Algebraic Numbers

By Andrew Hodges (P4403) on Saturday, June 09, 2001 - 06:03 pm:

Can all sines, cosines and tangents be expressed in surd form or simply as just rational numbers, like sin30, sin60, sin45 and sin54?

By David Loeffler (P865) on Saturday, June 09, 2001 - 10:47 pm:

It depends on what you mean by "all".

It is possible to divide all numbers into two classes. Algebraic numbers are nice numbers like 1/3 or 2^1/2 or (1+sqrt(5))^1/72 which are the roots of polynomials (equations like x⁵ + x³ - 7x + 8 = 0).
Anything you can get to in a finite number of steps from integers by multiplying, dividing, adding, subtracting and taking nth roots (where n is an integer; 2^sqrt(2) is not algebraic) will be algebraic.

Transcendental numbers are all other numbers. For example,

p and e are both transcendental.

In answer to your question, it has been proved that the sine of 1 radian is transcendental. (If you don't know about radians, they're a unit of angles. 1 radian is equal to 180/p degrees.) So not all sines are algebraic.

However, any angle which is a rational number of degrees is the root of some (probably very nasty) polynomial, since by De Moivre's theorem we can express cos nx as a polynomial in cos x for any integer n. Hence
if x = p/q (in degrees), we can set n=360q, so cos (360q) = 1 is a polynomial in cos (p/q). So cos p/q is a root of some polynomial with integer coefficients. And if cos x is algbraic, so is sin x.

(I've covered quite a lot of ground here, so if you don't know about De Moivre's theorem or anything else I've mentioned feel free to say so.)

By Andrew Hodges (P4403) on Sunday, June 10, 2001 - 11:09 am:

Yes, I've covered De Moivre's theorem - the polynomial you derive would be the real part of the expansion of (cos x + isin x)ⁿ wouldn't it? Wouldn't it be easier to use radians when using De Moivre's theorem, expressing angles as

p/x, as Euler's theorem e^ix = cos x + isin x works with radians as the natural angle to use? Thus, any angle which is p multiplied by (a rational number of degrees) would result in a polynomial with integer coefficients. What about angles with not transcendental but merely irrational values, e.g. sin(2^1/2 ). Can these be expressed as the root of a polynomial? Could this polynomial have irrational coefficients instead of integer ones?

p is also transcendental, so can the sine of 1 radian be expressed in terms of p?

By David Loeffler (P865) on Sunday, June 10, 2001 - 11:00 pm:

Yes, you're right, it probably makes more sense in radians :-)

As for sin (sqrt(2)) I am not sure. Perhaps it is algebraic, but I think it is very unlikely.

You see, algebraic numbers are really quite rare. It can be shown that there are many, many more transcendental numbers than algebraic ones. In fact, if you pick a number between 0 and 1 entirely at random the probability that it is algebraic is 0. So it doesn't look very likely that sin(sqrt(2)) is algebraic, or that sin 1 is expressible in terms of pi.

As for polynomials with non-integer coefficients, you have to be quite careful because p, for example, satisfies the vacuous equation x-p = 0.

If you had algebraic coefficients, then it can (I think) be shown that it wouldn't yield any polynomials with solutions which weren't algebraic anyway. To prove this seems to be very difficult, but I will try to give you some idea of what's involved. (Sorry to drag this discussion so far off its original topic.)

(This is getting close to the boundaries of my knowledge anyway, so if anybody better-informed reads this and thinks I'm talking rubbish please say so.)

If you take an algebraic number a, it will possess a number of algebraic conjugates. These are the othe roots of the polynomial of lowest degree having integer coefficients and a as a root.
This polynomial is then of the form
(x-a₁)(x-a₂)(x-a₃)¼ = 0.

If we expand out these brackets, we will get something like
xⁿ-(a₁+a₂+a₃¼)x^n-1+(a₁a₂+a₁ a₃¼+a₂a₃¼)x^n-2¼+-a₁a₂a₃

Now these must be the same as the coefficients of the original polynomial (divided through by the leading term) so they are all rational.

