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Use a proof by contradiction to show that there is no rational number r for which pi3 +r+1= 0.

Use a proof by contradiction to show that there is no rational number r for which pi3 +r+1= 0.

Algebra: Structure And Method, Book 1
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
Chapter9: Systems Of Linear Equations
Section9.4: The Addition-or-subtraction Method
Problem 21WE
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Use a proof by contradiction to show that there is no rational number r for which p3 +r+1 = 0. Solution: Suppose by way of contradiction that a/b is a rational root, where a and b are integers and this fraction is in lowest terms (that is, a and b have no common divisor greater than 1). Plug this proposed root into the equation to obtain +1= 0. Multiply through by b to obtain a + ab? + b = 0. If a and b are both odd, then the left-hand side is the sum of three odd mumbers and therefore must be odd. If a is odd and b is even, then the left-hand side is odd + even + even, which is again odd. Similarly, if a is even and b is odd, then the left-hand side is even + even + odd, which is again odd. Because the fraction a/b is in simplest terms, it cannot happen that both a and b are even. Thus in all cases, the left-hand side is odd, and therefore cannot equal 0. This contradiction shows that no such root exists.
Transcribed Image Text:Use a proof by contradiction to show that there is no rational number r for which p3 +r+1 = 0. Solution: Suppose by way of contradiction that a/b is a rational root, where a and b are integers and this fraction is in lowest terms (that is, a and b have no common divisor greater than 1). Plug this proposed root into the equation to obtain +1= 0. Multiply through by b to obtain a + ab? + b = 0. If a and b are both odd, then the left-hand side is the sum of three odd mumbers and therefore must be odd. If a is odd and b is even, then the left-hand side is odd + even + even, which is again odd. Similarly, if a is even and b is odd, then the left-hand side is even + even + odd, which is again odd. Because the fraction a/b is in simplest terms, it cannot happen that both a and b are even. Thus in all cases, the left-hand side is odd, and therefore cannot equal 0. This contradiction shows that no such root exists.
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