Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) E R if and only if ad= bc. Arrange the proof of the given statement in correct order to show that R is an equivalence relation. (Prove the given relation is reflexive first, and then symmetric and transitive.) Rank the options below. Hence, R is reflexive. If ((a, b), (c, d)) E E Rand ((c, d), (e, f) E ER, then ad bc and cf=de. Hence, R is transitive. Multiplying these equations gives acdf = bcde, and since all these numbers are nonzero, we have af be, so ((a, b), (e, f)) Є E R. If ((a, b), (c, d)) € ЄR then ad= bc, which also means that cb= da, so ((c, d), (a, b)) Є Є R. Hence, R is symmetric. ((a, b), (a, b)) € € R because ab = ba. Since R is reflexive, symmetric, and transitive, it is an equivalence relation. ་ ་ Let G be a graph with v vertices and e edges. Let M be the maximum degree of the vertices of G, and let m be the minimum degree of the vertices of G. Show that 2e/v≤ M. Step 1 Step 2 The degree of each vertex is ≤ M, and there are v vertices. Hence, 2e s vM. Therefore, 2e/v ≤ M. Step 3 The sum of the degrees of the vertices must be ≤ to VM. We know that 2e is the sum of the degrees of the vertices. Step 4 The sum of the degrees of the vertices must be ≥ VM. Reset
Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) E R if and only if ad= bc. Arrange the proof of the given statement in correct order to show that R is an equivalence relation. (Prove the given relation is reflexive first, and then symmetric and transitive.) Rank the options below. Hence, R is reflexive. If ((a, b), (c, d)) E E Rand ((c, d), (e, f) E ER, then ad bc and cf=de. Hence, R is transitive. Multiplying these equations gives acdf = bcde, and since all these numbers are nonzero, we have af be, so ((a, b), (e, f)) Є E R. If ((a, b), (c, d)) € ЄR then ad= bc, which also means that cb= da, so ((c, d), (a, b)) Є Є R. Hence, R is symmetric. ((a, b), (a, b)) € € R because ab = ba. Since R is reflexive, symmetric, and transitive, it is an equivalence relation. ་ ་ Let G be a graph with v vertices and e edges. Let M be the maximum degree of the vertices of G, and let m be the minimum degree of the vertices of G. Show that 2e/v≤ M. Step 1 Step 2 The degree of each vertex is ≤ M, and there are v vertices. Hence, 2e s vM. Therefore, 2e/v ≤ M. Step 3 The sum of the degrees of the vertices must be ≤ to VM. We know that 2e is the sum of the degrees of the vertices. Step 4 The sum of the degrees of the vertices must be ≥ VM. Reset
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:Let R be the relation on the set of ordered pairs of positive integers such that ((a, b), (c, d)) E R if and only if ad= bc. Arrange the proof of the given statement in correct order to show that R is an equivalence relation.
(Prove the given relation is reflexive first, and then symmetric and transitive.)
Rank the options below.
Hence, R is reflexive.
If ((a, b), (c, d)) E E Rand ((c, d), (e, f) E ER, then ad bc and cf=de.
Hence, R is transitive.
Multiplying these equations gives acdf = bcde, and since all these numbers are nonzero, we have af be, so ((a, b), (e, f)) Є E R.
If ((a, b), (c, d)) € ЄR then ad= bc, which also means that cb= da, so ((c, d), (a, b)) Є Є R.
Hence, R is symmetric.
((a, b), (a, b)) € € R because ab = ba.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation.
་
་
Transcribed Image Text:Let G be a graph with v vertices and e edges. Let M be the maximum degree of the vertices of G, and let m be the
minimum degree of the vertices of G.
Show that 2e/v≤ M.
Step 1
Step 2
The degree of each vertex is ≤ M, and there are v
vertices.
Hence, 2e s vM.
Therefore, 2e/v ≤ M.
Step 3
The sum of the degrees of the vertices must be ≤
to VM.
We know that 2e is the sum of the degrees of the
vertices.
Step 4
The sum of the degrees of the vertices must be ≥
VM.
Reset
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