1. **Problem Statement:** We have sets: - $A = \{ -3, -2, 0, 2, 4, 5, 6, 7, 9 \}$ - $B = \{ 1, 3, 5, 8, 9, 0, + \}$ - $C = \{ -, !, w, w, x, z, y \}$ (note $w$ appears twice, so effectively $C = \{ -, !, w, x, z, y \}$) We need to: (a) Define functions $f: A \to B$, $g: B \to C$, and $h: C \to C$ with properties about one-to-one and onto. (b) Use these functions to find values of expressions. --- 2. **(a)(i) Define $f: A \to B$ and determine if it is one-to-one or onto.** - Since $|A|=9$ and $|B|=7$, $f$ cannot be onto (onto requires every element of $B$ to be an image). - To be one-to-one, each element of $A$ must map to a unique element in $B$. - Since $|A| > |B|$, by pigeonhole principle, $f$ cannot be one-to-one. Define $f$ as: $$ \begin{array}{c|ccccccccc} x & -3 & -2 & 0 & 2 & 4 & 5 & 6 & 7 & 9 \\ f(x) & 1 & 3 & 5 & 8 & 9 & 0 & + & 1 & 3 \\ \end{array} $$ - Here, $f(-3) = 1$ and $f(7) = 1$ show $f$ is not one-to-one. - Not all elements of $B$ are images (e.g., $+$ is used, but some elements like $5$ appear once), so $f$ is not onto. **Answer:** $f$ is neither one-to-one nor onto. --- 3. **(a)(ii) Define $g: B \to C$ and determine if it is one-to-one or onto.** - $|B|=7$, $|C|=6$ (unique elements). - To be one-to-one, each element of $B$ must map to a unique element in $C$. - Since $|B| > |C|$, $g$ cannot be one-to-one. - To be onto, every element of $C$ must be an image. Define $g$ as: $$ \begin{array}{c|ccccccc} x & 1 & 3 & 5 & 8 & 9 & 0 & + \\ g(x) & - & ! & w & x & z & y & - \\ \end{array} $$ - $g(1) = -$ and $g(+) = -$ shows $g$ is not one-to-one. - All elements of $C$ appear as images, so $g$ is onto. **Answer:** $g$ is onto but not one-to-one. --- 4. **(a)(iii) Define $h: C \to C$ which is one-to-one and onto but not identity.** - $|C|=6$. - A bijection (one-to-one and onto) is a permutation of $C$. - Define $h$ by swapping some elements: $$ \begin{array}{c|cccccc} x & - & ! & w & x & z & y \\ h(x) & ! & - & x & w & y & z \\ \end{array} $$ - Each element maps uniquely (one-to-one). - All elements of $C$ appear as images (onto). - $h(x) \neq x$ for all $x$ (no element maps to itself). **Answer:** $h$ is bijection but not identity. --- 5. **(b)(i) Find $h(g(f(9)))$** - Compute $f(9)$: from $f$, $f(9) = 3$. - Compute $g(f(9)) = g(3)$: from $g$, $g(3) = !$. - Compute $h(g(f(9))) = h(!)$: from $h$, $h(!) = -$. **Answer:** $h(g(f(9))) = -$. --- 6. **(b)(ii) Find $h^{-1}(g(8))$** - Compute $g(8)$: from $g$, $g(8) = x$. - Find $h^{-1}(x)$: from $h$, $h(w) = x$, so $h^{-1}(x) = w$. **Answer:** $h^{-1}(g(8)) = w$. --- 7. **(b)(iii) Find $h^{-1}(h^{-1}(w))$** - Find $h^{-1}(w)$: from $h$, $h(x) = w$, so $h^{-1}(w) = x$. - Find $h^{-1}(x)$: from above, $h^{-1}(x) = w$. **Answer:** $h^{-1}(h^{-1}(w)) = w$. --- **Summary:** - $f$ is neither one-to-one nor onto. - $g$ is onto but not one-to-one. - $h$ is bijection but not identity. - $h(g(f(9))) = -$. - $h^{-1}(g(8)) = w$. - $h^{-1}(h^{-1}(w)) = w$.