√2 is an example of which type of number?

√2 is an example of which type of number?

• A. Irrational Number
• B. Rational Number
• C. Natural Numbers
• D. Integers

Answer: (A)

The key idea is whether a square root can be written as a ratio of two integers. For sqrt(2), there is no way to express it as p/q with integers p and q (q ≠ 0) without breaking the rules of whole-number ratios. A classic argument shows why: suppose sqrt(2) were equal to p/q in lowest terms. Then 2 q^2 = p^2, so p^2 is even, which means p is even. Let p = 2k; substituting gives 2 q^2 = (2k)^2 = 4k^2, so q^2 = 2k^2, which makes q even as well. Now p and q share a factor of 2, contradicting the assumption that the fraction was in lowest terms. This contradiction means sqrt(2) cannot be written as a fraction of integers, so it is not a rational number. Since natural numbers and integers are all rational (they can be expressed as fractions with denominator 1, or as integers themselves), sqrt(2) is not a natural number or an integer either. It sits outside the rational numbers as an irrational number.

The key idea is whether a square root can be written as a ratio of two integers. For sqrt(2), there is no way to express it as p/q with integers p and q (q ≠ 0) without breaking the rules of whole-number ratios. A classic argument shows why: suppose sqrt(2) were equal to p/q in lowest terms. Then 2 q^2 = p^2, so p^2 is even, which means p is even. Let p = 2k; substituting gives 2 q^2 = (2k)^2 = 4k^2, so q^2 = 2k^2, which makes q even as well. Now p and q share a factor of 2, contradicting the assumption that the fraction was in lowest terms. This contradiction means sqrt(2) cannot be written as a fraction of integers, so it is not a rational number.

Since natural numbers and integers are all rational (they can be expressed as fractions with denominator 1, or as integers themselves), sqrt(2) is not a natural number or an integer either. It sits outside the rational numbers as an irrational number.