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Solution (a) R = {(a,b) = (Z>0)²: Ak € Z such that a o. There can exist no real number c such that a o satisfy (a,b) € R. Then there is no k € Z such that a

Solution (a) R = {(a,b) = (Z>0)²: Ak € Z such that a o. There can exist no real number c such that a o satisfy (a,b) € R. Then there is no k € Z such that a

Advanced Engineering Mathematics
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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Are there any other ways to prove transitivity in this question Thanks :)
Solution (a) R = {(a,b) = (Z>0)²: Ak € Z such that a <kr < b or b < kn <a}. (b) We must prove reflexivity, symmetry, and transitivity. Reflexivity. Let a E Z>o. There can exist no real number c such that a <c<a (whether or not c is a multiple of л). So (a, a) E R. Therefore R is reflexive. Symmetry. Suppose a, b € Z>o satisfy (a,b) E R. Then there is no k € Z such that a<kл<bor b<kл <a. But, swapping that around to b<kл<a or a <kл <b, this is exactly the same condition as (b, a) E R. So R is symmetric. Transitivity. I will prove the contrapositive. Let a, b, c be natural numbers such that (a, c) R. We wish to prove that (a, b) & R or (b, c) & R. I By assumption, there exists an integer k such that a < kл < с or с <kл <a. Without loss of generality, a < kл < с. Because л is irrational, kл is not an integer, so it doesn't equal Therefore it is either less than or greater than b. But if kл <b, then a <kл<b so (a,b) & R, whereas if b < kë, then b <kл < c so (b,c) R. This proves the contrapositive.
Transcribed Image Text:Solution (a) R = {(a,b) = (Z>0)²: Ak € Z such that a <kr < b or b < kn <a}. (b) We must prove reflexivity, symmetry, and transitivity. Reflexivity. Let a E Z>o. There can exist no real number c such that a <c<a (whether or not c is a multiple of л). So (a, a) E R. Therefore R is reflexive. Symmetry. Suppose a, b € Z>o satisfy (a,b) E R. Then there is no k € Z such that a<kл<bor b<kл <a. But, swapping that around to b<kл<a or a <kл <b, this is exactly the same condition as (b, a) E R. So R is symmetric. Transitivity. I will prove the contrapositive. Let a, b, c be natural numbers such that (a, c) R. We wish to prove that (a, b) & R or (b, c) & R. I By assumption, there exists an integer k such that a < kл < с or с <kл <a. Without loss of generality, a < kл < с. Because л is irrational, kл is not an integer, so it doesn't equal Therefore it is either less than or greater than b. But if kл <b, then a <kл<b so (a,b) & R, whereas if b < kë, then b <kл < c so (b,c) R. This proves the contrapositive.
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