, 2003, 10:04am

on 09/01/03 at 09:59:55, Sir Col wrote:

Greetings, and welcome to the forum, ultrafiler!

I'm afraid that your proof doesn't quite work, in that the assumption at the heart of your proof is, in fact, what we are trying to prove.

To summarise your proof:
There exists a polynomial for cos(n^o) in terms of cos(1^o). If cos(1^o) was rational, then cos(n^o) would be rational for all n. As cos(20^o) is irrational, cos(1^o) cannot be rational.

My objection is two fold. It does not demonstrate/prove that, (i) there exists a polynomial for cos(n^o) in terms of cos(1^o), and, (ii) cos(20^o) is irrational.

Of course (ii) could quickly be resolved by picking cos(30^o) instead. However, (i) is neither trivial nor simple, and as I've already mentioned, is pretty much equivalent to the fact that cos(1^o) is irrational in the first place. In other words, it is one that needs to be proved to show that cos(1^o) is irrational.

Quite possibly. Like I said, I'm more than a little rusty. But I thought that had been shown earlier in the thread.

Title: Re: cos(1 deg) = irrational
Post by Icarus on Sep 1^st, 2003, 1:37pm

I think that Sir Col was assuming you were trying to prove it from scratch, since in my post showing that cos n^o is a polynomial of cos 1^o, it is only a very short way from there to the full theorem.

I assume that you proved this independently, and only bothered to give here the difference between your proof and mine.

And Sir Col is wrong about your showing that cos 20^o is irrational. cos 20^o is known not to be constructable (it is the key to the proof that it is impossible to trisect angles in contruction). And since as you said, all rationals are constructable, cos 20^o must therefore be irrational.

But that seems a very hard way to go about it. All you need is a single integer degree angle whose cosine is irrational. And the values cos 30^o = [sqrt]3/2 and cos 45^o = [sqrt]2/2 are both much better known as irrational numbers.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Sep 1^st, 2003, 2:09pm

Isn't that what I said? ???

The point I was trying to make was that it is no more obvious that cos(20^o) is irrational than it is that cos(1^o) is irrational. The proof for its irrationality is by no means trivial, which was why I presented the challenge... "Prove that cos(20^o) is irrational."

The problem with elementary proofs, like this, is that they are easy/hard depending on the results that are assumed to be true, and upon which you present the proof. The question is always, how much do I need to prove. For example, I used Euler's Formula in my proof, but perhaps I should have proved that. Then, I might have to show that ([sqrt]3+1)/(2[sqrt]2) is irrational, and so on. Where does one stop in proving something from 'first principles'. At the end of the day, the result that cos(1^o) is irrational, is itself fundamental and might be used as an assumption in a more difficult problem.

Title: Re: cos(1 deg) = irrational
Post by Icarus on Sep 1^st, 2003, 2:21pm

It sounded to me like you were saying he had not shown cos 20^o is irrational.

But he did, starting from a couple of results from construction theory.

Admittedly, cos 20^o is not constructable is a considerably more advanced result than the one he derived from it: cos 1^o is irrational. But (to the best of my recollection), the irrationality of cos 1^o is not used in showing cos 20^o is not constructable. Which means that making use of this result is valid.

I.e. Ultrafilter's proof works. It's just not a good way to go about it.

Title: Re: cos(1 deg) = irrational
Post by ultrafilter on Sep 1^st, 2003, 2:33pm

Actually, I'd say that my proof is pretty good if you already have the results it depends on. If not, then it is pretty bad.

That cos(n^o) can be expressed as a rational polynomial of cos(1^o) is a fairly standard result, isn't it? If not, it shouldn't be too hard to do by induction, using the formula from mathworld.

(btw, is the discussion in this forum limited to actual Putnam problems, or is any theoretical math ok?)

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Sep 1^st, 2003, 3:15pm

I agree, your proof is original, clever, and subtle, in that it is true to say that cos(1^o) being rational implies that a cosine, known to be irrational, is rational; which is a contradiction, hence cos(1^o) must be irrational. However, the opposite: the assumption that cos(1^o) is irrational, tells us nothing about other angles.

Excuse me digressing, but the previous objections I made were rooted in an issue that I have with mathematics and the way that we teach it. After my students present proof, I sometimes complain that they cannot assume a particular aspect to be true, as it is not a trivial result. They normally ask the question, "Well, what can I assume is true?" Good question!

As I have worked with intelligent school students for many years now, you begin to conclude that part of the problem is not with their solution, but the question. The proofs we learn, as mathematicians, and the proofs that I, in turn, expect my students to reproduce are often nothing more than jumping through hoops like performing animals.

For example, one of my objections to your proof, that cos(20^o) is irrational is by no means trivial, is reasonable. But why can't you assume it? As I mentioned in a previous post, is it any more reasonable for me to use Euler's formula, e^{[smiley=i.gif]x}=cos(x)+[smiley=i.gif]sin(x), without proof?

Equally, why couldn't I have proved the original problem by saying, "cos(x^o) is rational iff x=90k, 90k[pm]30, where k is integer. Therefore, it follows that cos(1^o) is irrational, as 1 is not a member of x." This is a well known fact, and is taught to most students as they are first introduced to the cosine graph; not in such explicit terms though. Of course, the formal proof of this 'fact' is definitely not trivial!

Anyway, back to my new challenge...

Prove that cos(20^o) is irrational.

(In reference to your question, I assume by the fact that this section of the forum has recently had "pure math" appended to the title implies that general mathematical problems (of a pure nature) are equally encouraged now – thereby widening the scope of posted challenges – but I am no authority. Perhaps one of the moderators could clarify?)

Title: Re: cos(1 deg) = irrational
Post by Icarus on Sep 1^st, 2003, 6:49pm

It seems to be the standard here that any problem of ~ Putnam level would be acceptable. Simpler math problems would be better suited to the main forums (a large number of them, as well as several that are harder than the Putnam level are to be found there).

on 09/01/03 at 15:15:28, Sir Col wrote:

Equally, why couldn't I have proved the original problem by saying, "cos(x^o) is rational iff x=90k, 90k[pm]30, where k is integer. Therefore, it follows that cos(1^o) is irrational, as 1 is not a member of x."

Say WHAT?? The reason you can't prove the original problem in that way is because your premise is false! Cosine takes on rational values for infinitely many angles not included in this set. It's possible that these are the only INTEGER-degree angles for which it is rational. But I would hardly call that "well-known". I doubt many people are even aware that these are the only multiples of 15^o for which the cosine is rational, though it is easy to deduce from the angle difference formulas and the standard values.

Quote:

This is a well known fact, and is taught to most students as they are first introduced to the cosine graph; not in such explicit terms though.

