Solution to Let R be the relation on Z (the set of integers) defined by (x, y) … - Sikademy
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Archangel Macsika

Let R be the relation on Z (the set of integers) defined by (x, y)  R iff x 2 + y2 = 2k for some integers k  0. Answer questions 13 to 15 by using the given relation R. Question 13 Which one of the following is an ordered pair in R? 1. (1, 0) 2. (2, 9) 3. (3, 8) 4. (5, 7) Question 14 R is symmetric. Which one of the following is a valid proof showing that R is symmetric? 1. Let x, y  Z be given. Suppose (x, y)  R then x 2 + y2 = 2k for some k  0. ie y 2 + x2 = 2k for some k  0. thus (x, y)  R. 2. Let x, y  Z be given. Suppose (x, y)  R then x 2 + y2 = 2k for some k  0. ie y 2 + x2 = 2k for some k  0. thus (y, x)  R. 3. Let x, y  Z  be given. Suppose (x, y)  R then x 2 + y2 = 2k for some k  0. thus (y, x)  R. 4. Let x, y  Z be given. Suppose (x, x)  R then x 2 + y2 = 2k for some k  0. ie y 2 + x2 = 2k for some k  0. thus (y, y)  R.

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Question 13

Taking into account that 1^2+0^2=1, 2^2+ 9^2=85, 3^2+ 8^2=73 are odd, and 5^2+ 7^2=74=2\cdot 37 is even, we conclude that only (5,7)\in R.


Question 14

Valid proof is the following:


2. Let x, y\in\Z be given. Suppose (x, y) \in R, then x^2 + y^2 = 2k for some k \ge 0, i.e. y^2 + x^2 = 2k for some k\ge 0. Thus (y, x) \in R.


Answer: 2


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Question ID: mtid-5-stid-8-sqid-2899-qpid-1598