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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

Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00th Edition
ISBN:9780395977224
Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Chapter10: Inequalities
Section10.7: Graphing Linear Inequalities
Problem 13OE
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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 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
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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