We are asked to list the elements of the following sets: (a) {$x: x$ is an integer greater than $-2$ but less than $9$}. (b) {$x: x$ is an integer} $\cap$ {$0, \sqrt{3}, \pi, 2i$}. (c) {$x^2 + 1: x \in A$} where $A = \{-2, -1, 0, 1, 2\}$. (d) {$\sqrt{x + 2}: x \in B$} where $B = \{-3, -4, 0, 1, 2\}$.

Discrete MathematicsSet TheorySet OperationsIntegersReal NumbersComplex Numbers
2025/4/15

1. Problem Description

We are asked to list the elements of the following sets:
(a) {x:xx: x is an integer greater than 2-2 but less than 99}.
(b) {x:xx: x is an integer} \cap {0,3,π,2i0, \sqrt{3}, \pi, 2i}.
(c) {x2+1:xAx^2 + 1: x \in A} where A={2,1,0,1,2}A = \{-2, -1, 0, 1, 2\}.
(d) {x+2:xB\sqrt{x + 2}: x \in B} where B={3,4,0,1,2}B = \{-3, -4, 0, 1, 2\}.

2. Solution Steps

(a) The integers greater than 2-2 but less than 99 are 1,0,1,2,3,4,5,6,7,8-1, 0, 1, 2, 3, 4, 5, 6, 7, 8.
(b) The set {x:xx: x is an integer} is the set of all integers. We are looking for the intersection of this set with the set {0,3,π,2i0, \sqrt{3}, \pi, 2i}.
The only integer in the set {0,3,π,2i0, \sqrt{3}, \pi, 2i} is 00. 3\sqrt{3} and π\pi are irrational numbers and 2i2i is an imaginary number. Therefore, the intersection is {00}.
(c) We have A={2,1,0,1,2}A = \{-2, -1, 0, 1, 2\}. We need to find the set {x2+1:xAx^2 + 1: x \in A}. We compute x2+1x^2 + 1 for each xx in AA.
If x=2x = -2, x2+1=(2)2+1=4+1=5x^2 + 1 = (-2)^2 + 1 = 4 + 1 = 5.
If x=1x = -1, x2+1=(1)2+1=1+1=2x^2 + 1 = (-1)^2 + 1 = 1 + 1 = 2.
If x=0x = 0, x2+1=(0)2+1=0+1=1x^2 + 1 = (0)^2 + 1 = 0 + 1 = 1.
If x=1x = 1, x2+1=(1)2+1=1+1=2x^2 + 1 = (1)^2 + 1 = 1 + 1 = 2.
If x=2x = 2, x2+1=(2)2+1=4+1=5x^2 + 1 = (2)^2 + 1 = 4 + 1 = 5.
Therefore, the set is {5,2,1,2,55, 2, 1, 2, 5} which is equal to {1,2,51, 2, 5}.
(d) We have B={3,4,0,1,2}B = \{-3, -4, 0, 1, 2\}. We need to find the set {x+2:xB\sqrt{x + 2}: x \in B}. We compute x+2\sqrt{x + 2} for each xx in BB.
If x=3x = -3, x+2=3+2=1=i\sqrt{x + 2} = \sqrt{-3 + 2} = \sqrt{-1} = i.
If x=4x = -4, x+2=4+2=2=i2\sqrt{x + 2} = \sqrt{-4 + 2} = \sqrt{-2} = i\sqrt{2}.
If x=0x = 0, x+2=0+2=2\sqrt{x + 2} = \sqrt{0 + 2} = \sqrt{2}.
If x=1x = 1, x+2=1+2=3\sqrt{x + 2} = \sqrt{1 + 2} = \sqrt{3}.
If x=2x = 2, x+2=2+2=4=2\sqrt{x + 2} = \sqrt{2 + 2} = \sqrt{4} = 2.
Therefore, the set is {i,i2,2,3,2i, i\sqrt{2}, \sqrt{2}, \sqrt{3}, 2}.

3. Final Answer

(a) {1,0,1,2,3,4,5,6,7,8-1, 0, 1, 2, 3, 4, 5, 6, 7, 8}
(b) {00}
(c) {1,2,51, 2, 5}
(d) {i,i2,2,3,2i, i\sqrt{2}, \sqrt{2}, \sqrt{3}, 2}

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