Given A = {0.(8), Sqrt(0.(1)), Sqrt(3), Sqrt(25)/5, 4.(123)}, How Many Elements Are In The Intersection Of A And The Set Of Irrational Numbers (R \ Q)?

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In the realm of mathematics, understanding the nature of numbers and their classifications is fundamental. Numbers can be categorized into various sets, such as rational numbers (Q) and real numbers (R). The set of irrational numbers is represented as R \ Q, which means all real numbers that are not rational. This article delves into the intersection of a specific set A with the set of irrational numbers. Set A is defined as A = {0.(8), sqrt(0.(1)), sqrt(3), sqrt(25)/5, 4.(123)}. Our goal is to determine the number of elements in the set A ∩ R \ Q, which represents the elements that are both in set A and are irrational numbers. This exploration will involve converting repeating decimals to fractions, simplifying square roots, and classifying each element in set A as either rational or irrational. Through this process, we will gain a deeper understanding of number classification and set theory, crucial concepts in mathematical analysis.

Defining Set A and Its Elements

To begin our exploration, let's define set A and examine its elements individually. Set A is given as A = {0.(8), sqrt(0.(1)), sqrt(3), sqrt(25)/5, 4.(123)}. Each element in this set represents a specific number, and to determine the intersection with the set of irrational numbers, we must classify each element as either rational or irrational. This classification involves understanding the properties of rational and irrational numbers and applying mathematical principles to simplify and analyze each element.

1. 0.(8)

The first element in set A is 0.(8), which is a repeating decimal. To classify this number, we need to convert it into a fraction. Repeating decimals can be expressed as rational numbers because they represent a ratio of two integers. Let x = 0.(8). This means x = 0.8888... To convert this repeating decimal to a fraction, we multiply x by 10, giving us 10x = 8.8888... Now, we subtract the original equation from this new equation: 10x - x = 8.8888... - 0.8888..., which simplifies to 9x = 8. Solving for x, we get x = 8/9. Since 8/9 is a fraction where both the numerator and the denominator are integers, 0.(8) is a rational number. Therefore, 0.(8) ∉ R \ Q.

2. √(0.(1))

The second element is √(0.(1)), the square root of a repeating decimal. First, we need to convert 0.(1) into a fraction. Let y = 0.(1), which means y = 0.1111... Multiplying by 10, we get 10y = 1.1111... Subtracting the original equation, 10y - y = 1.1111... - 0.1111..., simplifies to 9y = 1. Solving for y, we get y = 1/9. Now, we can rewrite the element as √(1/9). The square root of 1/9 is 1/3, since √(1/9) = √1 / √9 = 1/3. The number 1/3 is a fraction of two integers, thus it is a rational number. Hence, √(0.(1)) ∉ R \ Q.

3. √3

The third element is √3, the square root of 3. To determine if √3 is rational or irrational, we need to consider whether 3 is a perfect square. A perfect square is an integer that can be expressed as the square of another integer. Since 3 is not a perfect square, its square root cannot be expressed as a simple fraction. The square root of 3 is a non-repeating, non-terminating decimal, which means it is an irrational number. Thus, √3 ∈ R \ Q.

4. √(25)/5

The fourth element is √(25)/5. We first simplify the square root of 25, which is 5. So, the element becomes 5/5, which simplifies to 1. The number 1 can be expressed as a fraction 1/1, where both the numerator and denominator are integers. Therefore, 1 is a rational number. Consequently, √(25)/5 ∉ R \ Q.

5. 4.(123)

The fifth element is 4.(123), a repeating decimal. To convert this to a fraction, let z = 4.(123), which means z = 4.123123123... We multiply by 1000 to shift the repeating part to the left of the decimal point: 1000z = 4123.123123... Subtracting the original equation, 1000z - z = 4123.123123... - 4.123123..., simplifies to 999z = 4119. Solving for z, we get z = 4119/999. Since 4119/999 is a fraction with integer numerator and denominator, 4.(123) is a rational number. Therefore, 4.(123) ∉ R \ Q.

Identifying Elements in A ∩ R \ Q

Now that we have classified each element in set A, we can identify the elements that are also in the set of irrational numbers (R \ Q). We found that:

  • 0.(8) is rational.
  • √(0.(1)) is rational.
  • √3 is irrational.
  • √(25)/5 is rational.
  • 4.(123) is rational.

Therefore, the only element in set A that is also an irrational number is √3. This means the intersection of set A and the set of irrational numbers, A ∩ R \ Q, contains only one element.

Determining the Number of Elements in A ∩ R \ Q

Based on our analysis, the set A ∩ R \ Q contains only the element √3. Therefore, the number of elements in A ∩ R \ Q is 1. This corresponds to option B in the original question. The process of classifying numbers as rational or irrational is crucial in various mathematical contexts, including algebra, calculus, and real analysis. Understanding these classifications allows for precise mathematical reasoning and problem-solving.

Conclusion

In conclusion, by examining each element of set A and classifying it as either rational or irrational, we determined that the intersection of A and the set of irrational numbers (R \ Q) contains only one element, √3. This exercise demonstrates the importance of understanding number classifications and how to manipulate and simplify mathematical expressions to reveal their nature. The ability to distinguish between rational and irrational numbers is a fundamental skill in mathematics, enabling further exploration of more complex mathematical concepts and problems. Through this detailed analysis, we have not only answered the specific question but also reinforced the underlying principles of number theory and set theory.