Question
Let R be a relation on N × N defined by:
$(\text{a, b})\text{ R }(\text{c, d})\Leftrightarrow\text{a}+\text{d}=\text{b}+\text{c}$ for all $(\text{a, b}),(\text{c, d})\in\text{N}\times\text{N}$
Show that:
$(\text{a},\text{b})\text{ R }(\text{c, d})\text{ and (c, d) R (e, f)}$
$\Rightarrow(\text{a, b})\text{ R (e, f)}$ for all $(\text{a, b}),(\text{c, d}),(\text{e, f})\in\text{N}\times\text{N}$

Answer

We have,
$(\text{a, b})\text{ R }(\text{c, d})\Leftrightarrow\text{a}+\text{d}=\text{b}+\text{c}$ for all $(\text{a, b}),(\text{c, d})\in\text{N}\times\text{N}$
Now,
(a, b) R (c, d) and (c, d) R (e, f)
⇒ a + d = b + c and c + f = d + e
⇒ a + d + c + f = b + c + d + e [Adding]
⇒ a + f = b + e
⇒ (a, b) R (e, f)

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