(This is why they are called conjugates; the situation is analogous to the complex conjugate, which is unique complex number y for a given x such that x+y and xy are both real.)

Right. Consider a general polynomial a₁ + b₁ x + c₁ x² + d₁ x³ ...
where the coefficients are algebraic. Each of these coefficients will have algebraic conjugates. So we can write out a whole list of polynomials where we take every possible combination of algebraic conjugates of the coefficients.

For example, supppose we start with x² + sqrt(2)x + sqrt(3). Each of sqrt(2) and sqrt(3) has a single algebraic conjugate, -sqrt(2) and -sqrt(3) respectively. So our other polynomials are x² - sqrt(2)x + sqrt(3), x² + sqrt(2)x - sqrt(3) and x² - sqrt(2)x - sqrt(3).

Now try multiplying these four polynomials together. What do you notice?

I'll save you some leg work: the answer is 4-12x² +5x⁴ -6x⁶ +x⁸ . This has x² + sqrt(2)x + sqrt(3) as a factor, so any roots of the latter are roots of the former. And it has integer coefficients. So the roots of this polynomial are all algebraic.

(You could have found that out a lot quicker by using the quadratic formula, but what if the original polynomial was x²⁰ + sqrt(2)x + sqrt(3)=0?)

You can probably guess where this is going now: if you write out a massive list of all the polynomials obtained from your original one by conjugation and multiply them all together, you obtain a polynomial with rational coefficients. The reason the coefficients are rational is that they consist of sums of products of conjugates exactly like those in the (x-a₁)... polynomial above, which are known to be rational.

So any root of a polynomial with algebraic coefficients is algebraic, which is the result we want.

By Michael Doré (Michael) on Monday, June 11, 2001 - 12:46 am:

Just quickly on sin(Ö2) - there is a useful result called Lindemann's theorem which says that if a₁, a₂, ..., a_n are distinct algebraic numbers and b₁, b₂, ..., b_n are non-zero algebraic numbers then:

b₁exp(a₁)+b₂exp(a₂)+¼+b_nexp(a_n) ¹ 0

Now imagine that sin(Ö2) is algebraic and a root of P(x)=a₀+a₁ x+¼+a_r x^r where a_i are integral and a_r ¹ 0. Then by Euler's formula we have:

a₀+a₁(exp(iÖ2) -exp(-iÖ2))/(2i)+¼+a_r((exp(iÖ2) -exp(-iÖ2))/(2i))^r=0

When you expand this out you see there exist integral constants b_-r, b_r-1, ..., b_r so that:

b_-rexp(-r iÖ2) +b_-(r-1)exp(-(r-1)i Ö2)+¼+b_r exp(r iÖ2)=0

But b_r ¹ 0 so Lindemann's theorem is violated which is a contradiction therefore sin(Ö2) is transcendental.

By David Loeffler (P865) on Monday, June 11, 2001 - 09:14 pm:

Ah. Thanks Michael. I always wondered exactly how it was proved that sin 1 was transcendental. I suppose that this extends immediately to sin a where a is any algebraic number.

What about sin(p/Ö2)? Now that looks quite hard!

By Michael Doré (Michael) on Monday, June 11, 2001 - 10:29 pm:

Yes, you're right we need Gelfond-Schneider for that one. Gelfond-Schneider says:

If a, b are algebraicwith a not equal to 0, 1 and b irrational then a^b is transcendental.

Now if sin(p/Ö2) is algebraic then since 2isin(p/Ö2)=exp(pi/Ö2)-exp(pi/Ö2)^-1 you quickly find that exp(pi/Ö2) is algebraic. Hence by Gelfond-Schneider, exp(pi/Ö2)^Ö2 is transcendental, i.e. exp(pi)=-1 is transcendental which is a contradiction establishing the result.