Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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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.
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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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