Solution to Determine whether the relation R on the set of all real numbers is reflexive, symmetric, … - Sikademy
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Archangel Macsika

Determine whether the relation R on the set of all real numbers is reflexive, symmetric, antisymmetric, and/or transitive, where (x, y) ∈ R if and only if a) x + y = 0. b) x = ±y. c) x - y is a rational number.

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Part a

Consider the relation R : x+ y = 0, for \space x,y \in R .

Reflexive:

For 1 \in R, 1+1=2 \not=0.

Therefore, R is not reflexive.

Symmetric:

For x,y \in R, x+y=0=y+x.

Therefore, R is symmetric.

Anti-symmetric:

For -1,1 \in R, -1+1=0 \space and \space 1+(-1)=0 \space but \space -1 \not=1

Therefore, R is not anti-symmetric.

Transitive:

For -1,1 \in R, 1+(-1)=0 \space and \space (-1)+1=0

That is (1,-1) \in R and (-1,1) \in R but (1,1) \not \in R

Therefore, R is not transitive. 


Part b

Consider the relation R : ±y = 0, for \space x,y \in R .

Reflexive:

For x \in R, x=x \implies x=±x, so (x,x) \in R

Therefore, R is reflexive.

Symmetric:

For x,y \in R, x=±y \implies y=±x.

Therefore, R is symmetric.

Anti-symmetric:

For (-1,1) \in R, (1,-1) \in R; -1 = ±1 \space and \space 1=±(-1) \space but \space -1 \not=1

Therefore, R is not anti-symmetric.

Transitive:

For (x,y) \in R, (y,z) \in R \\ x= ±y \space and \space y=±z

That is x= ±z

Therefore, R is transitive. 


Part c

Consider the relation R:x-y is a rational number for x,y \in R

Reflexive: For x \in R, x-x=0 is a rational number, so (x,x) \in R .

Therefore, R is reflexive.

Symmetric: For (x,y) \in R ,

x-y is rational

y -x is rational

(y,x) \in R

Therefore, R is symmetric.

Anti-symmetric:

For (1,2) \in R and (2,1) \in R ,

That is, 1— 2 is rational and 2 is rational, but 2 \not =1 .

Therefore, R is not anti-symmetric.

Transitive:

For (x,y) \in R and (y,z) \in R, then x— y is rational and y—z is rational

Implies that x—z=x—y+y—z

As the sum of two rational numbers is rational follows that x — z is rational.

Therefore, R is transitive. 


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