R={(a,b)€z ,a+b =even } prove that …
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Sia ? 6 years, 5 months ago
R is reflexive , as 2 divides a-a = 0
Now (a,b){tex}\in {/tex}R implies (a-b) is divided by 2{tex}\Rightarrow{/tex} (b-a) is also divided by 2
Hence, (b,a){tex}\in {/tex}R
Hence, R is symmetric.
Now ,Let a,b,c {tex}\in{/tex}Z
If (a,b) {tex}\in {/tex}R
And (b,c) {tex}\in {/tex}R
Then a-b and b-c divided by 2
Therefore a-b is even and b-c is even
{tex}\Rightarrow{/tex}a-b +b-c is even, as sum of two even numbers is even
{tex}\Rightarrow{/tex}(a-c) is even
So a-c is divided by 2
{tex}\Rightarrow{/tex}(a,c) {tex}\in {/tex}R
Hence it is transitive.
Therefore R is reflexive,symmetric and transitive
So R is an equivalence relation
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