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14. multiselect select the numbers that are part of the set of rational numbers.
$-\frac{11}{sqrt{9}}$
$\frac{1}{sqrt{2}}$
$sqrt{-16}$
$\frac{sqrt3{-4096}}{sqrt{16}}$
$sqrt3{4}$
0.333...
15.use math tools explain how you could use a calculator to determine if $sqrt{8}$ expressed as a decimal ever terminates.
justify conclusions determine whether each statement is true or false. if the statement is false, give a counterexample or explain your reasoning.
16. the product of a non-zero rational number and an irrational number is rational.
17. expressing $sqrt{2}$ as the ratio $\frac{sqrt{2}}{1}$ means the $sqrt{2}$ is a rational number.
18. the product of two irrational numbers is irrational.

Question

Accepted Answer

### Question 14
# Explanation:
## Step 1: Analyze $-\frac{11}{\sqrt{9}}$
Simplify $\sqrt{9} = 3$, so $-\frac{11}{3}$ is a fraction of integers, hence rational.
## Step 2: Analyze $\frac{1}{\sqrt{2}}$
$\sqrt{2}$ is irrational, so $\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$ is irrational (product of non - zero rational and irrational is irrational).
## Step 3: Analyze $\sqrt{- 16}$
The square root of a negative number is not a real number, so it's not in the set of real rational numbers.
## Step 4: Analyze $\frac{\sqrt[3]{-4096}}{\sqrt{16}}$
$\sqrt[3]{-4096}=-16$ (since $(-16)^3=-4096$) and $\sqrt{16} = 4$. Then $\frac{-16}{4}=-4$, which is an integer (and thus rational).
## Step 5: Analyze $\sqrt[3]{4}$
$\sqrt[3]{4}$ cannot be expressed as a fraction of integers, so it's irrational.
## Step 6: Analyze $0.333\cdots$
$0.333\cdots=\frac{1}{3}$, which is a fraction of integers, so it's rational.
# Answer: $-\frac{11}{\sqrt{9}}$, $\frac{\sqrt[3]{-4096}}{\sqrt{16}}$, $0.333\cdots$

### Question 15
# Explanation:
## Step 1: Calculate $\sqrt{8}$ using a calculator
First, recall that $\sqrt{8}=2\sqrt{2}\approx2.8284271247\cdots$
## Step 2: Observe the decimal expansion
A terminating decimal has a finite number of digits after the decimal point. When we use a calculator to find the decimal expansion of $\sqrt{8}$, we see that the decimal digits go on without repeating in a finite pattern (it's a non - terminating, non - repeating decimal because $\sqrt{2}$ is irrational and $2\sqrt{2}$ inherits this property). So we can conclude that $\sqrt{8}$ expressed as a decimal does not terminate.
# Answer: Use a calculator to find the decimal expansion of $\sqrt{8}$. A terminating decimal has a finite number of decimal places. The decimal expansion of $\sqrt{8}\approx2.8284271247\cdots$ has an infinite number of non - repeating decimal places, so it does not terminate.

### Question 16
# Explanation:
Let $r$ be a non - zero rational number ($r=\frac{a}{b},a,b\in\mathbb{Z},b
eq0,a
eq0$) and $s$ be an irrational number. Assume $r\times s = t$ is rational. Then $s=\frac{t}{r}$. But the quotient of two rational numbers is rational, which contradicts the fact that $s$ is irrational. For example, let $r = 2$ (rational, non - zero) and $s=\sqrt{2}$ (irrational). Then $r\times s=2\sqrt{2}$, which is irrational. So the statement is false.
# Answer: False. Counterexample: Let the non - zero rational number be $2$ and the irrational number be $\sqrt{2}$. Their product $2\times\sqrt{2} = 2\sqrt{2}$ is irrational.

### Question 17
# Explanation:
A rational number is defined as a number that can be expressed as $\frac{p}{q}$, where $p$ and $q$ are integers and $q
eq0$. In the expression $\frac{\sqrt{2}}{1}$, the numerator $\sqrt{2}$ is not an integer. So just writing $\sqrt{2}$ as $\frac{\sqrt{2}}{1}$ does not make it rational because the definition of a rational number requires the numerator and denominator to be integers (with non - zero denominator). Since $\sqrt{2}$ is irrational, the statement is false.
# Answer: False. A rational number must be expressible as $\frac{p}{q}$ where $p$ and $q$ are integers ($q
eq0$). In $\frac{\sqrt{2}}{1}$, $\sqrt{2}$ is not an integer, so $\sqrt{2}$ is not rational.

### Question 18
# Explanation:
We can find a counterexample. Let the two irrational numbers be $\sqrt{2}$ and $\sqrt{2}$. Their product is $\sqrt{2}\times\sqrt{2}=2$, which is a rational number. So the statement "The product of two irrational numbers is irrational" is false.
# Answer: False. Counterexample: Let the two irrational numbers be $\sqrt{2}$ and $\sqrt{2}$. Their product $\sqrt{2}\times\sqrt{2} = 2$ is rational.

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QUESTION IMAGE

Question

Explanation:

Question 14

Step 1: Analyze \(-\frac{11}{\sqrt{9}}\)

Step 2: Analyze \(\frac{1}{\sqrt{2}}\)

Step 3: Analyze \(\sqrt{- 16}\)

Step 4: Analyze \(\frac{\sqrt[3]{-4096}}{\sqrt{16}}\)

Step 5: Analyze \(\sqrt[3]{4}\)

Step 6: Analyze \(0.333\cdots\)

Step 1: Calculate \(\sqrt{8}\) using a calculator

Step 2: Observe the decimal expansion

Answer:

Question 15