Question

Prove that $\sqrt{3}$ is irrational.

Answer 1

Hint: Consider $\sqrt{3}$ as rational. Prove that they are not coprime i.e. they have common factors other than 1. Take, $\sqrt{3}=\dfrac{a}{b}$, where a and b are co-prime. Finally, you will get 3 divides ${{a}^{2}}$and 3 divides ${{b}^{2}}$.

“Complete step-by-step answer:”
We have to prove that $\sqrt{3}$ is irrational.
Let us assume the opposite that $\sqrt{3}$ is rational.
Hence, $\sqrt{3}$can be written in the form $\dfrac{a}{b}$.
Where a and b$\left( b\ne 0 \right)$ are co-prime, they have no common factor other than 1.
Hence, $\sqrt{3}=\dfrac{a}{b}$
$\Rightarrow a=\sqrt{3}b$
Now, squaring on both sides

& {{\left( \sqrt{3}b \right)}^{2}}={{a}^{2}} \\\ & 3{{b}^{2}}={{a}^{2}} \\\ & \therefore \dfrac{{{a}^{2}}}{3}={{b}^{2}} \\\ \end{aligned}$$ Hence, 3 divides $${{a}^{2}}$$. By theorem: If p is a prime number, and p divides $${{a}^{2}}$$, then p divides a, where a is a positive number. $$\therefore $$3 also divides a …………………………….(1) Hence, we can say, $$\dfrac{a}{3}=c$$, where c is some integer. $$\therefore a=3c$$ We know, a=3c and $$3{{b}^{2}}={{a}^{2}}$$ Substituting the value of a $$\begin{aligned} & 3{{b}^{2}}={{\left( 3c \right)}^{2}} \\\ & \Rightarrow 3{{b}^{2}}=9{{c}^{2}} \\\ & \therefore {{b}^{2}}=3{{c}^{2}} \\\ & \dfrac{{{b}^{2}}}{3}={{c}^{2}} \\\ \end{aligned}$$ Hence, 3 divides $${{b}^{2}}$$. By theorem: If p is a prime number and p divides $${{b}^{2}}$$, then p divides b, where b is a positive number. $$\therefore $$3 also divides b …………………………...(2) By (1) and (2) $$\Rightarrow 3$$ divides both a & b. Hence, 3 is a factor of 3. $$\therefore $$ a and b are not coprime as they have a factor 3 other than 1. $$\therefore $$ Hence, our assumptions are wrong. $$\therefore $$$$\sqrt{3}$$ is irrational. Note: By proving that 3 divides both $${{a}^{2}}$$and $${{b}^{2}}$$, we come to the conclusion that they are divisible by 3 other than 1. So, they are co-prime.