Theorem: There is no rational number x such that 1 < x < 2. Proof: We use proof by contradiction. So suppose there are rational numbers between 1 and 2, and let x be the smallest one. Because x is rational, we can write x = where a and b are integers. Now let y = *1. Then y = , so y is rational. a+b %3D The assumption that x > 1 gives x +1 > 2, which implies that y > 1. The assumption that x > 1 also gives 2x > x +1, which implies that y < x. But now y is a rational number between 1 and 2 which is smaller than x. This contradicts the assumption that x was the smallest one. So the theorem is true using proof by contradiction. D.
Theorem: There is no rational number x such that 1 < x < 2. Proof: We use proof by contradiction. So suppose there are rational numbers between 1 and 2, and let x be the smallest one. Because x is rational, we can write x = where a and b are integers. Now let y = *1. Then y = , so y is rational. a+b %3D The assumption that x > 1 gives x +1 > 2, which implies that y > 1. The assumption that x > 1 also gives 2x > x +1, which implies that y < x. But now y is a rational number between 1 and 2 which is smaller than x. This contradicts the assumption that x was the smallest one. So the theorem is true using proof by contradiction. D.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:lere's an incorrect theorem with an incorrect proof.
Theorem: There is no rational number x such that 1 < x < 2.
Proof: We use proof by contradiction. So suppose there are rational numbers between 1 and 2, and let x be the smallest one. Because x is
rational, we can write x = where a and b are integers. Now let y = 1. Then y = ", so y is rational.
%3D
a+b
%3D
The assumption that x > 1 gives x +1 > 2, which implies that y > 1. The assumption that x > 1 also gives 2x > x +1, which implies that
y < x.
But now y is a rational number between 1 and 2 which is smaller than x. This contradicts the assumption that x was the smallest one. So the
theorem is true using proof by contradiction. D.
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