Exercise 1.4.9 (Challenging): A number x is algebraic if x is a root of a polynomial with integer coefficients, in other words, anx"+an_₁x²¹+...+a₁x+ao=0 where all an € Z. a) Show that there are only countably many algebraic numbers. b) Show that there exist non-algebraic (transcendental) numbers (follow in the footsteps of Cantor, use the uncountability of R).

Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter8: Polynomials
Section8.2: Divisibility And Greatest Common Divisor
Problem 35E
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Please solve all parts of exercise 1.4.9 with detailed explanations. Take your time. 

Exercise 1.4.9 (Challenging): A number x is algebraic if x is a root of a polynomial with integer coefficients,
in other words, anx" +an_1x²-1 +...+a₁x+ag=0 where all an € Z.
a) Show that there are only countably many algebraic numbers.
b) Show that there exist non-algebraic (transcendental) numbers (follow in the footsteps of Cantor, use the
uncountability of R).
Hint: Feel free to use the fact that a polynomial of degree n has at most n real roots.
Exercise 1.4.10 (Challenging): Let F be the set of all functions f: R→ R. Prove |R|< |F| using Cantor's
Theorem 0.3.34.*
Transcribed Image Text:Exercise 1.4.9 (Challenging): A number x is algebraic if x is a root of a polynomial with integer coefficients, in other words, anx" +an_1x²-1 +...+a₁x+ag=0 where all an € Z. a) Show that there are only countably many algebraic numbers. b) Show that there exist non-algebraic (transcendental) numbers (follow in the footsteps of Cantor, use the uncountability of R). Hint: Feel free to use the fact that a polynomial of degree n has at most n real roots. Exercise 1.4.10 (Challenging): Let F be the set of all functions f: R→ R. Prove |R|< |F| using Cantor's Theorem 0.3.34.*
Expert Solution
Step 1

A number x is said to be algebraic number of x is a root of a polynomial with integer coefficients.

In other words, anxn+an-1xn-1++a0=0, where all an.

In this solution we will prove that there are only countably many algebraic numbers.

Also we will prove that there exists non algebraic numbers.

In this solution, we will use the fact that x1+x2++xn=N has finitely many positive integer solution for all N.

Also we will use the fact that countable union of countable sets is countable.

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