Rational vs. Irrational Practice worksheet with ten math problems on operations and properties of rational and irrational numbers.

Let's go through each problem on the worksheet "Rational vs. Irrational Practice" step by step, solving and explaining each one.

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1. Find the product, then determine if the factors and product are rational or irrational.

$$
\sqrt{6} \cdot 3
$$

- Product:
$$
\sqrt{6} \cdot 3 = 3\sqrt{6}
$$
- Factors:
- $\sqrt{6}$: Irrational (since 6 is not a perfect square)
- $3$: Rational
- Product: $3\sqrt{6}$ → Irrational

> ✔ Answer:
> Product = $3\sqrt{6}$, Factors: $\sqrt{6}$ (irrational), $3$ (rational); Product: irrational

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2. Fill in the blank.

> The product of a non-zero rational number and an irrational number is always __________.

- This is a key property:
- Rational × Irrational = Irrational, if the rational number is non-zero.
- Example: $2 \cdot \sqrt{2} = 2\sqrt{2}$ → irrational
- But $0 \cdot \sqrt{2} = 0$ → rational, but 0 is excluded here ("non-zero")

> ✔ Answer: irrational

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3. Finish the sentence.

> The sum of a rational number and an irrational number is always __________.

- Fact: Rational + Irrational = Irrational
- Example: $1 + \sqrt{2}$ → irrational
- Proof sketch: Suppose $r + i = q$, where $r$ rational, $i$ irrational, $q$ rational. Then $i = q - r$, which would be rational (difference of rationals) — contradiction.

> ✔ Answer: irrational

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4. Find the sum, then determine if the addends and sum are rational or irrational.

$$
\sqrt{24} + 7
$$

- Simplify $\sqrt{24} = \sqrt{4 \cdot 6} = 2\sqrt{6}$ → Irrational
- $7$: Rational
- Sum: $2\sqrt{6} + 7$ → Irrational (rational + irrational = irrational)

> ✔ Answer:
> Sum = $2\sqrt{6} + 7$, Addends: $\sqrt{24}$ (irrational), $7$ (rational); Sum: irrational

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5. Find the product, then determine if the factors and product are rational or irrational.

$$
\sqrt{6} \cdot 0
$$

- Product: $\sqrt{6} \cdot 0 = 0$
- $\sqrt{6}$: Irrational
- $0$: Rational
- Product: $0$ → Rational

> ✔ Answer:
> Product = $0$, Factors: $\sqrt{6}$ (irrational), $0$ (rational); Product: rational

> 🔍 Note: This shows that rational × irrational can be rational — only when the rational number is zero.

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6. Find the sum, then determine if the addends and sum are rational or irrational.

$$
-0.37 + \frac{1}{2}
$$

- Convert to decimals:
$-0.37 + 0.5 = 0.13$
- $-0.37$: Rational (terminating decimal)
- $\frac{1}{2} = 0.5$: Rational
- Sum: $0.13$: Rational

> ✔ Answer:
> Sum = $0.13$, Addends: both rational; Sum: rational

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7. Prove this statement with two examples:

> The product of two rational numbers is always rational.

- Proof idea: A rational number is a ratio of integers: $\frac{a}{b}, \frac{c}{d}$, $b,d \ne 0$
- Product: $\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{bd}$ → still a ratio of integers → rational

#### Example 1:
$$
\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8} \quad \text{(rational)}
$$

#### Example 2:
$$
0.5 \cdot 0.25 = 0.125 = \frac{1}{8} \quad \text{(rational)}
$$

> ✔ Answer: Yes, the product of two rational numbers is always rational. Examples show it.

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8. Find the difference. Then determine if each number (including the answer) is rational or irrational.

$$
\sqrt{24} - 54
$$

- $\sqrt{24} = 2\sqrt{6}$ → Irrational
- $54$: Rational
- Difference: $2\sqrt{6} - 54$ → Irrational (irrational minus rational = irrational)

> ✔ Answer:
> Difference = $2\sqrt{6} - 54$, Numbers: $\sqrt{24}$ (irrational), $54$ (rational); Difference: irrational

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9. Determine if the statement is true or false. If it’s false, prove it with an example.

> The product of a rational number and an irrational number will always be irrational.

- False!
- Counterexample: $0 \cdot \sqrt{2} = 0$, and $0$ is rational
- So, even though $\sqrt{2}$ is irrational, multiplying by $0$ (rational) gives a rational result.

> ✔ Answer: False
> Counterexample: $0 \cdot \sqrt{2} = 0$ → rational

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10. Find the product, then determine if the factors and product are rational or irrational.

$$
-\frac{1}{5} \cdot 7
$$

- Product: $-\frac{1}{5} \cdot 7 = -\frac{7}{5} = -1.4$
- $-\frac{1}{5}$: Rational
- $7$: Rational
- Product: $-\frac{7}{5}$: Rational

> ✔ Answer:
> Product = $-\frac{7}{5}$, Factors: both rational; Product: rational

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## ✔ Final Summary Answers:

| Problem | Answer |
|--------|--------|
| 1 | Product: $3\sqrt{6}$, $\sqrt{6}$: irrational, $3$: rational, Product: irrational |
| 2 | irrational |
| 3 | irrational |
| 4 | Sum: $2\sqrt{6} + 7$, $\sqrt{24}$: irrational, $7$: rational, Sum: irrational |
| 5 | Product: $0$, $\sqrt{6}$: irrational, $0$: rational, Product: rational |
| 6 | Sum: $0.13$, both addends rational, Sum: rational |
| 7 | True: rational × rational = rational. Examples: $\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}$, $0.5 \cdot 0.25 = 0.125$ |
| 8 | Difference: $2\sqrt{6} - 54$, $\sqrt{24}$: irrational, $54$: rational, Difference: irrational |
| 9 | False. Counterexample: $0 \cdot \sqrt{2} = 0$ (rational) |
| 10 | Product: $-\frac{7}{5}$, both factors rational, Product: rational |

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