Answer

(A) `rarr` (r ); (B) `rarr` (q); (C) `rarr` (p); (D) `rarr` (p,q,r,s)

Explanation

(A) `because R={(x,y):xlty,x,yinN}`
`xcancelltxtherefore(x,x)cancelinR`
So, R is not reflexive.
Now, `(x,y)inRimpliesxltycancelimpliesyltximplies(y,x)cancelinR`
`therefore` R is not symmetric.
Let (x, y) `in R` and (y, z) `in` R
`implies x lt y and y lt z implies x lt z implies (x,z) in R`
`therefore R` is transitive.
(B) `because S={(x,y):x+y=10,x,yinN}`
`therefore x+x=10implies2x=10impliesx=5`
So, each element of N is not related to itself by the relation x + y = 10.
`therefore S` is not reflexive.
Now, `(x,y)inSimpliesx+y=10impliesy+x=10`
`implies (y,x)inS`
`therefore S` is symmetric relation.
Now, let `(3,7)inS and (7,3)inSimplies(3,3)cancelinS`
`therefore` S is not transitive.
(C ) `because T={(x,y):x=yorx-y=1,x,yinN}`
`therefore x = x`
So, `(x,x)inT,AAx inN`
`therefore` T is reflexive.
Let `(3,2)inT and3-2=1`
`cancelimplies2-3=-1implies(2,3)cancelinT`
`therefore` T is not symmetric.
Now, let `(3,2) in T and (2,1) in T`
`therefore 3-2=1and2-1=1`
Then, `(3,1)cancelinT" "[because3-1=2ne1]`
`therefore` T is not transitive.
(D) `U={(x,y):x^(y)=y^(x),x,yinN}`
`because x^(x)=x^(x)`
`therefore (x,x)inU`
`therefore U` is reflexive.
Now, `(x,y)inUimpliesx^(y)=y^(x)`
`implies y^(x)=x^(y)implies(y,x)inU`
`therefore` U is symmetric.
Now, let `(x,y)inUand(y,z)inU`
`impliesx^(y)=y^(x)andy^(z)=z^(y)`
Now, `(x^(y))^(z)=(y^(x))^(z)`
`implies (x^(z))^(y)=(y^(z))^(x)implies(x^(z))^(y)=(z^(y))^(x)`
`implies (x^(z))^(y)=(z^(x))^(y)impliesx^(z)=z^(x)implies(x,z)inU`
`therefore U` is transitive.
Hence, U is an equivalence relation.