Question #207343

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.


1
Expert's answer
2021-07-19T10:46:55-0400

Question 13

Taking into account that 12+02=1,22+92=85,32+82=731^2+0^2=1, 2^2+ 9^2=85, 3^2+ 8^2=73 are odd, and 52+72=74=2375^2+ 7^2=74=2\cdot 37 is even, we conclude that only (5,7)R.(5,7)\in R.


Question 14

Valid proof is the following:


2. Let x,yZx, y\in\Z be given. Suppose (x,y)R(x, y) \in R, then x2+y2=2kx^2 + y^2 = 2k for some k0k \ge 0, i.e. y2+x2=2ky^2 + x^2 = 2k for some k0k\ge 0. Thus (y,x)R.(y, x) \in R.


Answer: 2


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