Let [A=1,2,3,4,5].

Question

Let $$\mathrm{A}=\{1,2,3,4,5\}$$. Let $$\mathrm{R}$$ be a relation on $$\mathrm{A}$$ defined by $$x \mathrm{R} y$$ if and only if $$4 x \leq 5 \mathrm{y}$$. Let $$\mathrm{m}$$ be the number of elements in $$\mathrm{R}$$ and $$\mathrm{n}$$ be the minimum number of elements from $$\mathrm{A} \times \mathrm{A}$$ that are required to be added to R to make it a symmetric relation. Then m + n is equal to :

Accepted Answer

$$\begin{aligned}
& A=\{1,2,3,4,5\} \
& x R y \Leftrightarrow 4 x \leq 5 y \
& 4 x \leq 5 y \quad \Rightarrow \quad \frac{x}{y} \leq \frac{5}{4} \quad \Rightarrow \frac{x}{y} \leq 1.25
\end{aligned}$$

$$\begin{aligned}
& R=\{(1,2),(1,3),(1,4),(1,5),(1,1),(2,2),(2,3),(2,4), \
& (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\} \
& \therefore \quad n(R)=m=16
\end{aligned}$$

Elements to be added to $R$ to make it symmetric

$$(1,2) \in R \quad \Rightarrow \quad(2,1)$$ should be added

Similarly, $$(3,1),(4,1),(5,1),(3,2),(4,2),(5,2),(4,3), (5,3)$$

$$\therefore 9$$ elements should be added

$$\[\begin{array}{ll} \therefore & n=9 \ \therefore & m+n=25 \end{array}$$\]

Solutions · 1