A complex number x0 C is called algebraic if there is a nonz

A complex number x0 C is called algebraic if there is a non-zero polynomial f(x) with coefficients in Q such that f(x0) = 0.

(1) Prove that every rational number is algebraic.

(2) Prove that 2 is algebraic. (It can be shown that 2 is not rational.)

Solution

Solution :

( 1 )

Let q be any rational number. q=a/b, where a and b are integers.

bq - a = 0

bx - a is the nonzero polynomial p(x) with integer coefficients such that q is a root of p(x) = 0.

So, every rational number is an algebraic number.

( 2 )

Let\'s suppose 2 is a rational number. Then we can write it 2 = a/b where a, b are whole numbers, b not zero.

We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.

From the equality 2 = a/b it follows that 2 = a²/b², or a² = 2 · b². So the square of a is an even number since it is two times something.

From this we know that a itself is also an even number. Why? Because it can\'t be odd; if a itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don\'t believe me!

Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don\'t need to know what k is; it won\'t matter. Soon comes the contradiction.

If we substitute a = 2k into the original equation 2 = a²/b², this is what we get:

This means that b² is even, from which follows again that b itself is even. And that is a contradiction!!!

WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that 2 is rational) is not correct. Therefore 2 is not rational.