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Question

Let Z be the set of integers. Show that the relation R={(a,b):a,bZ and a+b is even } is an equivalence relation on Z.

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Solution

The given relation R on the set of integers Z is reflexive as (x,x)R as 2x is even for all xZ.
Again the relation is symmetric as if (x,y)R(y,x)R since x+y is even gives y+x is also even for all x,yZ.
Again this relation R is transitive as if (x,y)R and (y,z)R this gives (x,z)R as x+y=2k1.....(1) [ Since x+y is even ] and y+z=2k2.....(2) [ Since y+z is even] [ For k1 and k2 being integers]
Now adding (1) and (2) we get
x+2y+z=2(k1+k2)
or, x+z=2(k1+k2y)
This gives x+z is even.
Hence this relation is also transitive.
So the relation is equivalance relation.

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