1. **State the problem:** We have two relations $R_1$ and $R_2$ from set $A = B = \{1,2,3,4\}$ to itself. 2. **Define the relations:** - $R_1 = \{(x,y) \mid x \in A, y \in B, x = y^2 + 1\}$ - $R_2 = \{(x,y) \mid x \in A, y \in B, x \text{ is even and } y \text{ is odd}\}$ 3. **Recall the definition of a function:** A relation $R$ from $A$ to $B$ is a function if every element $x \in A$ is related to exactly one element $y \in B$. 4. **Analyze $R_1$:** - For each $x \in A$, find $y$ such that $x = y^2 + 1$. - Check $x=1$: $1 = y^2 + 1 \Rightarrow y^2 = 0 \Rightarrow y=0$ (not in $B$). - $x=2$: $2 = y^2 + 1 \Rightarrow y^2=1 \Rightarrow y=\pm 1$; only $y=1$ in $B$. - $x=3$: $3 = y^2 + 1 \Rightarrow y^2=2$ no integer $y$ in $B$. - $x=4$: $4 = y^2 + 1 \Rightarrow y^2=3$ no integer $y$ in $B$. - So $R_1 = \{(2,1)\}$ only. - Since not every $x \in A$ has a corresponding $y$, $R_1$ is **not a function**. 5. **Analyze $R_2$:** - $x$ must be even: $x \in \{2,4\}$. - $y$ must be odd: $y \in \{1,3\}$. - So $R_2 = \{(2,1), (2,3), (4,1), (4,3)\}$. - For $x=2$, there are two $y$ values (1 and 3), so $x=2$ relates to multiple $y$. - Hence, $R_2$ is **not a function** because $x=2$ has more than one image. **Final conclusion:** Neither $R_1$ nor $R_2$ represents a function from $A$ to $B$.