Answer :
Certainly! Let's go through each part of the problem step-by-step.
### Part (a): Determining if the decimals are terminating or non-terminating
To determine if the given rational numbers [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] have terminating decimals, we need to check the prime factors of their denominators in their simplest form:
1. [tex]$\frac{1}{2}$[/tex]:
- The denominator is [tex]$2$[/tex]. The prime factorization of [tex]$2$[/tex] is just [tex]$2$[/tex].
- Since the denominator only has the prime factor [tex]$2$[/tex], [tex]$\frac{1}{2}$[/tex] has a terminating decimal.
2. [tex]$\frac{2}{5}$[/tex]:
- The denominator is [tex]$5$[/tex]. The prime factorization of [tex]$5$[/tex] is just [tex]$5$[/tex].
- Since the denominator only has the prime factor [tex]$5$[/tex], [tex]$\frac{2}{5}$[/tex] has a terminating decimal.
Conclusion: Both [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] have terminating decimals.
### Part (b): Finding a rational number between [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex]
We can find a rational number between [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] by averaging them:
[tex]\[ \text{Rational between } \frac{1}{2} \text{ and } \frac{2}{5} = \frac{\frac{1}{2} + \frac{2}{5}}{2} \][/tex]
First, find a common denominator to add the fractions:
[tex]\[ \frac{1}{2} = \frac{5}{10}, \quad \frac{2}{5} = \frac{4}{10} \][/tex]
So,
[tex]\[ \frac{1}{2} + \frac{2}{5} = \frac{5}{10} + \frac{4}{10} = \frac{9}{10} \][/tex]
Now, average it:
[tex]\[ \frac{\frac{9}{10}}{2} = \frac{9}{10} \div 2 = \frac{9}{10} \times \frac{1}{2} = \frac{9}{20} \][/tex]
Thus, the rational number between [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] is:
[tex]\[ \frac{9}{20} \][/tex]
### Part (c): Additive inverse of [tex]$\frac{1}{2}$[/tex] and multiplicative inverse of [tex]$\frac{2}{5}$[/tex]
1. Additive Inverse of [tex]$\frac{1}{2}$[/tex]:
The additive inverse of a number is the number that, when added to the original number, results in zero. For [tex]$\frac{1}{2}$[/tex], the additive inverse is:
[tex]\[ -\frac{1}{2} \][/tex]
2. Multiplicative Inverse of [tex]$\frac{2}{5}$[/tex]:
The multiplicative inverse of a number is the number that, when multiplied by the original number, results in one. For [tex]$\frac{2}{5}$[/tex], the multiplicative inverse is:
[tex]\[ \frac{5}{2} \][/tex]
### Part (d): Verifying the closure properties of addition and multiplication
1. Addition:
We need to verify whether the sum of [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] is also a rational number.
[tex]\[ \frac{1}{2} + \frac{2}{5} = \frac{5}{10} + \frac{4}{10} = \frac{9}{10} \][/tex]
The result, [tex]$\frac{9}{10}$[/tex], is a rational number, so rationals are closed under addition.
2. Multiplication:
We need to verify whether the product of [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] is also a rational number.
[tex]\[ \frac{1}{2} \times \frac{2}{5} = \frac{1 \times 2}{2 \times 5} = \frac{2}{10} = \frac{1}{5} \][/tex]
The result, [tex]$\frac{1}{5}$[/tex], is a rational number, so rationals are closed under multiplication.
### Conclusion
- The decimals of [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] are both terminating.
- A rational number between [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] is [tex]$\frac{9}{20}$[/tex].
- The additive inverse of [tex]$\frac{1}{2}$[/tex] is [tex]$-\frac{1}{2}$[/tex], and the multiplicative inverse of [tex]$\frac{2}{5}$[/tex] is [tex]$\frac{5}{2}$[/tex].
- Using the given rational numbers, we verified that addition and multiplication of rational numbers follow the closure property. The sum is [tex]$\frac{9}{10}$[/tex], and the product is [tex]$\frac{1}{5}$[/tex], both of which are rational numbers.
