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## Here's the Solution to this Question

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.