1. **Problem:** Determine whether $\alpha$ is a mapping in each case. (a) $\alpha: \mathbb{N} \to \mathbb{N}$ defined by $\alpha(x) = -x$ for all $x \in \mathbb{N}$. - A mapping from $\mathbb{N}$ to $\mathbb{N}$ must assign each natural number to a natural number. - Since $-x$ is negative for any positive $x$, $\alpha(x)$ is not in $\mathbb{N}$ (natural numbers are non-negative). - Therefore, $\alpha$ is **not** a mapping. (b) $\alpha: \mathbb{R} \to \mathbb{R}$ defined by $\alpha(y) = \sqrt{y}$ for all $y \in \mathbb{N}$. - The domain is $\mathbb{R}$ but the function is defined only for $y \in \mathbb{N}$ (natural numbers). - For negative real numbers, $\sqrt{y}$ is not defined in real numbers. - Hence, $\alpha$ is **not** a mapping from $\mathbb{R}$ to $\mathbb{R}$ because it is not defined for all real numbers. **Final answers:** - (a) Not a mapping. - (b) Not a mapping. 2. **Problem:** Determine whether $\alpha: \mathbb{R} \to \mathbb{R}$ defined by $\alpha(x) = 3 - 4x$ is onto or one-to-one. - $\alpha(x) = 3 - 4x$ is a linear function. - To check if onto (surjective): For any $y \in \mathbb{R}$, solve $y = 3 - 4x$ for $x$: $$x = \frac{3 - y}{4}$$ - Since $x$ exists for every $y$, $\alpha$ is onto. - To check if one-to-one (injective): Suppose $\alpha(x_1) = \alpha(x_2)$, then $$3 - 4x_1 = 3 - 4x_2 \implies -4x_1 = -4x_2 \implies x_1 = x_2$$ - So $\alpha$ is one-to-one. **Final answer:** $\alpha$ is both onto and one-to-one. 3. **Problem:** Let $\alpha : A \to B$ and $\beta : B \to C$. If $\beta \circ \alpha$ is onto, show that $\beta$ is onto. - Since $\beta \circ \alpha$ is onto, for every $c \in C$, there exists $a \in A$ such that $$\beta(\alpha(a)) = c$$ - Let $b = \alpha(a) \in B$, then $\beta(b) = c$. - This shows for every $c \in C$, there exists $b \in B$ with $\beta(b) = c$. - Hence, $\beta$ is onto. 4. **Problem:** Let $\alpha : \mathbb{N} \to \mathbb{N}$ defined by $\alpha(n) = 2n$ and $\beta(n) = \begin{cases} \frac{n+1}{2}, & \text{if } n \text{ is odd} \\ \frac{n}{2}, & \text{if } n \text{ is even} \end{cases}$ (a) Show that $\beta$ and $\beta \circ \alpha$ are surjective but $\alpha$ is not. - $\alpha(n) = 2n$ maps $\mathbb{N}$ to even numbers only, so not all natural numbers are hit (odd numbers are missed). - So $\alpha$ is **not** surjective. - $\beta$ maps any natural number to a natural number: - If $n$ is odd, $\beta(n) = \frac{n+1}{2}$ is natural. - If $n$ is even, $\beta(n) = \frac{n}{2}$ is natural. - For any $m \in \mathbb{N}$, $\beta(2m) = m$, so $\beta$ is surjective. - $\beta \circ \alpha(n) = \beta(2n) = \frac{2n}{2} = n$, which hits every $n$, so $\beta \circ \alpha$ is surjective. (b) Show that $\alpha$ and $\beta \circ \alpha$ are injective but $\beta$ is not. - $\alpha(n) = 2n$ is injective because if $2n_1 = 2n_2$, then $n_1 = n_2$. - $\beta \circ \alpha(n) = n$ is the identity function, which is injective. - $\beta$ is not injective because $\beta(1) = 1$ and $\beta(2) = 1$, but $1 \neq 2$. 5. **Problem:** Let $\alpha : S \to T$ be a mapping. Prove that if $\alpha$ is invertible, then $\alpha^{-1}$ is also invertible. - If $\alpha$ is invertible, there exists $\alpha^{-1} : T \to S$ such that $$\alpha^{-1} \circ \alpha = \text{id}_S \quad \text{and} \quad \alpha \circ \alpha^{-1} = \text{id}_T$$ - To show $\alpha^{-1}$ is invertible, consider $\alpha$ as its inverse. - Then $$\alpha \circ \alpha^{-1} = \text{id}_T \quad \text{and} \quad \alpha^{-1} \circ \alpha = \text{id}_S$$ - So $\alpha$ is the inverse of $\alpha^{-1}$, proving $\alpha^{-1}$ is invertible. **Summary:** - (1) (a) Not a mapping, (b) Not a mapping. - (2) $\alpha$ is onto and one-to-one. - (3) If $\beta \circ \alpha$ is onto, then $\beta$ is onto. - (4a) $\beta$ and $\beta \circ \alpha$ surjective, $\alpha$ not. - (4b) $\alpha$ and $\beta \circ \alpha$ injective, $\beta$ not. - (5) If $\alpha$ invertible, $\alpha^{-1}$ invertible.