??? why would students be taught this when introduced to the cosine graph? I taught trig for many years and never saw any reason to bring in rational vs irrational values. For graphing you need to plot enough values to give them a feel for how the curve is shaped. (We had to do it all by hand, since we had no computers in the classrooms, and graphing calculators were still hideously expensive.) And of course you teach them the standard values (0, 30, 45, 60, 90...), but why bring up esoteric information about rational versus irrational values? ???

Quote:

Of course, the formal proof of this 'fact' is definitely not trivial!

Nor do I see that it is reasonable to use it (assuming we are refering to the restriction to integer x) in proving a statement that is a key part in proving it. This smacks of circular reasoning.

But Ultrafilter's proof does not require circular reasoning, in that proving the non-constructability of cos 20^o does not involve using the irrationality of cos 20^o or of other cosines (it does involve some other things from which the irrationality of cos 20^o is a trivial consequence, such as that it is algebraic with characteristic polynomial of degree > 1. But this is shown without reference to irrationality.

As to what can or cannot be used in proving a theorem. I would say it depends on 3 factors:
(1) Is the result you are trying to prove used in proving the theorem you want to use?

Clearly, if so - you can't use it in the proof!

(2) Do you have some understanding of the theorem you want to use? Have you seen it's proof? Do you have any idea of the logic involved in proving it?

If you answer no to any of these, you should at least be very hesistant to make use of it.

(3) Is your audience aware of the theorem you want to use?

Again, if the answer is no, you should either look for other ways, or aquaint your audience with the result as much as is reasonable before making use of it.

Quote:

Anyway, back to my new challenge...

Prove that cos(20^o) is irrational.

Why stop there: For which integers [smiley=n.gif] is cos [smiley=n.gif]^o rational?
Is it just [smiley=n.gif] [equiv] 0, [pm] 30 mod 90 ?

More generally, for which rationals [smiley=r.gif] is cos [smiley=r.gif][pi] rational?

Title: Re: cos(1 deg) = irrational
Post by Barukh on Sep 2^nd, 2003, 7:02am

on 09/01/03 at 18:49:56, Icarus wrote:

Icarus, your guess is right. There exists a theorem stating that if [smiley=r.gif] is rational, then only rational values of cos([smiley=r.gif][pi]) are 0, [pm]1/2, [pm]1. If you (or somebody else) are interested, I can try to find or reproduce it - the one I saw looked quite elementary.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Sep 2^nd, 2003, 8:47am

I'd definitely be interested, Barukh. Although I made a mistake, it's actually, n[equiv]0 mod 90 or n[equiv]90[pm]30 mod 180.

on 09/01/03 at 18:49:56, Icarus wrote:

??? why would students be taught this when introduced to the cosine graph? ... And of course you teach them the standard values (0, 30, 45, 60, 90...), but why bring up esoteric information about rational versus irrational values? ???

The way that I teach trigonometry, and I am sure many other high school teachers nowadays, is to begin with construction; for example, a 30^o right angle triangle with hypotenuse length, say, 10 cm, and get them to measure the opposite side. Either by further construction or by talking about the enlargement/scaling principle we would discuss, and work towards, the idea that an n degree right angle triangle with hypotenuse length 1 would be useful in determining the length of the opposite side in a triangle with any length hypotenuse. Hence we talk about the definition of sine, with reference to sine tables (which I used in school) or the button on scientific calculators, and define it as: the length of the opposite side in a unit right angle triangle. It is a small step then to introduce cosine, as the adjacent side, and finally to consider the unit circle and, of course, the sine/cosine graph. Natural questions relate to the 'nice' answer we got for the opposite side in a 30^o triangle (and the adjacent side in a 60^o triangle). The teacher would then usually reinforce the idea that other answers (generally not multiples of 30) are going to be approximations, because... the cosine/sine of those other angles will be irrational. Don't misunderstand me, I do not teach this explicitly, nor do I labour this point, but I would like to believe that my students take with them a vague impression of this 'fact'.

If I asked one of my newly initiated trigonometres if cos(1^o) is irrational, I would be delighted if their answer was, "Yes, becasue it's only multiples of 30 that give nice answers and not all of them work anyway; even cos(30^o) is irrational, but 0, 60, 90 are nice (rational)."

Is that a bad answer? I don't think so, as it conveys a much more general truth than a specific reference to one known irrational result. However, mathematicians become more sophisticated and recongnise the importance of rigour. This means building on smaller truths. But my question still stands, how far back do we need to go before it becomes a 'complete' proof?

One of the reasons, in my view, that many logical and intelligent people reject mathematics is because they find the subjective choice of assumptions that can be made, and those which can't be made, illogical. I have learned the 'rules', and enjoy solving mathematical problems, but it doesn't necessarily make it right.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Sep 2^nd, 2003, 2:28pm

I am really struggling with this latest problem... can it be done without an exhaustive approach?

We have already demonstrated that there exists a polynomial for cos(x) in terms of cox(x/n).

If cos(x/n) were rational, so too would be cos(x). If it is known that cos(x) is irrational, it follows that cos(x/n) is also irrational. In other words, the cosine of all the factors of x will also be irrational.

As cos(45), cos(30), and cos(20), can be shown to be irrational, it follows that cos(x) is irrational for x=1,2,3,4,5,6,10,15,20,30,45.

That leaves an awful lot of gaps.

Playing with addition formula seems to be fruitless. Consider cos(59+1)=cos(60)=cos(59)cos(1)–sin(59)sin(1); it is hardly intuitive that the RHS is rational.

It would seem that the irrationality of cos(x), will have to investigated for quite a few special cases. :-/

Title: Re: cos(1 deg) = irrational
Post by Icarus on Sep 2^nd, 2003, 3:58pm

My method of teaching was similar, except I never saw any need to bring in questions of rationality. The natural occurence of 30^o, 45^o, and 60^o in applications were sufficient to show why they need to learn these values. They are sufficient to get a fair idea of how to draw the graph of an affine function of cosine or sine. I know you are blessed with teaching a "higher grade" of student than I was (I was teaching mostly people who put off taking high school level mathematics until they were in college - which should give you a hint about their abilities), but if I brought up rational versus irrational values to my students I would of just confused them even more.

Barukh - I vaguely recall having seen that result somewhere, but if I ever investigated the proof, I have completely forgotten it.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Sep 2^nd, 2003, 4:41pm

on 09/02/03 at 15:58:53, Icarus wrote:

if I brought up rational versus irrational values to my students I would of just confused them even more.

Who says I don't confuse them!? After all, it's part of my job description as a teacher of mathematics. ::) Imagine being at a party and the conversation going like this. Guest: "So what do you do?" Me: "Oh, I teach mathematics." Guest: "Excellent, I really enjoyed maths at school – it made a lot of sense to me." It just wouldn't be natural, would it? ;)

By the way, I think that Barukh's result is the only (or best) way to solve this problem.

