Let R be a relation over the set N×N and it is defined by (a$$,$$b)R(c$$,$$d)⇒$$a+d=b+c$$. Then, R is

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Let R be a relation over the set N×N and it is defined by (a$$,$$b)R(c$$,$$d)⇒$$a+d=b+c$$. Then, R is

Infinity Learn · Accepted Answer

We have (a$$,$$b)R(a$$,$$b) for all (a$$,$$b)∈N×N As $$a+b=b+a$$ . Hence, R is reflexive R is symmetric as we have (a$$,$$b)R(c$$,$$d)⇒    $$a+d=b+c$$⇒    $$d+a=c+b$$⇒    $$c+b=d+a$$⇒    $$c+b=d+a$$⇒    (c$$,$$d)R(a$$,$$b)  Now, let (a$$,$$b)R(c$$,$$d) and (c$$,$$d)R(e$$,$$f)  Then, by definition of R , we have $$a+d=b+c$$ and $$c+f=d+eHence$$ by addition, we get $$a+d+c+f=b+c+d+e$$ or  $$a+f=b+e$$ Thus, (a$$,$$b)R(c$$,$$d) and (c$$,$$d)R(e$$,$$f)⇒$$(a$$,$$b)R(e$$,$$f)Therefore$$, R is transitive.

$Let R be a relation over the set N \times N and it is defined by (a, b) R (c, d) \Rightarrow a + d = b + c . Then, R is$

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Let R be a relation over the set N×N and it is defined by (a,b)R(c,d)⇒a+d=b+c. Then, R is

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$Let R be a relation over the set N \times N and it is defined by (a, b) R (c, d) \Rightarrow a + d = b + c . Then, R is$