Exercises with multiplicative and additive inverse of number - Free Printable

Problem Analysis:

The task involves solving several problems related to multiplicative inverses and additive inverses of numbers. Let's break down each part step by step.

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#### 1. Write the multiplicative inverse of numbers: \(10, 5, -20, -40, 100, 7\)

The multiplicative inverse of a number \(x\) is a number \(y\) such that:
\[
x \cdot y = 1
\]
For any nonzero number \(x\), the multiplicative inverse is given by:
\[
y = \frac{1}{x}
\]

- For \(10\):
\[
\text{Multiplicative inverse} = \frac{1}{10}
\]

- For \(5\):
\[
\text{Multiplicative inverse} = \frac{1}{5}
\]

- For \(-20\):
\[
\text{Multiplicative inverse} = \frac{1}{-20} = -\frac{1}{20}
\]

- For \(-40\):
\[
\text{Multiplicative inverse} = \frac{1}{-40} = -\frac{1}{40}
\]

- For \(100\):
\[
\text{Multiplicative inverse} = \frac{1}{100}
\]

- For \(7\):
\[
\text{Multiplicative inverse} = \frac{1}{7}
\]

Final Answer:
\[
\boxed{\frac{1}{10}, \frac{1}{5}, -\frac{1}{20}, -\frac{1}{40}, \frac{1}{100}, \frac{1}{7}}
\]

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#### 2. Write the multiplicative inverse of numbers: \(\frac{1}{2}, \frac{7}{3}, 1\frac{2}{3}, -3\frac{7}{10}, -1\frac{2}{9}\)

First, convert mixed numbers to improper fractions:
- \(1\frac{2}{3} = \frac{3 \cdot 1 + 2}{3} = \frac{5}{3}\)
- \(-3\frac{7}{10} = \frac{-3 \cdot 10 - 7}{10} = \frac{-37}{10}\)
- \(-1\frac{2}{9} = \frac{-1 \cdot 9 - 2}{9} = \frac{-11}{9}\)

Now, find the multiplicative inverse of each number:
- For \(\frac{1}{2}\):
\[
\text{Multiplicative inverse} = \frac{1}{\frac{1}{2}} = 2
\]

- For \(\frac{7}{3}\):
\[
\text{Multiplicative inverse} = \frac{1}{\frac{7}{3}} = \frac{3}{7}
\]

- For \(\frac{5}{3}\):
\[
\text{Multiplicative inverse} = \frac{1}{\frac{5}{3}} = \frac{3}{5}
\]

- For \(\frac{-37}{10}\):
\[
\text{Multiplicative inverse} = \frac{1}{\frac{-37}{10}} = -\frac{10}{37}
\]

- For \(\frac{-11}{9}\):
\[
\text{Multiplicative inverse} = \frac{1}{\frac{-11}{9}} = -\frac{9}{11}
\]

Final Answer:
\[
\boxed{2, \frac{3}{7}, \frac{3}{5}, -\frac{10}{37}, -\frac{9}{11}}
\]

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#### 3. Write the multiplicative inverse of numbers: \(0.5, 0.2, -1.2, 2.45, -3.05, 0.008\)

Convert decimals to fractions (if needed) and find the multiplicative inverse:
- For \(0.5\):
\[
0.5 = \frac{1}{2}, \quad \text{Multiplicative inverse} = \frac{1}{\frac{1}{2}} = 2
\]

- For \(0.2\):
\[
0.2 = \frac{1}{5}, \quad \text{Multiplicative inverse} = \frac{1}{\frac{1}{5}} = 5
\]

- For \(-1.2\):
\[
-1.2 = -\frac{6}{5}, \quad \text{Multiplicative inverse} = \frac{1}{-\frac{6}{5}} = -\frac{5}{6}
\]

- For \(2.45\):
\[
2.45 = \frac{245}{100} = \frac{49}{20}, \quad \text{Multiplicative inverse} = \frac{1}{\frac{49}{20}} = \frac{20}{49}
\]

- For \(-3.05\):
\[
-3.05 = -\frac{305}{100} = -\frac{61}{20}, \quad \text{Multiplicative inverse} = \frac{1}{-\frac{61}{20}} = -\frac{20}{61}
\]

- For \(0.008\):
\[
0.008 = \frac{8}{1000} = \frac{1}{125}, \quad \text{Multiplicative inverse} = \frac{1}{\frac{1}{125}} = 125
\]

Final Answer:
\[
\boxed{2, 5, -\frac{5}{6}, \frac{20}{49}, -\frac{20}{61}, 125}
\]

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#### 4. Is there a multiplicative inverse of zero?