Title: Re: cos(1 deg) = irrational
Post by Icarus on Sep 2^nd, 2003, 5:04pm

Well - Don't show it yet! Let us have a chance to work it out ourselves!

Title: Re: cos(1 deg) = irrational
Post by SWF on Sep 2^nd, 2003, 7:49pm

Elementary proof that cos(20^o) might be irrational ;):
[hide]cos(60^o)=1/2=cos(3*20^o). Use identity for cos(3x) (Sir Col gave it earlier) to get:

8x³-6x-1=0, when x=cos(20^o).

Try to find relatively prime integers p and q, such that x=p/q:

8p³-6pq²-q³=0

The first two terms are even integers, so q must be even, and p must be odd, since it is relatively prime to q. Substitute q by 2*u where u is some integer, and divide by 8:

p³-u³=3*p*u²

Remember that p is odd. If u is odd, left side of equation is even, and right side is odd which is not possible. If u is even, left side of equation is odd, and right side is even which also is not possible. Therefore, x cannot equal p/q.
[/hide]

Title: Re: cos(1 deg) = irrational
Post by Barukh on Sep 3^rd, 2003, 1:03am

on 09/02/03 at 17:04:44, Icarus wrote:

Well - Don't show it yet! Let us have a chance to work it out ourselves!

Icarus, as I didn't quite get whether your exclamation was for the last problem, or for the proof of the general theorem, I will present here the latter in the hidden form.
[hide]
Let [alpha] = [smiley=r.gif][pi]. Because [smiley=r.gif] is rational, the sequence of angles [alpha], 2[alpha], ..., 2ⁿ[alpha],... has repeated terms, and so does the sequence
cos([alpha]), cos(2[alpha]), ..., cos(2ⁿ[alpha]), ... (*)
Let cos([alpha]) = p/q (the fraction is reduced). Then cos(2[alpha]) = 2cos([alpha])² - 1 = (2p² - q²)/q². Because gcd(p,q) = 1, we have gcd(2p² - q², q²) <= 2. Thus, if q > 2, the sequence (*) will consist of (reduced) fractions with strictly increasing denominators, and therefore won't have repeated values. This contradiction shows that q must be equal to 1 or 2, and completes the proof.

A really simple and brilliant argument, I must admit (I don't know who's the author).
[/hide]

Title: Re: cos(1 deg) = irrational
Post by Barukh on Sep 3^rd, 2003, 1:31am

on 09/02/03 at 19:49:06, SWF wrote:

Elementary proof that cos(20^o) might be irrational

SWF, your proof is very good. I just wanted to note that in the last part of your proof the following criterion could be used:[hide] The rational roots of the equation with integral coefficients are also integral; and they divide the free term.[/hide]
This immediately shows that your derived cubic equation doesn't have rational roots (although it makes the proof less elementary... ::))

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Sep 3^rd, 2003, 10:06am

Great proof, SWF! I love the reasoning on the last part; my version was so much more tedious. Although worked slightly differently, I wrote the equivalent of: [hide]p³=u(u²+3pu)=ku, and argued that because HFC(p,u)=1, all the factors present in p must be in k, therefore, u=[pm]1. Similarly, from u=p(p²-3u²), we deduce that p=[pm]1. As this does not satisfy the equation, there can be no rational solution, p/q[/hide].

Barukh, I'm obviosuly being really stupid, but could you, or someone, please explain the central part of the proof?
::[hide]
Because gcd(p,q) = 1, we have gcd(2p² - q², q²) <= 2 (why?).
Thus, if q > 2 (why?),
the sequence (*) will consist of (reduced) fractions with strictly increasing denominators (why?)
[/hide]::

Title: Re: cos(1 deg) = irrational
Post by Barukh on Sep 3^rd, 2003, 11:20am

on 09/03/03 at 10:06:21, Sir Col wrote:

Barukh, I'm obviosuly being really stupid, but could you, or someone, please explain the central part of the proof?

Let me try:
[hide]
1) gcd(2p² - q², q²) = gcd(2p², q²) (the fundamental property of gcd, the basis of Euclid's algorithm for finding gcd).
gcd(p,q) = 1 implies gcd(2p², q²) = gcd(2, q²) <= 2.

2) We assume that q is greater than 2, and show that this leads to a contradiction.

3) Because of 1), the fraction (2p² - q²)/q² cannot be reduced by a factor greater than 2. Therefore, if q > 2, the reduced fraction will have the denominator at least q²/2 > q. Obviously, this relation will hold for every term in the sequence.
[/hide]

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Sep 3^rd, 2003, 3:28pm

Feels even more stupid now you've explained it so well and made it sound so straight forward. :-[

Many thanks, Barukh – you're right, it is a really simple and brilliant argument and adds some closure (formal proof) to one of the results I've had for many years. So, one down, only a few hundred more to go on my list now... ;)

Title: Re: cos(1 deg) = irrational
Post by Icarus on Sep 4^th, 2003, 5:24pm

I've resolutely avoided Barukh's posts, because I want to figure this out myself. So far, this is what I've come up with.

Going back to the polynomial expressions for [smiley=c.gif][smiley=o.gif][smiley=s.gif] [smiley=n.gif][smiley=x.gif], we see that they are all integer polynomials in [smiley=c.gif][smiley=o.gif][smiley=s.gif] [smiley=x.gif] with leading coefficient 2[supn][supminus][sup1] and constant term 0 or [pm]1.

let [smiley=x.gif] = ([smiley=r.gif]/[smiley=s.gif])[pi] with [smiley=r.gif] and [smiley=s.gif] relatively prime integers. Because they are relatively prime, [exists] [smiley=n.gif] such that [smiley=n.gif][smiley=r.gif] [equiv] 1 mod [smiley=s.gif]. Applying the polynomial formula tells me that [smiley=c.gif][smiley=o.gif][smiley=s.gif] [smiley=r.gif][pi]/[smiley=s.gif] is rational iff [smiley=c.gif][smiley=o.gif][smiley=s.gif] [pi]/[smiley=s.gif] is.

Applying the polynomial formula to -1 = [smiley=c.gif][smiley=o.gif][smiley=s.gif] [smiley=n.gif][pi]/[smiley=n.gif] gives me a polynomial equation in [smiley=c.gif][smiley=o.gif][smiley=s.gif] [pi]/[smiley=n.gif] with leading coefficient 2[supn][supminus][sup1] and constant term 0, 1, or 2. The rational root theorem says that if [smiley=c.gif][smiley=o.gif][smiley=s.gif] [pi]/[smiley=n.gif] is rational, it must be of the form [pm]2[supminus][supi] for some [smiley=i.gif] with 0 [le] [smiley=i.gif] [le] [smiley=n.gif] - 1.

