Question

If R (the set of real number) be the universal set and sets V={yϵR:0< y ≤3} and W={yϵR:2 ≤ y <4 } What is V l
For sets A and B , if A and B are subset of Z (the set of Integer) which of the following relations between the two subset is true? (a) (AuB)= A (b) (A\B)n(B\A)= empty set (c) (A\B)n(B\A)= Z (d) (A\B)u(B\A)= empty set
Which of the following pair of functions has f o g = g o f (a)f(y)=y 3 and g(y)=y √ 3 (b) f(y)=y^{5} and g(y)=3y+7$$ (c) f(y)=y 2 and g(y)=y+7 (d) f(y)=y 2 and g(y)=3y+7

Answer 1

Answer on Question #54790 – Math – Abstract Algebra

1. If $R$ (the set of real numbers) be the universal set and sets $V = \{ y \in R : 0 < y \leq 3 \}$ and $W = \{ y \in R : 2 \leq y < 4 \}$ What is $V \setminus W$?

2. For sets $A$ and $B$, if $A$ and $B$ are subsets of $Z$ (the set of integers) which of the following relations between the two subsets is true?

(a) $A \cup B = A$

(b) $(A \setminus B) \cap (B \setminus A) = \emptyset$

(c) $(A \setminus B) \cap (B \setminus A) = Z$

(d) $(A \setminus B) \cup (B \setminus A) = \emptyset$

3. Which of the following pair of functions has $f \circ g = g \circ f$

(a) $f(y) = y - 3$ and $g(y) = y \cdot \sqrt{3}$

(b) $f(y) = y^{\{5\}}$ and $g(y) = 3y + 7$

(c) $f(y) = y - 2$ and $g(y) = y + 7$

(d) $f(y) = y - 2$ and $g(y) = 3y + 7$

Solution

1. $V \setminus W = \{ y \in R : 0 < y < 2 \}$ is all elements from $V$, which do not belong to $W$.

2. $A \cup B$ means that we have objects from $A$ or $B$ (and they don't equal each other). That's why $A \cup B$ is not always equal to $A$.

$(A \setminus B) \cup (B \setminus A)$. We don't know whether $A \subset B$ or $B \subset A$. There are at least two elements: one from $A$, and the other from $B$. That's why $(A \setminus B) \cup (B \setminus A)$ is not empty.

$$(A \setminus B) \cap (B \setminus A)$$

We can see from the diagram that this intersection is empty.

It is not equal to $Z$.

The true relation is (b).

3. $f \circ g = f(y \cdot \sqrt{3}) = y\sqrt{3} - 3$ and $g \circ f = g(y - 3) = (y - 3)\sqrt{3} \circ y\sqrt{3} - 3\sqrt{3} \Rightarrow f \circ g \neq g \circ f$

$$f \circ g = f(3y + 7) = (3y + 7)^{\{5\}} \quad \text{and} \quad g \circ f = g(y^{\{5\}}) = 3y^{\{5\}} + 7 \Rightarrow f \circ g \neq g \circ f$$ $$f \circ g = f(y + 7) = y + 7 - 2 = y + 5 \quad \text{and} \quad g \circ f = g(y - 2) = y - 2 + 7 = y + 5 \Rightarrow$$ $$f \circ g = g \circ f$$$$f \circ g = f(3y + 7) = 3y + 7 - 2 = 3y + 5 \text{ and}$$ $$g \circ f = g(y - 2) = 3(y - 2) + 7 = 3y + 1 \Rightarrow f \circ g \neq g \circ f$$

Answer: 1. V\W={γ∈R: 0< y <2}

2. (b)

3. (c)

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