We are given three problems to solve. 1) Identify which of the four given terms, $\frac{1}{\sqrt{3}}$, $\frac{1}{(\sqrt{3})^2}$, $\frac{1}{(\sqrt{3})^3}$, $\frac{1}{(\sqrt{3})^4}$, are irrational. 2) Write $y^{-1}$, $y^0$, $y^2$ and $y^3$ in increasing order of size when $y < -2$. 3) Given that $-3 < x \le 3\frac{1}{3}$, (a) find the largest integer value of $x$, (b) find the smallest prime number value of $x$.

1. Problem Description

We are given three problems to solve.

1) Identify which of the four given terms, $\frac{1}{\sqrt{3}}$, $\frac{1}{(\sqrt{3})^2}$, $\frac{1}{(\sqrt{3})^3}$, $\frac{1}{(\sqrt{3})^4}$, are irrational.

2) Write $y^{-1}$, $y^0$, $y^2$ and $y^3$ in increasing order of size when $y < -2$.

3) Given that $-3 < x \le 3\frac{1}{3}$,

(a) find the largest integer value of $x$,

(b) find the smallest prime number value of $x$.

2. Solution Steps

1) A number is irrational if it cannot be expressed as a ratio of two integers.

$\frac{1}{\sqrt{3}}$ is irrational because $\sqrt{3}$ is irrational.

$\frac{1}{(\sqrt{3})^2} = \frac{1}{3}$, which is rational.

$\frac{1}{(\sqrt{3})^3} = \frac{1}{3\sqrt{3}}$, which is irrational.

$\frac{1}{(\sqrt{3})^4} = \frac{1}{9}$, which is rational.

Therefore, the irrational terms are $\frac{1}{\sqrt{3}}$ and $\frac{1}{(\sqrt{3})^3}$.

2) Let's analyze the given terms for $y < -2$. Let's pick $y=-3$ as an example.

$y^{-1} = \frac{1}{y} = \frac{1}{-3} = -\frac{1}{3}$

$y^0 = 1$

$y^2 = (-3)^2 = 9$

$y^3 = (-3)^3 = -27$

In increasing order, we have $y^3 < y^{-1} < y^0 < y^2$.

Therefore, the terms in increasing order are $y^3, y^{-1}, y^0, y^2$.

3)

(a) The given inequality is $-3 < x \le 3\frac{1}{3}$, which is $-3 < x \le \frac{10}{3}$.

Since $\frac{10}{3} = 3.333...$, the largest integer value of $x$ is

3. (b) We want to find the smallest prime number value of $x$, such that $-3 < x \le \frac{10}{3}$.

Prime numbers are numbers greater than 1 that have only two factors: 1 and themselves.

The prime numbers greater than $-3$ are $2$ and $3$.

Both $2$ and $3$ satisfy the inequality $-3 < x \le \frac{10}{3}$.

Therefore, the smallest prime number is

2.

3. Final Answer

1) $\frac{1}{\sqrt{3}}$, $\frac{1}{(\sqrt{3})^3}$

2) $y^3, y^{-1}, y^0, y^2$

3) (a) 3

(b) 2