Related Formula
A relation R is symmetric if (x,y) in R (y,x) in R.$$\text{A relation } R \text{ is symmetric if } (x,y) \in R \implies (y,x) \in R.$$
Core Logic
Check condition y ≥ (2x)/(3)$y \geq \frac{2x}{3}$ for each x in A = 2, 3, 5, 7, 9$x \in A = \{2, 3, 5, 7, 9\}$ to generate ordered pairs (x,y)$(x,y)$. Count the pairs to get $\ell$. Then find asymmetric pairs to get m$m$.
Step 1: Enumerate elements of R
x = 2 y ≥ 4/3 = 1.33 y in 2, 3, 5, 7, 9$x = 2 \implies y \geq 4/3 = 1.33 \implies y \in \{2, 3, 5, 7, 9\}$ (5 elements)
x = 3 y ≥ 6/3 = 2 y in 2, 3, 5, 7, 9$x = 3 \implies y \geq 6/3 = 2 \implies y \in \{2, 3, 5, 7, 9\}$ (5 elements)
x = 5 y ≥ 10/3 = 3.33 y in 5, 7, 9$x = 5 \implies y \geq 10/3 = 3.33 \implies y \in \{5, 7, 9\}$ (3 elements)
x = 7 y ≥ 14/3 = 4.66 y in 5, 7, 9$x = 7 \implies y \geq 14/3 = 4.66 \implies y \in \{5, 7, 9\}$ (3 elements)
x = 9 y ≥ 18/3 = 6 y in 7, 9$x = 9 \implies y \geq 18/3 = 6 \implies y \in \{7, 9\}$ (2 elements)
Total elements in R$R$ is = 5 + 5 + 3 + 3 + 2 = 18$\ell = 5 + 5 + 3 + 3 + 2 = 18$.
Step 2: Determine Missing Symmetric Pairs
We need to check which reverse pairs (y,x)$(y,x)$ are missing.
(2, 5) in R$(2, 5) \in R$, but (5, 2) R$(5, 2) \notin R$.
(2, 7) in R$(2, 7) \in R$, but (7, 2) R$(7, 2) \notin R$.
(2, 9) in R$(2, 9) \in R$, but (9, 2) R$(9, 2) \notin R$.
(3, 5) in R$(3, 5) \in R$, but (5, 3) R$(5, 3) \notin R$.
(3, 7) in R$(3, 7) \in R$, but (7, 3) R$(7, 3) \notin R$.
(3, 9) in R$(3, 9) \in R$, but (9, 3) R$(9, 3) \notin R$.
(5, 9) in R$(5, 9) \in R$, but (9, 5) R$(9, 5) \notin R$.
These are m = 7$m = 7$ pairs that need to be added.
Thus, + m = 18 + 7 = 25$\ell + m = 18 + 7 = 25$.
Pattern Recognition
When counting pairs for small sets, list them manually by rows. To make symmetric, any non-diagonal pair (x,y)$(x,y)$ that is present while (y,x)$(y,x)$ is absent counts as 1$1$ missing element.
Chapter Mix
Class 11 Maths: Sets and Relations