Question

Which of the following statements are true? Please justify the answers.
1. a≥b⇔-a≤-b is an absolute inequality.
2. If A=φ,B={1,2},C={-1,-2}, then A×B×C has 4 elements.
3. The argument of 1+√(3)i is π/3.
4. A linear equation over R can have at most one root in C\R.
5. |x₁-x₂|=|x₁|-|x₂|∀x₁,x₂∈R

Answer 1

1. True
2. False
3. True
4. False
5. False

Justifying.

1)

$a \geq b, | \cdot(-1);$

$-a \leq -b$

2)

$A×B×C = \emptyset×B×C = \emptyset$

3)

$z=1+\sqrt{3}i$ is a complex number;

$z=x+yi$ is a complex number in the general form;

if $x > 0 \implies Arg(z) = \arctan(\frac y x) = \arctan(\frac {\sqrt{3}} {1}) = \arctan( \sqrt{3}) = \frac \pi 3$

4)

C\R set does not consist of elements of R set(real numbers), so a linear equation over R cannot have root over C\R.

5)

if x₁ = -2 and x₂ = 1, then

|x₁ - x₂| = | -2 - 1 | = | -3| = 3

|x₁| - |x₂| = |-2| - |1| = 2 - 1 =1,

therefore

|x1 - x2| $\neq$ |x1| - |x2| $\forall$ x1, x2 $\in R$