In this exercise, you want to show that a relation RC A? is transitive + R" C R for all n EN= {1,2,3,...}. Here is a step-by-step guide: %3D

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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3.
In this exercise, you want to show that a relation RC A?
is transitive + R" C R for all n E N = {1,2, 3, .. .}. Here is a step-by-step guide:
(a) We are going to establish the direction first. We accomplish this by induction on
n. For the base case, we have n
1, but R'
= RCRdoes not even depend on
Transcribed Image Text:3. In this exercise, you want to show that a relation RC A? is transitive + R" C R for all n E N = {1,2, 3, .. .}. Here is a step-by-step guide: (a) We are going to establish the direction first. We accomplish this by induction on n. For the base case, we have n 1, but R' = RCRdoes not even depend on
whether R is transitive or not. Thus, the base case holds vacuously. For the inductive
case, suppose that RK C Runder the assumption that R is transitive. We want to show
that Rk+1 is also a subset or R. To do this, we want to show that if (a, c) E Rk+1, then
(a, c) E R also-call this property A. Thus, suppose that (a, c) E Rk+1 = Rk o R. This
means that there exists b e A such that (a, b) E R and (b, c) E R*. Explain briefly
why you can conclude that (a, c) e R also, hence showing that A holds. (Hint: You
are using induction for the proof.)
(b) We have established the direction above; we now want to establish the converse
direction . This direction is, in fact, very easy to establish. For, let (a, b), (b, c) E R.
Conclude that (a, c) E R from the assumption that R" CR for all n E N. (Hint: Use
Universal Instantiation on “R" C R for all n E N," specifying the precise value of n
that makes the proof work.)
Transcribed Image Text:whether R is transitive or not. Thus, the base case holds vacuously. For the inductive case, suppose that RK C Runder the assumption that R is transitive. We want to show that Rk+1 is also a subset or R. To do this, we want to show that if (a, c) E Rk+1, then (a, c) E R also-call this property A. Thus, suppose that (a, c) E Rk+1 = Rk o R. This means that there exists b e A such that (a, b) E R and (b, c) E R*. Explain briefly why you can conclude that (a, c) e R also, hence showing that A holds. (Hint: You are using induction for the proof.) (b) We have established the direction above; we now want to establish the converse direction . This direction is, in fact, very easy to establish. For, let (a, b), (b, c) E R. Conclude that (a, c) E R from the assumption that R" CR for all n E N. (Hint: Use Universal Instantiation on “R" C R for all n E N," specifying the precise value of n that makes the proof work.)
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