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Given p,q,r,s belongs to Z and the relation (p,q) R (r,s)

<=> (p+q) = (r+s) give a proof for each of the following
is R reflexive?
is R symmentric?
is R transitive?
explain me briefly

User HMarioD
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1 Answer

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Final answer:

The relation R is reflexive, symmetric, and transitive.

Step-by-step explanation:

To determine if the relation R is reflexive, we need to show that (p,q) R (p,q) for all (p,q) in Z. Since (p,q) R (r,s) if and only if (p+q) = (r+s), we can substitute (p,q) for both (r,s) and (p+q) for (r+s) in the relation to get (p+q) = (p+q), which is true for all (p,q) in Z. Therefore, R is reflexive.

To determine if the relation R is symmetric, we need to show that if (p,q) R (r,s), then (r,s) R (p,q). If (p,q) R (r,s), then (p+q) = (r+s). By rearranging the equation, we get (r+s) = (p+q), which implies (r,s) R (p,q). Therefore, R is symmetric.

To determine if the relation R is transitive, we need to show that if (p,q) R (r,s) and (r,s) R (u,v), then (p,q) R (u,v). If (p,q) R (r,s) and (r,s) R (u,v), then (p+q) = (r+s) and (r+s) = (u+v), respectively. By transitivity, we can substitute (p+q) for (u+v) and get (p,q) R (u,v). Therefore, R is transitive.

User DaveIt
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