Prove the following statement by contradiction. If a and b are rational numbers, b 0, and ris 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 that b O Suppose not. That is, suppose there exist rational numbers a and b 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 and b such that b O Suppose not. That is, suppose there exist rational numbers a and b such that b + 0, ris an irrational number, and a + br is rational. O Suppose not. That suppose there exist irrational numbers a and b such that b 0, ris an irrational number, and a + bris rational. Then by definition of rational, a-b-, and a + br where c, d, i, j, m, and n are ---Select--- ✓and-Select-- 0, r is a rational number, and a + br is rational. 0, r is a rational number, and a + br is irrational. ✓. Since b + 0, we also have that / * 0. Solving this equation for r and representing the result as a single quotient in terms of c, d, i, j, m, and n gives that substitution, Note that r-Select--a ratio of two integers because products and differences of integers --Select- integers. Also the denominator of r?0. Therefore, by definition of rational, it follows that r is-Select-, which contradicts the supposition. Hence the supposition is false and the given statement is true.
Prove the following statement by contradiction. If a and b are rational numbers, b 0, and ris 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 that b O Suppose not. That is, suppose there exist rational numbers a and b 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 and b such that b O Suppose not. That is, suppose there exist rational numbers a and b such that b + 0, ris an irrational number, and a + br is rational. O Suppose not. That suppose there exist irrational numbers a and b such that b 0, ris an irrational number, and a + bris rational. Then by definition of rational, a-b-, and a + br where c, d, i, j, m, and n are ---Select--- ✓and-Select-- 0, r is a rational number, and a + br is rational. 0, r is a rational number, and a + br is irrational. ✓. Since b + 0, we also have that / * 0. Solving this equation for r and representing the result as a single quotient in terms of c, d, i, j, m, and n gives that substitution, Note that r-Select--a ratio of two integers because products and differences of integers --Select- integers. Also the denominator of r?0. Therefore, by definition of rational, it follows that r is-Select-, which contradicts the supposition. Hence the supposition is false and the given statement is true.
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter10: Inequalities
Section10.2: Solving Inequalities
Problem 65WE
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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 that b = 0, r is a rational number, and a + br is rational.
O Suppose not. That is, suppose there exist rational numbers a and b such that b = 0, r is a rational number, and a + br is irrational.
Suppose not. That is, suppose there exist rational numbers a and b 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 and b such that b = 0, r is an irrational number, and a + br is rational.
O Suppose not. That is, suppose there exist irrational numbers a and b such that b + 0, r is an irrational number, and a + br is rational.
Then by definition of rational,
i
7'
where c, d, i, j, m, and n are ---Select---
a =
с
등
b =
C
= + 1/1/0
d
_r=
m
n
and a + br =
m
n
and ---Select---
. 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, and n gives that
Note that r ---Select--- a ratio of two integers because products and differences of integers
statement is true.
-Select---
integers. Also the denominator of r?
0. Therefore, by definition of rational, it follows that r is ---Select---, which contradicts the supposition. Hence the supposition is false and the given
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