Answer to Question #207345 in Discrete Mathematics for Kay

Question

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)

Answer 1

Given Relation is-

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

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

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

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

So R is not antisymmetric.

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

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

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

So This is antisymmetric.

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

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

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

So R is not antisymmetric.

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

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

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

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)