### Part (a): Determining if the decimals are terminating or non-terminating
To determine if the given rational numbers [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] have terminating decimals, we need to check the prime factors of their denominators in their simplest form:
1. [tex]$\frac{1}{2}$[/tex]:
- The denominator is [tex]$2$[/tex]. The prime factorization of [tex]$2$[/tex] is just [tex]$2$[/tex].
- Since the denominator only has the prime factor [tex]$2$[/tex], [tex]$\frac{1}{2}$[/tex] has a terminating decimal.
2. [tex]$\frac{2}{5}$[/tex]:
- The denominator is [tex]$5$[/tex]. The prime factorization of [tex]$5$[/tex] is just [tex]$5$[/tex].
- Since the denominator only has the prime factor [tex]$5$[/tex], [tex]$\frac{2}{5}$[/tex] has a terminating decimal.
Conclusion: Both [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] have terminating decimals.
### Part (b): Finding a rational number between [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex]
We can find a rational number between [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] by averaging them:
[tex]\[ \text{Rational between } \frac{1}{2} \text{ and } \frac{2}{5} = \frac{\frac{1}{2} + \frac{2}{5}}{2} \][/tex]
First, find a common denominator to add the fractions:
[tex]\[ \frac{1}{2} = \frac{5}{10}, \quad \frac{2}{5} = \frac{4}{10} \][/tex]
So,
[tex]\[ \frac{1}{2} + \frac{2}{5} = \frac{5}{10} + \frac{4}{10} = \frac{9}{10} \][/tex]
Now, average it:
[tex]\[ \frac{\frac{9}{10}}{2} = \frac{9}{10} \div 2 = \frac{9}{10} \times \frac{1}{2} = \frac{9}{20} \][/tex]
Thus, the rational number between [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] is:
[tex]\[ \frac{9}{20} \][/tex]
### Part (c): Additive inverse of [tex]$\frac{1}{2}$[/tex] and multiplicative inverse of [tex]$\frac{2}{5}$[/tex]
1. Additive Inverse of [tex]$\frac{1}{2}$[/tex]:
The additive inverse of a number is the number that, when added to the original number, results in zero. For [tex]$\frac{1}{2}$[/tex], the additive inverse is:
[tex]\[ -\frac{1}{2} \][/tex]
2. Multiplicative Inverse of [tex]$\frac{2}{5}$[/tex]:
The multiplicative inverse of a number is the number that, when multiplied by the original number, results in one. For [tex]$\frac{2}{5}$[/tex], the multiplicative inverse is:
[tex]\[ \frac{5}{2} \][/tex]
### Part (d): Verifying the closure properties of addition and multiplication
1. Addition:
We need to verify whether the sum of [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] is also a rational number.
[tex]\[ \frac{1}{2} + \frac{2}{5} = \frac{5}{10} + \frac{4}{10} = \frac{9}{10} \][/tex]
The result, [tex]$\frac{9}{10}$[/tex], is a rational number, so rationals are closed under addition.
2. Multiplication:
We need to verify whether the product of [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] is also a rational number.
[tex]\[ \frac{1}{2} \times \frac{2}{5} = \frac{1 \times 2}{2 \times 5} = \frac{2}{10} = \frac{1}{5} \][/tex]
The result, [tex]$\frac{1}{5}$[/tex], is a rational number, so rationals are closed under multiplication.
### Conclusion
- The decimals of [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] are both terminating.
- A rational number between [tex]$\frac{1}{2}$[/tex] and [tex]$\frac{2}{5}$[/tex] is [tex]$\frac{9}{20}$[/tex].
- The additive inverse of [tex]$\frac{1}{2}$[/tex] is [tex]$-\frac{1}{2}$[/tex], and the multiplicative inverse of [tex]$\frac{2}{5}$[/tex] is [tex]$\frac{5}{2}$[/tex].
- Using the given rational numbers, we verified that addition and multiplication of rational numbers follow the closure property. The sum is [tex]$\frac{9}{10}$[/tex], and the product is [tex]$\frac{1}{5}$[/tex], both of which are rational numbers.
