Why is the square root of 23 an irrational number
The Correct Answer and Explanation is:
The square root of 23 is an irrational number because it cannot be expressed as a ratio of two integers. In other words, there are no two integers aaa and bbb such that 23=ab\sqrt{23} = \frac{a}{b}23=ba, where b≠0b \ne 0b=0 and the fraction is in simplest form.
A number is considered irrational if its decimal expansion neither terminates nor repeats. For 23\sqrt{23}23, the decimal form is approximately 4.79583152…, and it continues infinitely without forming any repeating pattern. This non-repeating, non-terminating nature confirms that the number is irrational.
The rational root theorem, often used in number theory, indicates that if a square root of a natural number results in a rational number, then the original number must be a perfect square. Examples of perfect squares include 1, 4, 9, 16, 25, and so on. These numbers have whole numbers as their square roots: 16=4\sqrt{16} = 416=4, 25=5\sqrt{25} = 525=5, etc. However, 23 is not a perfect square. It lies between 16 and 25, whose square roots are 4 and 5 respectively. Therefore, 23\sqrt{23}23 must be between 4 and 5, but not equal to either.
Furthermore, a proof by contradiction can demonstrate the irrationality of square roots of non-perfect square integers. Suppose 23\sqrt{23}23 were rational, and equal to ab\frac{a}{b}ba in simplest terms. Squaring both sides would yield 23=a2b223 = \frac{a^2}{b^2}23=b2a2, or a2=23b2a^2 = 23b^2a2=23b2. This implies that a2a^2a2 is divisible by 23, so aaa must also be divisible by 23. Letting a=23ka = 23ka=23k, and substituting back gives (23k)2=23b2(23k)^2 = 23b^2(23k)2=23b2, which leads to 529k2=23b2529k^2 = 23b^2529k2=23b2, or b2=23k2b^2 = 23k^2b2=23k2. This implies that bbb is also divisible by 23, contradicting the assumption that ab\frac{a}{b}ba is in lowest terms. Hence, 23\sqrt{23}23 must be irrational.
