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

Last updated date: 13th Jul 2024
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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 ${{b}^{2}}$.

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
\begin{align} & {{\left( \sqrt{3}b \right)}^{2}}={{a}^{2}} \\ & 3{{b}^{2}}={{a}^{2}} \\ & \therefore \dfrac{{{a}^{2}}}{3}={{b}^{2}} \\ \end{align}
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{align} & 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{align}
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.