The problem consists of four true/false statements that need to be evaluated. i) The number $\pi^0$ is irrational. ii) The quadratic equation $x^2 = -4$ has no real solutions. iii) The rational expression $\frac{3x-1}{1-3x}$ when simplified is equal to $-1$. iv) $x^{5/3} = \sqrt[5]{x^3}$ for a real number $x$.

1. Problem Description

The problem consists of four true/false statements that need to be evaluated.

i) The number $\pi^0$ is irrational.

ii) The quadratic equation $x^2 = -4$ has no real solutions.

iii) The rational expression $\frac{3x-1}{1-3x}$ when simplified is equal to $-1$.

iv) $x^{5/3} = \sqrt[5]{x^3}$ for a real number $x$.

2. Solution Steps

i) $\pi^0 = 1$. Since 1 is a rational number, the statement that $\pi^0$ is irrational is false.

ii) $x^2 = -4$. If $x$ is a real number, then $x^2$ must be non-negative ($x^2 \ge 0$). Therefore, there is no real number $x$ such that $x^2 = -4$. The statement that the quadratic equation $x^2 = -4$ has no real solutions is true.

iii) $\frac{3x-1}{1-3x} = \frac{3x-1}{-(3x-1)}$. If $3x-1 \ne 0$, then $\frac{3x-1}{-(3x-1)} = -1$. The statement that the rational expression $\frac{3x-1}{1-3x}$ when simplified is equal to $-1$ is true (provided $x \ne \frac{1}{3}$).

iv) $x^{5/3} = \sqrt[3]{x^5}$

$x^{5/3} = \sqrt[5]{x^3}$ is false. $x^{5/3}$ is equal to $\sqrt[3]{x^5}$.

3. Final Answer

i) F

ii) T

iii) T

iv) F