Let R be a relation defined on the set of natural numbers N as (x,y) belongs to R such that x^y +y^x is even.show that R is an equivalence relation on N

							
R: x^y +y^x is even. 
we observe 4 cases:
	
x is even, y is even: clearly, x^y +y^x is even. hence this case is valid.
	
x is odd, y is odd: clearly, x^y +y^x is even. hence this case is valid.
	
x is even, y is odd: clearly, x^y +y^x is odd, bcoz x^y would be even but y^x would be odd. and even+odd= odd. hence this case is invalid.
	
x is odd, y is even: clearly, x^y +y^x is odd, bcoz x^y would be odd but y^x would be even. and even+odd= odd. hence this case is invalid.
now, (a,a) belongs to R coz a^a + a^a= 2a^a is obviously even. hence reflexive.
now, if (a,b) belongs to R then a^b + b^a is even or b^a + a^b is even so that (b, a) belongs to R. hence symmetric.
now, let (a,b) and (b,c) belong to R.
so a^b + b^a= 2m......(*)
and b^c + c^b= 2p.....(#)
from our 4 cases, we have either (a,b) are both odd or both even.
	
(a,b) are both odd: this implies that for (#) to be true, c also needs to be odd because b is odd. hence (a,c) are both odd. thus, a^c + c^a is even, so (a,c) belong to R.
	
(a,b) are both even: this implies that for (#) to be true, c also needs to be even because b is even. hence (a,c) are both even. thus, a^c + c^a is even, so (a,c) belong to R.
in both cases, (a,c) is seen to belong to R. hence transitive.
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