If [smiley=n.gif] is even, then [pi]/2 - [pi]/[smiley=n.gif] is a multiple of [pi]/[smiley=n.gif], and so its cosine must be rational as well. But
[smiley=c.gif][smiley=o.gif][smiley=s.gif]([pi]/2 - [pi]/[smiley=n.gif]) = [smiley=s.gif][smiley=i.gif][smiley=n.gif] [pi]/[smiley=n.gif] = [sqrt]( 1 - 4[supminus][supi] ) = [sqrt](4[supi] - 1) / 2[supi],
which is clearly irrational if [smiley=i.gif] > 0.

Hence, for the cosine of a rational multiple of [pi] to be rational, it must be 0 or [pm]2[supminus][supi] for some [smiley=i.gif] [ge] 0, and the denominator of the argument must be 2 or odd.

This is where I'm at so far. I just have to figure away to eliminate odd denominators > 3.

Title: Re: cos(1 deg) = irrational
Post by Icarus on Sep 29^th, 2003, 7:11pm

I finally gave up hope that I would have time to return to this one and figure the rest of it out myself. So I have read Barukh's post. This is certainly a lot easier than the method I was following!

Title: Re: cos(1 deg) = irrational
Post by BNC on Sep 29^th, 2003, 10:52pm

on 09/02/03 at 08:47:20, Sir Col wrote:

However, mathematicians become more sophisticated and recongnise the importance of rigour. This means building on smaller truths. But my question still stands, how far back do we need to go before it becomes a 'complete' proof?

On a side note, you may want to check the origins of proof (http://pass.maths.org.uk/issue7/features/proof1/index-gifd.html).

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Sep 30^th, 2003, 9:50am

Thanks for the link, BNC; it was an interesting article.

I must say that I still remain unconvinced and skeptical about what we are actually proving mathematically. I don't know if anyone has read the excellent book, Proof and Refutations, by Imre Lakatos? He uses the example of Descartes-Euler polyhedral formula: V+F=E+2, and he employs a Socratic method in which a teacher becomes involved in a discussion with some of his students about the validity of this result. After the students attempt to 'prove' the result, their ideas are continually refuted until most of them reluctantly come to the conclusion, along with the rest of us when we first discover, that the relation does not hold for all polyhedra.

It is a result we teach, but it is not true.

We teach that the angles in a triangle are equal to two right angles, but that isn't true.

We teach that 0.999...=1, but that... only joking! :P

Philosophically, so much of our reasoning is circular and is rooted in unprovable axioms – self evident, really? It was thought that Cantor, or someone after him, would prove the 'self-evident' continuum hypothesis; but it is now known to be unknowable in our system of logic and understanding.

Kronecker was no dummy, and he denounced the irrationals as an abomination in the face of the beauty of mathematics. He, and his peers, could intelligently justify a rejection of the infinite and methods of analysis.

Don't misunderstand me, I love mathematics. I enjoy proving results and solving problems, but... what are we building our mathematics on? Can we really be sure that the foundations are firm?

Title: Re: cos(1 deg) = irrational
Post by Icarus on Sep 30^th, 2003, 8:49pm

hoo-boy! What a can of worms to open, and it's late...

First of all, how much of this was intentional and how much was the result of limited understanding on the part of the author, I don't know, but the article takes a very simplistic and ill-defined approach to a number of things. (I expect that most if not all of it was an intentional simplification in order to make her particular point more readily available to her audience. But if so, I think she did them a disservice by stating the final paragraph as she has. And Sir Col's remarks are a clear indication of why it is a disservice. Perhaps she means to clarify the matter in the promised later articles.)

I'll start off with a point that I hope you are familiar with, but which she totally ignores: Euclid's first few definitions are nothing of the kind. There is indication in the Elements that Euclid himself knew this, and was only offering them as "heuristic descriptions" rather than true definitions. The problem is: how do you define "point"? Euclid defines them as "that which has no part". But this only begs the question of what does "having no part" mean? In order to define anything, you have to use terminology which is already defined. But if you are to avoid circularity, at some point you end up with some terms for which no predefined terms exist. These are called "primatives". Everything else is defined in terms of them, but for them, there is no definition. In Geometry, "point", "line", "plane", "between", "closer together than" are usually taken as primatives (though only "point" and "closer together than" need be - the rest can be defined from these two and the "common notions" of set theory).

The same thing hold true for theorems. The starting points this time are the axioms: A collection of relationships between the primatives. A more sophisticated view of primatives and axioms is that they are, taken together, a definition of themselves: that is, the axioms provide meaning to the primatives by describing how they relate to each other.

The article, and Sir Col, raise the question: are these axioms really true? While I suspect the author knows the answer, she leaves the waters muddied. This question has occupied the attention on many great mathematicians. When Gauss first realized the existence of hyperbolic geometry, he surrepticiously surveyed the angles formed by three well-separated peaks, to see if he could detect any difference between the sum of the angles and 180^o. He was trying to find out which was the "true" geometry. Hyperbolic (or Lobatchevskian) geometry was slow in gaining acceptance because mathematicians doubted it's "truthfulness".

Then Weierstrauss discovered the pseudosphere: a shape in Euclidean geometry which shows the same behavior as a strip cut from the Hyperbolic plane. He was followed by analytic descriptions from Poincare and Dedekind which exactly model hyperbolic geometry in its entirety. A simpler model for Euclidean geometry had already been found within hyperbolic geometry. This makes the answer to the question "which is the true geometry" clear: They are either both true, or both false. The truth of one implies the truth of the other!

But the two are identical in their axioms except for axiom 5. One demands it to be true, the other demands it to be false. How then can the truth of one imply the truth of the other? Isn't this contradictory?. The problem is not in either Euclidean or Hyperbolic geometry. It is not in the concept of proof. Where the problem lies is in the concept of "true".

The usage of "true" in both the article and in Sir Col's post is ill-defined. I have intentionally kept with this ill-defined usage in the statements above.

Obviously, I have more to say (when have I ever been short-winded)? But it's late and I have to work tomorrow, so I'm leaving it for now with this question: what does it mean - mathematically - for a statement to be "true"?

[ Sir Col - One last parting shot: I have never read the book by Lakatos, but quite frankly from your description I am not impressed. He pretends that a famous formula is applied more broadly than the theorem actually states, and then uses this artificially expanded scope to "disprove" it? This is chicanery, not mathematics! If his purpose is to show that you should be careful about exactly what it is you think you know, then I applaud. Descartes and Euler never claimed that particular formula applied to all polyhedra. Only to polyhedra topologically equivalent to a sphere (they stated it differently of course, but this is how I know it). But your description seems to indicate that he is claiming it shows a weakness in the idea of proof, and that it most certainly does not! ]

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Oct 1^st, 2003, 9:20am

I apologise, as I have clearly done a major disservice to Lakatos's work. You can read a little about him at St. Andrew's History of Mathematics (http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Lakatos.html).

