The question of whether √5 and √2
are irrational numbers is a common one in mathematics, and both can be proved to be irrational using similar methods. Here’s an explanation for √5:
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Prove that $$\sqrt{5}$$ is irrational.
Sol. Let us suppose that $$\sqrt{5}$$ is a rational number.
$$\sqrt{5}=\frac{\mathrm{a}}{\mathrm{b}}$$, where a and b are co-prime and $$\mathrm{b} \neq 0$$
Squaring on both sides, we get
Now,
$$
\begin{aligned}
(\sqrt{5})^2 & =\left(\frac{a}{b}\right)^2 \\
5 b^2 & =a^2….(1)\end{aligned}
$$
5 is a factor of $$\mathrm{a}^2$$
5 is a factor of a
Let
Substituting
$$\mathrm{a}=5 \mathrm{c}$$, where c is some integer.
$$a=5 \mathrm{c}$$ in (i), we get
$$
\begin{aligned}
5 b^2 & =(5 c)^2 \\
5 b^2 & =25 c^2 \\
b^2 & =5 c^2
\end{aligned}
$$
5 is a factor of $$\mathrm{b}^2$$
5 is factor of b
From eq. (ii) and (iii), we get
5 is a common factor of a and b
This contradicts the fact that a and b are co-prime. So, our assumption is wrong.
Hence, $$\sqrt{5}$$ is irrational.
