Problems with functions and relations Determine whether the

Problems with functions and relations. Determine whether the following relations are re- flexive, symmetric, transitive, antisymmetric or equivalence. A relation can be in more than one category, so list all the categories applicable.

(a) Given the set of positive integers, the relation R is defined such that xRy holds if |x y| < 1 is an even number.

(b) Given the set of positive integers, the relation R is defined such that xRy holds if x^2+y^2 is an even number

Solution

xRy holds if |x y| < 1 is an even number.

Since |x-y| cannot be negative only positive values it can take.

|x-y|<1 is equivalent to |x-y|=0

Or x =y

x=y is reflexive since x =x

x=y is transitive as x=y and y=z implies x=z

x = y is symmetric as x=y implies y =x

Thus the relation |x-y|<1 is an equivalence relation.

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xRy holds if x^2+y^2 is an even number

X^2+y^2 can be an even number if x and y are odd or both are even

For eg, x is odd means square of x is odd similarly y odd y^2 is odd so sum of x^2+y^2 will be even always.

xRy is reflexive as x^2+x^2 = 2x^2 is even for all x.

xRy is symmetric as if x^2+y^2 is even then y^2+x^2 is also even

xRy is transitive.

Reason: xRy implies both x and y are even or both odd.

Similarly yRz implies both y,z odd or both even.

In other words, x,y,z all 3 either odd or all even.

Hence x^2+z^2 is even for x and z

So transitive.

Thus equivalence relation.