Consider the expression

Yes, the sum of the two rational numbers \( \frac{5}{2} + \frac{3}{2} \) will produce a rational number. This is because the sum of two rational numbers (which can be expressed as the ratio of integers) is also a rational number; in this case, the result is \( \frac{8}{2} = 4 \), which is a rational number.

Prove that the product of the two rational numbers

4/3 and 9/2 is a rational number. Find the product and explain how you know it is a rational number. Explain your reasoning in 1-2 sentences

To find the product of the two rational numbers \( \frac{4}{3} \) and \( \frac{9}{2} \), we multiply them as follows:

\[
\frac{4}{3} \times \frac{9}{2} = \frac{4 \times 9}{3 \times 2} = \frac{36}{6} = 6.
\]

The product \( 6 \) is a rational number because it can be expressed as a ratio of integers (\( \frac{6}{1} \)). This proves that the product of two rational numbers is always a rational number.

Prove that the sum of

1/2 and√2 is irrational. Show your work and in 1-2 sentences explain why the sum is an irrational number.

To prove that the sum \( \frac{1}{2} + \sqrt{2} \) is irrational, we can use a proof by contradiction.

1. Assume that the sum \( \frac{1}{2} + \sqrt{2} \) is rational. Then, we can express it as \( r \), where \( r \) is a rational number. This gives us:

\[
\sqrt{2} = r - \frac{1}{2}
\]

2. Since \( r \) is rational and \( \frac{1}{2} \) is rational, their difference \( r - \frac{1}{2} \) must also be rational.

3. Therefore, if \( \sqrt{2} \) is rational, this implies that \( \sqrt{2} \) can be expressed as a ratio of two integers. However, it is known that \( \sqrt{2} \) is irrational, as it cannot be expressed as a fraction of two integers without leading to a contradiction (such as assuming it can be written as \( \frac{a}{b} \) in lowest terms and arriving at a contradiction when squaring both sides).

4. Since our assumption that \( \frac{1}{2} + \sqrt{2} \) is rational leads us to a contradiction, we conclude that the assumption must be false. Therefore, the sum \( \frac{1}{2} + \sqrt{2} \) is irrational.

In summary, we know the sum is irrational because it relies on the irrationality of \( \sqrt{2} \); since the result of adding a rational number (like \( \frac{1}{2} \)) to an irrational number (like \( \sqrt{2} \)) is always irrational, the sum \( \frac{1}{2} + \sqrt{2} \) must also be irrational.

Prove that the product of 3π and

2/5 is an irrational number. Find the product and explain why the value is irrational. Explain your reasoning in 1-2 sentences.

To prove that the product \( 3\pi \times \frac{2}{5} \) is irrational, we first compute the product:

\[
3\pi \times \frac{2}{5} = \frac{6\pi}{5}.
\]

Next, we need to show that \( \frac{6\pi}{5} \) is irrational. The number \( \pi \) is known to be an irrational number, meaning it cannot be expressed as a fraction of two integers. If we assume \( \frac{6\pi}{5} \) were rational, we could express it as a fraction \( \frac{p}{q} \), where \( p \) and \( q \) are integers.

This would imply that \( \pi = \frac{5p}{6q} \), which suggests \( \pi \) could be expressed as a ratio of integers, contradicting its known irrationality. Therefore, \( 3\pi \times \frac{2}{5} = \frac{6\pi}{5} \) is irrational.

In summary, the product \( \frac{6\pi}{5} \) is irrational because it is a non-zero multiple of the irrational number \( \pi \), and the product of a non-zero rational number and an irrational number is always irrational.

Describe why simplifying square roots into their most reduced radical form is important in mathematical problem-solving. Also, outline the steps involved in simplifying a square root to its simplest radical form.

### Importance of Simplifying Square Roots

Simplifying square roots into their most reduced radical form is important in mathematical problem-solving for several reasons:

1. **Clarity and Precision**: Simplified radicals are clearer and easier to interpret, which helps in understanding the results and concepts involved in mathematical problems.

2. **Ease of Calculation**: Working with simpler expressions makes it easier to perform further calculations, whether in algebra, geometry, or trigonometry.

3. **Standard Form**: In many mathematical contexts, particularly in higher-level mathematics, it’s expected to present answers in simplest form. This aids in communication and comparability of results.

4. **Error Reduction**: Simplification can help identify and reduce errors in calculations, as simpler forms are less prone to computational complexities or misinterpretation.

### Steps to Simplifying a Square Root to Its Simplest Radical Form

1. **Factor the Radicand**: Identify the number under the square root (the radicand) and factor it into prime factors.

2. **Pair the Factors**: Group the factors into pairs. Each pair of the same number can be taken out of the square root.

3. **Take Out the Pairs**: For every pair of identical factors, take one factor out of the square root.

4. **Multiply the Factors Outside**: Multiply all factors that have been taken outside the square root together. This will be the coefficient in front of the root.

5. **Keep the Remaining Factors Inside**: Any factors that could not be paired will remain inside the square root.

6. **Write the Final Result**: Combine the coefficient with the square root of the remaining factors to express the simplified radical form.

### Example

Let's simplify \( \sqrt{72} \):

1. **Factor the Radicand**: \( 72 = 2^3 \times 3^2 \).

2. **Pair the Factors**: From \( 2^3 \), we can take out one \( 2 \) (since there’s one pair of 2s), and from \( 3^2 \), we can take out one \( 3 \) (since there's a pair of 3s).

3. **Take Out the Pairs**: \( \sqrt{72} = \sqrt{2^2 \times 2 \times 3^2} = 2 \times 3 \times \sqrt{2} \).

4. **Multiply the Factors Outside**: \( 2 \times 3 = 6 \).

5. **Keep the Remaining Factors Inside**: The remaining \( \sqrt{2} \) stays inside.

6. **Write the Final Result**: Thus, \( \sqrt{72} = 6\sqrt{2} \).

This process results in the final answer in its simplest radical form.

make that shorter in a sentence

Simplifying square roots into their simplest radical form improves clarity and ease of calculation, with steps including factoring the radicand, pairing factors, taking out identical pairs, multiplying them, and expressing any remaining factors inside the square root.