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.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc2795211-6cf9-4e3f-a656-86b2e0dc6b6f%2F4ba35f7a-f06d-4cfc-8594-9fe7529f4a92%2F5ha418_processed.jpeg&w=3840&q=75)
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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