__A supplementary note on the irrationality of trigonometric functions__

Recall that rational numbers can be expressed in forms of simpliest ratio between two integers, in which the function involved is division (related to multiplication which is the inverse of division), which implies that rational numbers form a group under multiplication. However irrational numbers can't form a group ifself.

Denote a polynomial if and only if . For example, for , . We call a polynomial a polynomial with

**integral coefficient**, while calling a polynomial a

**real polynomial**. Our analysis prevent anything beyond real polynomial otherwise it will be far difficult.

Note that these symbols have their special meaning in

*group theory*like the lattice point , but this is boyond the scope of this passage.

Let be a irrational number, i.e., and . Then we say that if , then . The inverse of this statement may NOT be true.

Proof:

Assume , then where n is rational and k is non-negative integer, so as the concerned polynomial f. Therefore .

For example, we know that is irrational since while . Simply putting into f gives the irrationality of the desired number.

However, the same arguement is not applicable on tangent function because it's angle sum formula is NOT a polynomial but a rational function.

We have . For example, we have . Although by the same arguement it works, but we shall prove the irrationality of tangent function by rewriting it into a continued fraction so that for every rational x, tan x is irrational unless it trivially is. (See Wolfram Mathworld)

An interesting fact is that the statement is not equivalent, however

**it CAN be equivalent**. It should be easy for readers to find out its equivalent case when the inverse of f is considered, so it's proof is left for readers.

*Exercise.*Find the necessary condition that is equivalent to .

We start from easier case: linear and quadratic functions.

*Theorem.*For linear , rationality of f(x) is equivalence to the rationality of x. Since rational number is a equivalence class, irrationality of f(x) also implies the irrationality of x.

*Proof.*If , then for rational m and c. Inversely which is also linear, so the rationality of x implies the rationality of f(x). For since the coefficients are rational, we can apply the same procedure to show the irratioanlity relation.

*Theorem.*For quadratic f, the statement is equivalent only when , i.e., square root of a rational number.

*Proof.*left as exercise.

Sometimes problems concerning rational numbers are easier to be solved when rational numbers are replaced by integers. Then tools like congrunences can be applied easier.

For example, the statement " is irrational" is equivalent to " is insoluble in integers" because you can assume , which is equivalent to the second statement. Similarly we can rewrite the statement " is insoluble in rationals" (The

*method of Conic paramization*) in equation over the integral domain. (Exercise)

The final aim of this passage is to proof the irrationality of more numbers.

*Theorem.*is either integer (trivial case) or irrational.

Recall that it's equivalent to the statement is soluble in integers of simpliest ratio. It follows that and for any prime factor p of b. Now so that which contradicts our assumption.

In fact this is a generalization of the second (historical) proof.

Here's another proof given by G.H. Hardy:

Assume where a,b,c are integers while with 0 < b < c. Now so that . Let , we have where 0 < d < b, which is a contradiction.

Now a interesting problem arises: why the contradictory result from "simpliest fraction" shows that the statement is false?

You may say that axioms of arithmetics are parallel, therefore if A is equivalent to B and B is proved to be false, then A is false as well.

Meanwhile we can say the lack of simpliest solution led to the infinite descent of solution set which is eventually out of our range [of concern], and therefore such case would not happen and the statement is false.

Similar arguement can be make to prove the irrationality of .

*Theorem.*is either integer or irrational.

*Proof*. Exercise.

Now we extend numbers to algebraic numbers.

*Rational Root Theorem*For a polynomial with integral coefficients, rational roots, if exist, in forms of , then p and q must be integral factor of respectively.

*Proof.*Exercise. (It's trivial when the rational number is directly subbed into the polynomial. Otherwise Gauss' Lamma can be used.

As a special case of it, we have

*Theorem.*For polynomial with integral coefficient 1, its

**real root**is either integer or irrational.

*Example.*is irrational.

Let

is the original equation with root . Quickly check that is not a root of it, and therefore this polynomial has no integral zeros, so its roots so as the concerned number, must be irrational.

It provides us a powerful test for reducibility of polynomials or to test the existance of integral roots, in which we will discuss next time including Eisenstein's irreducible criterion and more tests on irrational numbers by infinite series.

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