1. **Problem statement:** We have a relation $f = \{(5nm, 3m + 3n) : m,n \in \mathbb{Z}\}$. We want to find two pairs $(m_1, n_1)$ and $(m_2, n_2)$ such that they have the same domain value but different range values. This will show $f$ is not a function. 2. **Understanding the domain and range:** The domain values are the first components $5nm$ and the range values are the second components $3m + 3n$. 3. **Goal:** Find $(m_1, n_1)$ and $(m_2, n_2)$ such that: $$5 n_1 m_1 = 5 n_2 m_2$$ but $$3 m_1 + 3 n_1 \neq 3 m_2 + 3 n_2$$ 4. **Simplify the domain equality:** $$5 n_1 m_1 = 5 n_2 m_2 \implies n_1 m_1 = n_2 m_2$$ 5. **Try pairs:** Let $(m_1, n_1) = (1, 2)$, then domain value is $5 \times 2 \times 1 = 10$ and range value is $3 \times 1 + 3 \times 2 = 3 + 6 = 9$. Let $(m_2, n_2) = (2, 1)$, then domain value is $5 \times 1 \times 2 = 10$ and range value is $3 \times 2 + 3 \times 1 = 6 + 3 = 9$. Here, range values are equal, so try another pair. 6. **Try $(m_1, n_1) = (1, 3)$:** Domain: $5 \times 3 \times 1 = 15$ Range: $3 \times 1 + 3 \times 3 = 3 + 9 = 12$ Try $(m_2, n_2) = (3, 1)$: Domain: $5 \times 1 \times 3 = 15$ Range: $3 \times 3 + 3 \times 1 = 9 + 3 = 12$ Range equal again, try different pairs. 7. **Try $(m_1, n_1) = (1, 0)$:** Domain: $5 \times 0 \times 1 = 0$ Range: $3 \times 1 + 3 \times 0 = 3$ Try $(m_2, n_2) = (0, 1)$: Domain: $5 \times 1 \times 0 = 0$ Range: $3 \times 0 + 3 \times 1 = 3$ Range equal again. 8. **Try $(m_1, n_1) = (1, 2)$:** Domain: $10$ Range: $9$ Try $(m_2, n_2) = (2, 1)$: Domain: $10$ Range: $9$ Same as before, try $(m_2, n_2) = (1, 5)$: Domain: $5 \times 5 \times 1 = 25$ Range: $3 + 15 = 18$ No match in domain. 9. **Try $(m_1, n_1) = (2, 3)$:** Domain: $5 \times 3 \times 2 = 30$ Range: $6 + 9 = 15$ Try $(m_2, n_2) = (3, 2)$: Domain: $5 \times 2 \times 3 = 30$ Range: $9 + 6 = 15$ Range equal again. 10. **Try $(m_1, n_1) = (1, 1)$:** Domain: $5$ Range: $3 + 3 = 6$ Try $(m_2, n_2) = (5, \frac{1}{5})$ but $n_2$ must be integer, so invalid. 11. **Try $(m_1, n_1) = (1, 4)$:** Domain: $20$ Range: $3 + 12 = 15$ Try $(m_2, n_2) = (2, 2)$: Domain: $5 \times 2 \times 2 = 20$ Range: $6 + 6 = 12$ Range values differ! 12. **Conclusion:** We found two pairs: $$ (m_1, n_1) = (1, 4), \quad (m_2, n_2) = (2, 2) $$ Both have domain value $20$ but range values $15$ and $12$ respectively. This shows $f$ is not a function because the same domain value maps to different range values.