First slide

Relations XII

Question

Let N denotes the set of all natural numbers. On NxN define R as follows: $(a, b), (c, d) \in N \times N$ (a, b) R (c, d) if ad(b + c) = bc(a + d), then

Moderate

A

R is reflexive
B

R is transitive
C

R is an equivalence relation
D

R is symmetric

Solution

$\begin{aligned} ad (b + c) = bc (a + d) \\ \Leftrightarrow \frac{b + c}{bc} = \frac{a + d}{ad} \\ \Leftrightarrow \frac{1}{b} + \frac{1}{c} = \frac{1}{a} + \frac{1}{d} \\ \Leftrightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{d} \end{aligned}$
R is reflexive
Let $(a, b) \in N \times N$
$\begin{aligned} \frac{1}{a} - \frac{1}{b} = \frac{1}{a} - \frac{1}{b} \\ \Rightarrow & (a, b) R (a, b) \end{aligned}$
R is reflexive.
R is symmetric.
Suppose $(a, b), (c, d) \in N \times N and (a, b) R (c, d)$
$\begin{array}{ll} \Leftrightarrow & \frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{d} \\ \Leftrightarrow & \frac{1}{c} - \frac{1}{d} = \frac{1}{a} - \frac{1}{b} \\ \Leftrightarrow & (c, d) R (a, b) \end{array}$
R is symmetric.
R is transitive.
Suppose $(a, b), (c, d), (e, f) \in N \times N$ and (a, b) R (c, d), (c, d) R (e,f)
$\begin{aligned} \Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{d} and \frac{1}{c} - \frac{1}{d} = \frac{1}{e} - \frac{1}{f} \\ \Rightarrow \frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{f} \Rightarrow (a, b) R (e, f) \end{aligned}$
R is transitive.
R is an equivalence relation.

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