The multiplicative inverse of a number \(x\) is defined as a number \(y\) such that:
\[
x \cdot y = 1
\]
If \(x = 0\), then:
\[
0 \cdot y = 1
\]
This equation has no solution because any number multiplied by zero is zero, not 1. Therefore, zero does not have a multiplicative inverse.

Final Answer:
\[
\boxed{\text{No}}
\]

---

#### 5. Write the additive inverse of numbers: \(4, 3, 1023, -2, -1, -100\)

The additive inverse of a number \(x\) is a number \(y\) such that:
\[
x + y = 0
\]
For any number \(x\), the additive inverse is given by:
\[
y = -x
\]

- For \(4\):
\[
\text{Additive inverse} = -4
\]

- For \(3\):
\[
\text{Additive inverse} = -3
\]

- For \(1023\):
\[
\text{Additive inverse} = -1023
\]

- For \(-2\):
\[
\text{Additive inverse} = -(-2) = 2
\]

- For \(-1\):
\[
\text{Additive inverse} = -(-1) = 1
\]

- For \(-100\):
\[
\text{Additive inverse} = -(-100) = 100
\]

Final Answer:
\[
\boxed{-4, -3, -1023, 2, 1, 100}
\]

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#### 6. Write the additive inverse of numbers: \(\frac{1}{2}, 1\frac{3}{7}, -2\frac{5}{6}, -\frac{4}{9}\)

First, convert mixed numbers to improper fractions:
- \(1\frac{3}{7} = \frac{7 \cdot 1 + 3}{7} = \frac{10}{7}\)
- \(-2\frac{5}{6} = \frac{-2 \cdot 6 - 5}{6} = \frac{-17}{6}\)

Now, find the additive inverse of each number:
- For \(\frac{1}{2}\):
\[
\text{Additive inverse} = -\frac{1}{2}
\]

- For \(\frac{10}{7}\):
\[
\text{Additive inverse} = -\frac{10}{7}
\]

- For \(\frac{-17}{6}\):
\[
\text{Additive inverse} = -\left(\frac{-17}{6}\right) = \frac{17}{6}
\]

- For \(\frac{-4}{9}\):
\[
\text{Additive inverse} = -\left(\frac{-4}{9}\right) = \frac{4}{9}
\]

Final Answer:
\[
\boxed{-\frac{1}{2}, -\frac{10}{7}, \frac{17}{6}, \frac{4}{9}}
\]

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#### 7. Write the additive inverse of numbers: \(0.3, 1.7, -1.3, -7.02, -0.143\)

The additive inverse of a number \(x\) is \(-x\):
- For \(0.3\):
\[
\text{Additive inverse} = -0.3
\]

- For \(1.7\):
\[
\text{Additive inverse} = -1.7
\]

- For \(-1.3\):
\[
\text{Additive inverse} = -(-1.3) = 1.3
\]

- For \(-7.02\):
\[
\text{Additive inverse} = -(-7.02) = 7.02
\]

- For \(-0.143\):
\[
\text{Additive inverse} = -(-0.143) = 0.143
\]

Final Answer:
\[
\boxed{-0.3, -1.7, 1.3, 7.02, 0.143}
\]

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#### 8. Is there an additive inverse of zero?

The additive inverse of a number \(x\) is defined as a number \(y\) such that:
\[
x + y = 0
\]
If \(x = 0\), then:
\[
0 + y = 0
\]
This equation is satisfied when \(y = 0\). Therefore, the additive inverse of zero is zero itself.

Final Answer:
\[
\boxed{\text{Yes}}
\]

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Final Summary of Answers:

1. \(\boxed{\frac{1}{10}, \frac{1}{5}, -\frac{1}{20}, -\frac{1}{40}, \frac{1}{100}, \frac{1}{7}}\)
2. \(\boxed{2, \frac{3}{7}, \frac{3}{5}, -\frac{10}{37}, -\frac{9}{11}}\)
3. \(\boxed{2, 5, -\frac{5}{6}, \frac{20}{49}, -\frac{20}{61}, 125}\)
4. \(\boxed{\text{No}}\)
5. \(\boxed{-4, -3, -1023, 2, 1, 100}\)
6. \(\boxed{-\frac{1}{2}, -\frac{10}{7}, \frac{17}{6}, \frac{4}{9}}\)
7. \(\boxed{-0.3, -1.7, 1.3, 7.02, 0.143}\)
8. \(\boxed{\text{Yes}}\)