Answer to Question #131380 in Algebra for Angel Harmon

Question

Find the square root of 7. Does it repeat? does it end? Is it a irrational or rational number?

Answer 1

If we use a calculator, we'll obtain $\sqrt{7} \approx 2.64575131106459.$

Let us first determine if the square root of 7 is rational or irrational. Let us assume it is rational, so

$\sqrt{7} = \dfrac{m}{n} \Rightarrow 7 = \dfrac{m^2}{n^2} \Rightarrow m^2 = 7n^2.$ m and n are integers, so $m^2$ is divisible by 7, so m is divisible by 7. Therefore,

$(7k)^2 = 7n^2 \Rightarrow 7k^2 = n^2.$ So n should be divisible by 7, and so on. We will obtain the same equality, but m and n can't decrease infinitely. Therefore, our assumption is incorrect, so $\sqrt{7}$ is irrational.

If the fraction ends, we can write it in the form $a_0.a_1a_2a_3\ldots a_n$ . But we can transform it into a rational number

$a_0.a_1a_2a_3\ldots a_n = \dfrac{a_0a_1a_2a_3\ldots a_n}{10^n}$ . However, we proved that $\sqrt{7}$ is irrational, so it can't be represented as a finite fraction.

If the fraction repeats, we call it periodical. Let it have a period of n numbers: $\sqrt{7} = X = a_0.(a_1\ldots a_n)$ . Let us prove it is a rational number.

$10^nX - X = a_0a_1\ldots a_n.(a_1\ldots a_n) - a_0.(a_1\ldots a_n) = a_0a_1\ldots a_n - a_0.$

$X(10^n-1) = a_0a_1\ldots a_n - a_0 \Longrightarrow X = \dfrac{a_0a_1\ldots a_n - a_0 }{10^n-1}.$

So our X is rational. However, we proved that $\sqrt{7}$ is irrational, so it can't be represented as a periodical fraction.