Question #207345

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.

Question 15

R is not antisymmetric. Which of the following ordered pairs can be used together in a 

counterexample to prove that R is not antisymmetric? (Remember that R is defined on Z

.)

1. (–1, 1) & (1, –1)

2. (5, 9) & (13, 15)

3. (8, 7) & (7, 8)

4. (3, 1) & (1, 3)


1
Expert's answer
2021-06-17T18:28:31-0400

Given Relation is-

R={(x,y);x2+y2=2k(x,y);x^2+y^2=2k } for some integers k.


1.(-1,1)&(1,-1)


(1)2+(1)2=2k    2=2k    k=1(-1)^2+(1)^2=2k\implies 2=2k\implies k=1


and (1,1)    (1)2+(1)2=2k    2=2k    k=1(1,-1)\implies (1)^2+(-1)^2=2k\implies 2=2k\implies k=1

So R is not antisymmetric.


2.(5,9)&(13,15)

(5,9)    (5)2+(9)2=2k    25+81=2k    k=68(5,9)\implies (5)^2+(9)^2=2k\implies 25+81=2k\implies k= 68


For (13,15)    (13)2+(15)2=2k    169+225=2k    k=19713,15) \implies (13)^2+(15)^2=2k\implies 169+225=2k\implies k=197


So This is antisymmetric.


3.(8,7)&(7,8)


for (8,7)    (8)2+(7)2=2k    113=2k    k=56.58,7)\implies (8)^2+(7)^2=2k\implies 113=2k\implies k=56.5


for (7,8)    (7)2+(8)2=2k    113=2k    k=56.5(7,8)\implies (7)^2+(8)^2=2k\implies 113=2k\implies k=56.5


So R is not antisymmetric.


4.(3,1)&(1,3)


for (3,1)    (3)2+(1)2=2k    10=2k    k=5(3,1)\implies (3)^2+(1)^2=2k\implies 10=2k\implies k=5


for (1,3)    (1)2+(3)2=2k    10=2k    k=5(1,3)\implies (1)^2+(3)^2=2k\implies 10=2k\implies k=5

So R is not antisymmetric.


So, The ordered pair Which are not antisymmetric are (1,-1)&(-1,1), (8,7)&(7,8) and (3,1)&(1,3)


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