a) Prove that A B = A B , where A and B are any two sets. [2] b) Let A = \a, b\ , B = \b, c\ and C = \c, d\ . Find A (B C) and A (B C) . [2] c) Test the periodicity of the function f(x) = x and find…

a) Prove that $A-\overline{B}=A\cap B$

Proof:

We start by recalling the definitions of the sets involved:

• $A-\overline{B}$: This is the set of all elements that are in $A$ but not in $\overline{B}$ (the complement of $B$).
• $A\cap B$: This is the set of all elements that are in both $A$ and $B$.

Now, let us prove the equality $A-\overline{B}=A\cap B$ step-by-step.

Step 1: Express $A-\overline{B}$ in terms of $A$ and $B$

By definition, $A-\overline{B}$ consists of all elements $x$ such that: $$x\in A\text{and}x\notin \overline{B}.$$

Since $\overline{B}$ is the complement of $B$, $x\notin \overline{B}$ is equivalent to $x\in B$. Therefore: $$A-\overline{B}=\{x\mid x\in A\text{ and }x\in B\}=A\cap B.$$

Step 2: Verify the equality using set theory laws

We can also verify this using De Morgan's laws and the distributive property of sets. Recall that: $$\overline{B}=U-B,$$ where $U$ is the universal set.

Thus: $$A-\overline{B}=A-(U-B)=A\cap B.$$

This follows from the fact that: $$A-(U-B)=A\cap B.$$

Conclusion:

We have shown that: $$A-\overline{B}=A\cap B.$$

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b) Let $A=\{a,b\}$, $B=\{b,c\}$ and $C=\{c,d\}$. Find $A\times (B\cup C)$ and $A\times (B\cap C)$.

Step 1: Compute $B\cup C$ and $B\cap C$

Given: $$B=\{b,c\},C=\{c,d\}.$$

• Union of $B$ and $C$: $$B\cup C=\{b,c,d\}.$$

• Intersection of $B$ and $C$: $$B\cap C=\{c\}.$$

Step 2: Compute $A\times (B\cup C)$

Given: $$A=\{a,b\},B\cup C=\{b,c,d\}.$$

The Cartesian product $A\times (B\cup C)$ is the set of all ordered pairs $(x,y)$ where $x\in A$ and $y\in (B\cup C)$. Thus: $$A\times (B\cup C)=\{(a,b),(a,c),(a,d),(b,b),(b,c),(b,d)\}.$$

Step 3: Compute $A\times (B\cap C)$

Given: $$A=\{a,b\},B\cap C=\{c\}.$$

The Cartesian product $A\times (B\cap C)$ is the set of all ordered pairs $(x,y)$ where $x\in A$ and $y\in (B\cap C)$. Thus: $$A\times (B\cap C)=\{(a,c),(b,c)\}.$$

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c) Test the periodicity of the function $f(x)=\cos \pi x$ and find its period.

Step 1: Recall the definition of periodicity

A function $f(x)$ is periodic if there exists a positive number $T$ such that for all $x$ in the domain of $f$: $$f(x+T)=f(x).$$ The smallest such $T$ is called the fundamental period of $f$.

Step 2: Test for periodicity

Consider the function: $$f(x)=\cos \pi x.$$

We know that the cosine function $\cos \theta$ has a fundamental period of $2\pi$. That is: $$\cos (\theta +2\pi )=\cos \theta .$$

Let us set $\theta =\pi x$. Then: $$f(x+T)=\cos \pi (x+T)=\cos (\pi x+\pi T).$$

For $f(x+T)=f(x)$, we require: $$\cos (\pi x+\pi T)=\cos (\pi x).$$

This holds if: $$\pi T=2\pi n,\text{for some integer }n\ge 1.$$

Thus: $$T=2n.$$

The smallest positive $T$ occurs when $n=1$: $$T=2.$$

Step 3: Verify the period

Let us verify $T=2$: $$f(x+2)=\cos \pi (x+2)=\cos (\pi x+2\pi )=\cos \pi x=f(x).$$

Thus, $f(x)$ is periodic with period $2$.

Conclusion:

The function $f(x)=\cos \pi x$ is periodic with a fundamental period of 2.