Again, I've also failed to express myself clearly. One of my main areas of interest is the history of mathematical thought and philosophy. I am no expert and I do not claim to understand the all of the mathematics – in that respect, I am an enthusiastic novice – but with regard the life and works of the mathematicians I am well versed. I selected the Descartes-Euler relation as one example of many: interestingly, Euler abandonded the formula when he was unable to deal with the counter-examples.

My point, which I made badly, was that if there is one thing we can learn from history, other than we never learn from history, is that something remains true until it is no longer true.

That sounds like I'm stating the obvious, but it is in fact quite subtle and has everything to do with perception. I believe ideas to be true, because I respect the minds behind the concept, and I can (eventually) be persuaded of their truth. I depend on the insights and brilliance of the giants of intellect that appear so rarely: it is estimated that there are around 100 revolutionary great thinkers in each century. I do not decide truth, because I am an intellectual pygmy on the grand scale of things.

Once in a while someone, who gains respect from his/her peers arrives on the scene, and they challenge our thinking. Among many of the great thinkers I have had the privilege to meet, through their written works, Lakatos is one who has obtained special respect.

What Lakatos does, through his book, is challenge the methodology of mathematics. He presents a case for a system of observation, proof, and refutation; constantly refining and improving ideas through creativity and criticism. He challenges the ideas of formalism and the dogmas of logical positivism, and questions the notion that mathematics is merely the accumulation of established truths; ultimately, he calls for a more organic approach to the way in which we learn and discover mathematics.

Title: Re: cos(1 deg) = irrational
Post by Icarus on Oct 1^st, 2003, 8:48pm

on 10/01/03 at 09:20:26, Sir Col wrote:

Without reading his work myself, I cannot say one way or another.

Quote:

Again, I've also failed to express myself clearly. One of my main areas of interest is the history of mathematical thought and philosophy. I am no expert and I do not claim to understand the all of the mathematics – in that respect, I am an enthusiastic novice – but with regard the life and works of the mathematicians I am well versed. I selected the Descartes-Euler relation as one example of many: interestingly, Euler abandonded the formula when he was unable to deal with the counter-examples.

I am not intimately familiar with Euler's life, but I do not believe that this is correct. Euler did not "abandon the formula". He had no need of doing such a thing. It was proved quite well for the polyhedra for which it was stated. I find it far far far more likely that what Euler did was either give up trying to extend the class of polyhedra on which it was known to hold to the fullest extent, or he gave up trying to generalize the formula to apply to all polyhedra.

"Abandoning the formula" because there are polyhedra for which it does not hold would be like abandoning the formula 0^x = 0 because you have the case 0⁰ = 1. You don't simply abandon the whole thing because of exceptions. Instead you state explicitly where it is known to hold: [forall] x [in] [bbr] with x [ne] 0, 0^x = 0.

Quote:

My point, which I made badly, was that if there is one thing we can learn from history, other than we never learn from history, is that something remains true until it is no longer true.

That sounds like I'm stating the obvious, but it is in fact quite subtle and has everything to do with perception. I believe ideas to be true, because I respect the minds behind the concept, and I can (eventually) be persuaded of their truth. I depend on the insights and brilliance of the giants of intellect that appear so rarely: it is estimated that there are around 100 revolutionary great thinkers in each century. I do not decide truth, because I am an intellectual pygmy on the grand scale of things.

I do not believe ideas to be true just because someone who is considered an "intellectual giant" says they are. I accept things as true only on the basis of evidence. (This does not mean that I reject what people say until I have understood it myself. I have neither the time nor the capacity to follow every line of reasoning. But I hold all results I have not understood at "2nd level". I accept them, but not without reservation.)

I am not asking anyone to decide what is true or what is not. I am asking for what it means to be true. Why is 1 = 1 true and 1 = 2 not true?

You say that the Descartes-Euler formula is false. I say it matters first of all exactly what you are stating as the Descartes-Euler formula. If you say "For all polyhedra, V+F = E+2", then yes, that is false. It is also not the Descartes-Euler formula, as you have applied it more broadly than they ever did. This falsehood is the result of bad teaching, not the result of broken mathematics. If you say "For all polyhedra homeomorphic to a sphere (or a different, still workable restiction), V+F = E+2", then you are wrong in saying it is not true.

The case for 180^o in a triangle is similar, but it is again late, so I will expound on it another time.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Oct 2^nd, 2003, 10:24am

on 10/01/03 at 20:48:18, Icarus wrote:

I do not believe ideas to be true just because someone who is considered an "intellectual giant" says they are. I accept things as true only on the basis of evidence.

Where did you acquire the art of reasoning and discernment? I believe it is a cultural indoctrination. We think in words, pictures, and numbers, based on our experiences, and this methodology is disseminated to future generations. To think outside out of how we think, is almost a contradiction. Our standards, absolutes, and truths, are determined by the conditioning of our mental faculties.

Although I am passionate about mathematics, I share an equal love for language and etymology. Sad as it may sound, I get quite excited when I discover the origin, or root, of a particular word. I'm generally satisfied to know that a word like, geometry, comes from the Greek, geo=earth, and metron=measure; that is, earth measure, which helps us understand the original practical aspect of the science. However, how far back do we go? Where did those Greek words come from? Am I interested? If not, why not? This is analogous with my concern with what we do in mathematics. As I keep saying, I enjoy what I do, but I recognise that I am deluding myself with a form of semi-truth. I accept on the one hand that I am not really getting to the heart of the matter, but on the other hand, I don't care. If I asked someone to explain where the word, geometry, came from, and they answered, "From the dictionary," should I be concerned? To reject that type of response, and accept the, geo-metron, response is still falling short of what is actually going on.

I am not seeking approval, but I am trying to explain where my apparent anti-establishment attitude is coming from. It is not to get a reaction, neither is it a primitive naivity, it is by considering mathematics and learning on a philosophical level. Unfortunately for me, it is one of those revelations that fascinates on the one level, yet abhors on another.

Title: Re: cos(1 deg) = irrational
Post by towr on Oct 2^nd, 2003, 1:53pm

I don't really see what the problem with proof would be in mathematics. You have primitives and axiom, and everything what is proven in mathematics is true given those primitives and axioms.

The problems with logical positivism arises when you deal with physical reality, and observation, and you suddenly find yourself lacking both the necessary primitives and axioms to proof things. You can only guess at things to model reality with as best you can.
As soon as you abstract from the reality that problem disappears, you can just examine where your (assumed) primitives and axioms lead you. Which is quite valid, and can help you discover more about similar systems, which can in turn give new insights, which you may, if you so wish, apply to new models of reality.

One of the problems with positivism in science is that you can't proof general statements, since you would need to exhaustively search your domain to check your model against it. The only meaningfull thing you can do is try to disprove general claims, by finding counter examples. But then you still won't end up with models which are true, just models which haven't been proven false.
So the problem you have is, how much credence should you give such models? None of them can be proven, but some are instinctively 'better' than others. What gives a good measure that drives you closer to the truth?
One of the popular ways to prove a model/theory is better than another these days is to find the most unlikely event implied by the theory, and then try to disprove it (which should be easy if it's an event that seems quite unlikely to be true).

