Explanation Given information: Define a relation Q on R as follows: For all real numbers x and y, x Q y ⇔ x - y is rational. Calculation: Let us consider the set below. Q = { x , y ∈ R | x − y is rational } Reflexive: The relation Q is reflexive if ( a , a ) ∈ Q for every element a ∈ R . We note that Q contains ( x , x ) because x − x = 0 and 0 is rational (since 0 = 0 1 ) , thus Q is reflexive. x R x It is reflexive Symmetric: The relation Q on a set R Is symmetric if ( b , a ) ∈ Q whenever ( a , b ) ∈ Q . Let us assume that ( a , b ) ∈ Q . By definition of Q : a − b is rational By the definition of rational, there exists integers c and d such that: a − b = c d Multiply each side by − 1: − ( a − b ) = − c d Distributive property: b − a = − c d This last inequality then implies that b − a is irrational since − c d is rational (as − c is an integer and d is an integer), which then implies ( b , a ) ∈ Q and thus Q is symmetric

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Chapter 8.2, Problem 18ES

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To justify whether the given relation is reflexive, symmetric, transitive, or none of these.

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Chapter 8.2, Problem 17ES

Chapter 8.2, Problem 19ES

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Chapter 8 Solutions

WEBASSIGN F/EPPS DISCRETE MATHEMATICS

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To justify whether the given relation is reflexive, symmetric, transitive, or none of these.

Solution Summary: The author explains that the relation Q on a set R is reflexive if (a,b,)in Q for every element.

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To justify whether the given relation is reflexive, symmetric, transitive, or none of these.

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Chapter 8 Solutions