In which set(s) of numbers would you find the number \sqrt{62} rational number irrational number natural number real number integer whole number
The Correct Answer and Explanation is:
The number 62\sqrt{62}62 is an irrational number and also a real number. Let’s break this down step by step:
- Irrational Number:
A rational number is one that can be expressed as the quotient of two integers (i.e., pq\frac{p}{q}qp, where ppp and qqq are integers and q≠0q \neq 0q=0). On the other hand, an irrational number cannot be written as such a fraction. The number 62\sqrt{62}62 is an irrational number because 62 is not a perfect square (i.e., there are no integers nnn such that n2=62n^2 = 62n2=62). Since its decimal expansion does not terminate or repeat, it cannot be expressed as a fraction. - Real Number:
Real numbers include both rational and irrational numbers, as well as integers, whole numbers, and natural numbers. Because 62\sqrt{62}62 is not imaginary, it is part of the real number set.
Now, let’s analyze the other sets mentioned:
- Natural Number:
Natural numbers are the set of positive integers used for counting, starting from 1. Since 62\sqrt{62}62 is not a whole number, it is not a natural number. - Whole Number:
Whole numbers are all the non-negative integers (0, 1, 2, 3, …). 62\sqrt{62}62 is not an integer, so it is not a whole number. - Integer:
Integers include all positive and negative whole numbers, as well as zero. Since 62\sqrt{62}62 is not an integer (it’s not a whole number and not negative), it is not an integer.
Conclusion:
The number 62\sqrt{62}62 is an irrational number and a real number, but not a natural number, whole number, or integer.