One of the things Lakatos (and Kuhn) are well known for in this context is that they state that there are research programs, with a hard core that doesn't change much for long times, and looser hypothesis that get adapted over time. And every once in a while there's a scientific revolution that sparks rivaling research programs which may drive the other to extinction.

(To be honest, "philosophy of science" by Bechtel explains it much better, and more thoroughly. And it's thin enough for even me to read :P)

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Oct 2^nd, 2003, 3:27pm

Thanks for your thoughts and book recommendation, towr. After a tricky search on the internet I've located copy of Bechtel, and one should be with me within a couple of days.

I know what you mean about radical thinking, or being the better orator, driving one methodolgy to extinction; as you cited, Kuhn vs Popper is a classic example. The right person doesn't always win, or sometimes the right person wins, but for the wrong reasons.

Title: Re: cos(1 deg) = irrational
Post by Icarus on Oct 2^nd, 2003, 7:50pm

on 10/02/03 at 10:24:24, Sir Col wrote:

I do not claim that I independently rediscover anything here. My abilities of reasoning and discernment were indeed trained into me. But on the other hand, I kept at that training because I could follow the sense of it myself. I never accepted "because that's the way it's done!" as the answer to the question "why do it that way?"

I have great respect for those who first uncovered these results I find so interesting, who first searched laboriously through unmapped country to find the things of beauty that I am privileged to come to by well-built roads. But I trace the path of the roads myself to see if they are leading in the direction I was told.

Turning aside from towr's well-stated comments on the philosophy of science back to the foundations of mathematics, maybe tonight I can finally figure out how to say what I want before it gets too late.

What was learned from the discovery of hyperbolic geometry, and was driven home by the development of formalism was this: any mathematical theory is dependent upon its axioms and primatives. And these are NOT statements that are "self-evidently true", but simply the assumptions - the definitions upon which the theory is based. By choosing any set of primatives and axioms, you can develop a mathematical theory from them, and the only measure of the validity - the "truthfulness" of the resulting theory is whether or not it is contradictory. All non-contradictory theories are equally valid.

Is the sum of the angles of a triangle equal to 180^o? In Euclidean geometry, yes. In Lobatchevskian geometry, no. Without reference to the geometry in use, the question is meaningless. Usually, that reference is implicit, but it still must be there for the question to even make sense.

If I say "the sum of the angles of a triangle is 180^o", it should now be evident that what I am really saying is "if the axioms of Euclidean geometry are used, then the sum of the angles of a triangle is 180^o". Which is true.

If I say "the sum of the angles of a triangle is less than 180^o", what I am really saying is "if the axioms of Hyperbolic geometry are used, then the sum of the angles of a triangle is less than 180^o". Which is also true.

Since, when we first teach students (or whatever is was that towr calls them) about geometry, we do not want to muddy the waters with needless complications, we teach them Euclidean geometry. Everything we say is within the context of Euclidean geometry. So when we say that the sum of the angles of a triangle is 180^o, what we have said is wholly true.

And when we say that V+F = E+2, what we say is also true, provided we didn't pretend it holds for more than can be shown.

It follows that there is no such thing as an "unprovable axiom". Axioms are where proofs start. Within the theory all axioms are true by definition. (If the theory is contradictory, they are also all false. But such theories are worthless.)

Your charge of circular logic is one that needs to be backed up. Any proof involving circular logic is invalid, and mathematicians go to great lengths to avoid this. Search carefully, and if you can find an example, bring it to light. But if you do, I suspect further investigation will show that the result can and has been proven without the use of results dependent on it.

I am surprised any one would think the continuum hypothesis "self-evident". It is rather surprising that it apparently independent of the standard axioms of set theory, but that is because it involves the real numbers, a set with very strong structure built into it. The idea that even with all that structure, one can either assume to be true or to be false that there exists sets strictly larger than the natural numbers, but strictly smaller than the full reals, is hard to believe.

However, there is nothing "self-evident" about the hypothesis itself. Without knowing its history, I would assume when first confronted with it that one could either prove it to be true, or to be false, but there is nothing to suggest to me which would be the case.
(The axiom of choice is another matter: when stated properly, it seems so obvious that it must be provable.)

If you would like to come to a better understanding of the foundations of mathematics, and the basis of proof, I would recommend that you study formalism. I personally am very fond of Bourbaki's Elements of Set Theory, though some find it very dry, and others pan his approach as cumbersome. But any well written formalism will do.

Of course you have already stated Lakatos has raised objections to formalism, and others have been critical of the formalist movement as well. I do not disagree (I can't entirely agree either because I only have a limited knowledge of their objections). That is beside the point. Formalism has limitations, but it is a wonderful taskmaster for defining the basis of mathematical theory and of proof.

For example, Bourbaki defines a proof as being a list of statements satisfying the condition that every statement in the list is either an instance of an axiom, or is a statement "B" which is preceded somewhere in the list by statements of the form "A" and "A [implies] B" (syllogism). A theorem is any statement that appears in a proof. As you can see, this clearly prevents the possibility of circular proof.

Of course, such a "proof" would be prohibitively long for real mathematicians. So he then proceeds to develop metamathematical methods of proving the existance of full proofs from less strenuous requirements. For instance, if you have a list wherein each statement is either a theorem from some pre-existing proof or a syllogism, then there exists a full proof containing all the statements in this list. This is fairly easy to justify: just replace each instance of a theorem in your list with its proof.

Bourbaki continues in this vein to validate all the "standard" practices in proofs - from proof by contradiction (he follows standard logic, so this is possible) to trans-finite induction.

By the formalist ideal, if two mathematicians are in dispute over whether or not a particular "proof" is valid, all they need do is convert it into a full proof. If this is possible, the "proof" was valid. If not, it wasn't. (In practice, even the formalists admitted that it isn't so easy.)

Kronecker & co. may have felt the irrationals to be an abomination, but while they could rightfully turn their backs on analysis and apply their energies to the parts of mathematics they liked, I heartily disagree that they could intelligently reject analysis and the infinite as false or unfit for mathematics. To do so, they would have to show that these ideas lead to actual contradictions. This they and their successors have entirely failed to do.

Quote:

but... what are we building our mathematics on? Can we really be sure that the foundations are firm?

The foundations of a mathematical theory come in three parts: the logic on which it is based, the basis of proof, and the axioms.

While other logics are possible, the standard symbolic logic of Aristotle is both simple and powerful. Other logics are studied using it, so I have no fear of it being baseless.

