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Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter5: Rings, Integral Domains, And Fields
Section5.4: Ordered Integral Domains
Problem 16E: If x and y are positive rational numbers, prove that there exists a positive integer n such that...
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What is r

Prove the following statement by contradiction. If a and b are rational numbers, b # 0, and r is an irrational number, then a + br is irrational. Proof by contradiction: Select an appropriate statement to start the proof. O suppose not. That is, suppose there exist irrational numbers a and b such thatb + 0, r is a rational number, and a + br is rational. O Suppose not. That is, suppose there exist irrational numbers a and b such thatb + 0, r is an irrational number, and a + br is rational. O Suppose not. That is, suppose there exist rational numbers a andb such that b + 0, r is an irrational number, and a + br is irrational. O Suppose not. That is, suppose there exist rational numbers a andb such that b + 0, r is an irrational number, and a + br is rational. O Suppose not. That is, suppose there exist rational numbers a andb such that b + 0, r is a rational number, and a + br is irrational. Then by definition of rational, a = b = and a + br = where c, d, i, j, m, and n are integers and d # 0,j# 0, and n # 0 v Since b # 0, we also have that i + 0. By substitution, + Solving this equation for r and representing the result as a single quotient in terms of c, d, i, j, m, andn gives that r = Note that r is a ratio of two integers because products and differences of integers are integers. Also the denominator of r 0. Therefore, by definition of rational, it follows thatris rational , which contradicts the supposition. Hence the supposition is false and the given statement is true. Need Help? Read It
Transcribed Image Text:Prove the following statement by contradiction. If a and b are rational numbers, b # 0, and r is an irrational number, then a + br is irrational. Proof by contradiction: Select an appropriate statement to start the proof. O suppose not. That is, suppose there exist irrational numbers a and b such thatb + 0, r is a rational number, and a + br is rational. O Suppose not. That is, suppose there exist irrational numbers a and b such thatb + 0, r is an irrational number, and a + br is rational. O Suppose not. That is, suppose there exist rational numbers a andb such that b + 0, r is an irrational number, and a + br is irrational. O Suppose not. That is, suppose there exist rational numbers a andb such that b + 0, r is an irrational number, and a + br is rational. O Suppose not. That is, suppose there exist rational numbers a andb such that b + 0, r is a rational number, and a + br is irrational. Then by definition of rational, a = b = and a + br = where c, d, i, j, m, and n are integers and d # 0,j# 0, and n # 0 v Since b # 0, we also have that i + 0. By substitution, + Solving this equation for r and representing the result as a single quotient in terms of c, d, i, j, m, andn gives that r = Note that r is a ratio of two integers because products and differences of integers are integers. Also the denominator of r 0. Therefore, by definition of rational, it follows thatris rational , which contradicts the supposition. Hence the supposition is false and the given statement is true. Need Help? Read It
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