Answer
See explanation
Work Step by Step
We are asked to prove:
7a. There is only one element in B that is an identity for $+$.
7b. There is only one element in B that is an identity for $\cdot$.
We’ll prove both using **basic Boolean algebra properties** and the **definition of identity elements**.
---
### ✅ Part 7a: Uniqueness of Identity for \(+\)
Suppose there are two identity elements for \(+\), say \(e\) and \(e'\), such that for all \(a \in B\):
\[
a + e = a \quad \text{and} \quad a + e' = a
\]
Let’s prove \(e = e'\).
Take \(a = e'\). Then:
\[
e' + e = e' \quad \text{(since } e \text{ is an identity)}
\]
Now take \(a = e\). Then:
\[
e + e' = e \quad \text{(since } e' \text{ is an identity)}
\]
So:
\[
e' = e' + e = e + e' = e
\Rightarrow \boxed{e = e'}
\]
✅ Hence, the identity for \(+\) is **unique**.
---
### ✅ Part 7b: Uniqueness of Identity for \(\cdot\)
Suppose there are two identity elements for \(\cdot\), say \(f\) and \(f'\), such that for all \(a \in B\):
\[
a \cdot f = a \quad \text{and} \quad a \cdot f' = a
\]
Take \(a = f'\). Then:
\[
f' \cdot f = f' \quad \text{(since } f \text{ is an identity)}
\]
Take \(a = f\). Then:
\[
f \cdot f' = f \quad \text{(since } f' \text{ is an identity)}
\]
So:
\[
f' = f' \cdot f = f \cdot f' = f
\Rightarrow \boxed{f = f'}
\]
✅ Hence, the identity for \(\cdot\) is **unique**.
---
### ✅ Final Answer:
\[
\boxed{
\begin{aligned}
\textbf{(a)} &\ \text{There is only one identity element for } + \text{ in } B. \\
\textbf{(b)} &\ \text{There is only one identity element for } \cdot \text{ in } B.
\end{aligned}
}
\]