The concept of proof as given by Bourbaki (but which in essence was first established by Thales of Miletus) is easily justified as logically sound, being only an expression of Aristotlean logic itself. The more sophisticated concepts of proof actually used have been carefully tied back to this basic concept and have thus been justified as well. A proffered proof needs to be examined carefully to make sure that it is correctly constructed, but if so, there is no need to fear that the method itself is invalid.

This leaves the axioms. But there is no need for axioms to meet some nebulous requirement to be "true". Within their theory, all axioms are by definition "true". The complete truth stated by a theorem is not contained in the theorem statement itself, but rather in stating that the theorem follows from the axioms.

The only condition to be placed on axioms is that the theory they produce not be contradictory. That is, that is impossible in the theory to both prove a statement "A" and its negation "[lnot]A". This is the place where modern mathematics is most vulnerable to attack. Just because we know of no contradictions now does not mean there isn't one out there, waiting to pounce. The best approach we have for insuring that a theory is not contradictory is to build a model of it within another theory. But this only works if the new theory isn't contradictory either.

However, with that said, I (like most other mathematicians) am not worried. The lion has already pounced once, and it was just a matter of redefining to avoid his claws. Should he pounce again, I do not doubt that another work-around will be possible. I look forward to what we will learn from the experience. Last time was a real eye-opener in several areas, even if the cure has left us unable to do some of the things we would like to do.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Oct 3^rd, 2003, 1:52pm

I have re-read your post three times now, Icarus, to give it the respect that it deserves.

I quite agree with most of what you say. I do, however, still have some reservations.

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My abilities of reasoning and discernment were indeed trained into me. But on the other hand, I kept at that training because I could follow the sense of it myself.

One of my concerns is that our ability to judge the methods of reasoning we are taught is already too entrenched in earlier learned methods. In other words, when we reach the stage where we are able to critically consider the methods we use to critically consider ideas, we can only use those taught methods. An eye can only, at best, see a mere reflection of itself.

Let me explain... I accept that the system of logic and mathematics is consistent; what we can ever know within that system is another question. Our axioms are simple, self-evident truths, created by earlier intelligent generations. A young child, because of their lack of mathematical experience, is not able to make a well informed judgement about whether or not we have truly selected the optimum axioms, and ones which will lead us to the best generalisations. By the time we reach a point where we can make a truly intelligent judgement about their optimality, our world has been shaped by them; it is all we know.

I think that William Wordsworth summed it up very well.

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Mathematics is an imaginary world created out of pure intelligence.

As I said, I take great pleasure in 'doing' mathematics. Using the definitions, rules, and theorems, to manipulate ideas and solve problems. However, I would still contest that we are doing nothing more than building towers with the toy bricks we have been presented with. Hence my rhetorical question, "What are we actually proving?"

Title: Re: cos(1 deg) = irrational
Post by Icarus on Oct 3^rd, 2003, 5:46pm

on 10/03/03 at 13:52:50, Sir Col wrote:

Our axioms are simple, self-evident truths, created by earlier intelligent generations.

Apparently, my wordiness has utterly obscured my point, if you still think this after three readings.

Our axioms are not, nor need to be, "simple self-evident truths". Nor do they have to be those handed down from earlier generations. In the last ~125 years mathematics has moved away from the concept that its theories are expressions of "truth" -at least directly.

The truths of mathematics are no longer such things as "vertical angles are equal", but rather "Under the rules of syllogism, the axioms of Euclidean geometry imply that vertical angles are equal." (Yes, the vertical angle theorem holds in other geometries, but that is another truth.)

Any statement may be taken as the axiom of a mathematical theory. In that theory it is true. In other theories, it is false. In some it is both - but these theories are worthless and discarded.

So while mathematics still is engaged in a hunt for truth, it is not the theorems that are the real truth we find, but rather their dependence on the axioms.

This is not "handed down". We can can consider: what if it took 3 points to determine any line? Suppose our geometry possesses an asymmetry, so that vertical angles in one direction are larger than their opposite? Such ideas would have been scoffed at as preposterous not long ago. Now the only reason not to study them is simply that the results may not be worth the while.

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However, I would still contest that we are doing nothing more than building towers with the toy bricks we have been presented with. Hence my rhetorical question, "What are we actually proving?"

I suggest you look more carefully at these "toys". They appear to be stronger than titanium with perfectly tight non-slip interlocking edges. You may call them toys, but I would trust any building built with them by a contractor who uses them with care.

And I believe your rhetorical question has been non-rhetorically answered quite well.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Oct 4^th, 2003, 9:04am

on 10/03/03 at 17:46:12, Icarus wrote:

Apparently, my wordiness has utterly obscured my point, if you still think this after three readings.

Icarus, I deeply respect your expertise. In fact, and I hope you don't find this embarrassing, you are one of the most experienced mathematicians I have ever met – albeit via the internet. Apart from the internet, and books and journals I read, I have no face-to-face contact with passionate mathematicians. However, much of what we have been discussing lies outside of the objective security of mathematics. We have been philosophising concepts of truth, both axiomatic and intuititive. Consequently, it is likely that we will differ on some counts.

For example, two fundamental questions that challenge those of us serious about what we study are, (i) do numbers exist, or are they creation of man? (ii) does the continuum exist?

I'm sure that both of us, like most mathematicians, agree that numbers exist independently of man. Question (ii), is where many mathematicians begin to disagree. No one can prove the existence of the continuum, or the infinite. You can define it, but that is not the same. Physicists still hypothesise the divisibility of space ad infinitum; confer tiny strings and Planck unit.

The work of Gödel and Cohen, especially, has demonstrated that the starting point in mathematics (axiomatic truth), will never be sufficient to prove everything.

There is little doubt that the infinite, in analysis, produces results that are too accurate to be ignored, but no one will ever prove the existence of the infinite.

Kroncker et alii, for all their disgraceful treatment of Cantor, should be respected as mathematicians for starting with the axiom: "God made the integers, all the rest is the work of man."

The question that motivated Gödel throughout his life was, "What is proof?" He recognised the importance of this question, he showed the world that the answer will stupify.

Title: Re: cos(1 deg) = irrational
Post by Icarus on Oct 4^th, 2003, 7:06pm

Here is definitely where we part ways: What exactly do you mean by "exist"? Do you think that these intellectual concepts have some sort of nebulous existance beyond the conception of those who think of them? (I include not only man, but animals and any other thinking beings who may exist.) If so, please point out to me where "one" may be found.

Do numbers exist? I say yes, but not apart from the conception of those who think of them. Does the continuum exist? Of course it does - in the exact same way! There is only one test for existance for any constructed mathematical object: The theory in which it appears must not be contradictory. If this holds, then it is foolishness to deny its existance! Within that theory, it exists. And as long as that theory is non-contradictory, the theory has as much truth and "vitality" as any other.

Kronecker et al. had every right to avoid the infinite in their mathematics. But they were fools in demanding others should avoid it as well. "Infinity" (your pick as to which) has every bit as much as existance as "one".

