N is the set of natural numbers. The relation R is defined on \( \Large N \times N \) as follows: \( \Large \left(a,\ b\right)R \left(c,\ d\right) \Leftrightarrow a+d=b+c \) is:
Correct Answer: Description for Correct answer:
(a) \( \Large \left(a,\ b\right) R \left(a,\ b\right) \Leftrightarrow \ a+b=b+a \)
Therefore, R is reflexive
(b) \( \Large \left(a,\ b\right) R \left(c,\ d\right) => a+d = b+c \)
=> \( \Large c+b = d+a \)
=> \( \Large \left(c,\ d\right) R \left(a,\ b\right) \)
Therefore, R is symmetric
(c) \( \Large \left(a,\ b \right) R \left(c,\ d\right)\ and\ \left(c,\ d\right) R \left(e,\ f \right) \)
=> \( \Large a+d=b+c\ and \ c + f=d+e \)
=> \( \Large a+b+c+f = b+c+d+e \)
=> \( \Large a+f = b+e \)
=> \( \Large \left(a,\ b\right) R \left(e,\ f\right) \)
Therefore, R is transitive
Thus R is an equivalence relation \( \Large N \times N \)
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