I did not need Gödel to tell me mathematics cannot prove everything. It is obvious that mathematics can only prove things about mathematics. Thus any great and eternal truths of the Universe must come from another source. Alas that none exists with absolute certainty (for those of a religious bent (everyone is, but some deny that their religion is a religion), this uncertainty is rooted in the human condition).

Title: Re: cos(1 deg) = irrational
Post by Barukh on Oct 5^th, 2003, 8:09am

on 10/04/03 at 19:06:28, Icarus wrote:

Kronecker et al. had every right to avoid the infinite in their mathematics. But they were fools in demanding others should avoid it as well...

Wow, it's too much, I think. Although I generally agree with Icarus (the content, not the way), I wouldn't call Kronecker et al fools in any respect. Maybe, they were not too far-sighted and did not anticipated that mathematical foundations would not be univeral anymore (how many axiomatic set theories there exist today?).

Let me also remind that Hilbert - who was the founder of the Formalists school - for the proof of the consistency of arithmetics (the central part of his famous programme) suggested to use the "finitary" methods that were just the methods of Intuitionsits, whose forerunner was Kronecker.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Oct 5^th, 2003, 8:35am

It's funny that you should mention Hilbert, as I was thinking about metamathematics and his system of absolute proof of consistency (through finitistic methods), when I mentioned Kronecker. However, like so many things in mathematics, I don't know enough about it to speak with any authority.

I hear, and have certainly been moved on in my thinking by the persuasive arguments, but I still remain quite befuddled with all of this. I think my fundamental objection is the demand for consistency even at the cost of detachment from meaningful concepts.

As my level of mathematics is unlikely to ever go beyond recreational/elementary/applied fields, I am probably never going to fully understand abstract higher mathematics.

Title: Re: cos(1 deg) = irrational
Post by Icarus on Oct 5^th, 2003, 1:53pm

I make no aspersions on their intelligence. And I am not sure to what extent they were actually guilty of attempting to prevent other mathematicians from working in areas they disparaged. But to whatever extent that was, I would call them foolish. Wisdom and Intelligence are not the same thing.

I do not know if Kronecker and his disciples ever engaged in this sort of a practice. I do know that this foolishness has reared its ugly head before and since. Nicholas Bourbaki was not an actual mathematician, but rather the pen name used by a group of young french mathematicians to publish mathematical results that would have resulted in repression had they been published in their own names, because the mathematical "elite" in France at the time did not feel these were fit areas for research.

The appellation of "foolish" is not an indictment against their mathematics, but rather against their personal behavior.

Consistency is not an external demand in mathematics. It is a requirement fundamental to all we do. To be inconsistent is to be completely meaningless for the mathematician, because anything can be proved from a contradiction.

On the other hand, just because we are able to create theories on any consistent foundation, even when the foundation of one theory may be directly in opposition to the foundation of another, does not mean that these concepts we introduce are meaningless. Rather, these theories allow us to explore new worlds of possible meanings.

Euclidean geometry was the valley we grew up in. We wandered it and became intimately familiar with all of its glades and dells. But we kept exploring its boundaries until one day we came to the top of a rise. Lo & behold! Beyond the rise lay the new valley of hyperbolic geometry, very similar to the one we knew, but with distinct differences as well. And so we proceeded to explore it.

But its existance and our new knowledge of it does not effect in any way the original valley. Euclidean geometry is still there. Still the same strong useful theory it has ever been. The only change is now we know that it isn't everything there is.

And by continuing to explore this metaphorical world, we discover that there are other areas of it that look nothing like our comfortable little valley: barren deserts, frozen wastelands, majestic mountains, wide oceans. Because these are like nothing we have seen before does not mean that they should be rejected, or that they are worthless. To abandon my burgeoning metaphor for a simple simile, I live in a region which 150 years ago was called "the great American desert". As explorers of European descent first came to the region, they noticed that it got significantly less rainfall than did lands in Europe or the eastern parts of North America. The flora mostly consisted of tough drought-resistant grasses. It was quickly written off as unfit for habitation. Today, of course, it is known as the most productive farmland anywhere in the world.

The point is, just because we invent theories that have no grounding in our current experience does not mean that these theories are meaningless. It merely means that we must explore them before we can comprehend their meaning.

Title: Re: cos(1 deg) = irrational
Post by Sir Col on Oct 5^th, 2003, 2:52pm

on 10/05/03 at 13:53:35, Icarus wrote:

I would add the word, evil, to their description. The extent, rather, the depth to which Leopold Kronecker, and his colleagues, sunk to stop free thinkers is abominable. After attempting to destroy Karl Weierstrass's career, and failing due to his strength of character, Kronecker turned all his destructive energies on the life of the more vulnerable Georg Cantor. He ensured that Cantor would never achieve his lifelong dream of joining the teaching staff at the University of Berlin, he rubbished Cantor's reputation and thwarted every opportunity to publish in respected journals, and continued to undermine the brave mathematician who was seeking to help mathematicians to move into new exciting territories.

I could strongly recommend a book, called The Mystery of Aleph, by Amir D. Aczel. It is partly a biographical account of the life of Cantor and his works, but equally an historical tour of the concepts of actual infinity (as opposed to potential) leading to the continuum problem and beyond.

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The point is, just because we invent theories that have no grounding in our current experience does not mean that these theories are meaningless. It merely means that we must explore them before we can comprehend their meaning.

I know what you're saying. Before non-Euclidean geometries became established, no one really suspected, or could be sure, that they related to real space. It was a purely abstract adventure, and it taught us a valuable lesson.

I'm almost there, Icarus. I am beginning to see/appreciate the benefits of abstract exploration. Actually I'm reading a super book at the moment, called Gödel's Proof, by James R. Newman and Ernest Negel (with a new forward by Douglas Hofstadter). Coincindentally, it has taken me amazingly close to everything that has been discussed here.

Title: Re: cos(1 deg) = irrational
Post by Icarus on Oct 5^th, 2003, 8:26pm

on 10/05/03 at 14:52:01, Sir Col wrote:

I would add the word, evil, to their description...

Yes, I agree. Both foolhardy and evil.

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I'm almost there, Icarus. I am beginning to see/appreciate the benefits of abstract exploration.

The important thing is to understand the position. Then even if you still disagree, you will be disagreeing on the basis of what you know, not on the basis of not following the argument.

Title: Re: cos(1 deg) = irrational
Post by david martinez on Dec 15^th, 2004, 6:42pm

this result is a consequence of the irreducibility of the n-th cyclotomic polynomial

Title: Re: cos(1 deg) = irrational
Post by Icarus on Dec 16^th, 2004, 6:02pm

True, but when you replace the jargon with the actual proof, this is a rather difficult way to go about it. The proofs offered by Sir Col and I in the first & second replies are far more elementary.