tisanurnr9c

2023-02-24

Prove that $3$ is an irrational number.

Jayden Landry

Beginner2023-02-25Added 7 answers

Let us suppose that $\sqrt{3}$ is a rational number.

Then there are positive integers a and b such that $\sqrt{3}=\frac{a}{b}$, where a and b are co-prime, meaning their HCF is 1.

$3=\frac{a}{b}\Rightarrow 3=\frac{{a}^{2}}{{b}^{2}}\Rightarrow 3{b}^{2}={a}^{2}\Rightarrow 3divides{a}^{2}[\because 3divides3{b}^{2}]\Rightarrow 3dividesa.....................\left(i\right)\Rightarrow a=3cforsomeintegerc\Rightarrow {a}^{2}=9{c}^{2}\Rightarrow 3{b}^{2}=9{c}^{2}[\because {a}^{2}=3{b}^{2}]\Rightarrow {b}^{2}=3{c}^{2}\Rightarrow 3divides{b}^{2}[\because 3divides3{c}^{2}]\Rightarrow 3dividesb..............................\left(ii\right)$

We can see that a and b share at least 3 as a common factor from (1) and (2).

The fact that a and b are co-prime, on the other hand, contradicts this and indicates that our hypothesis is incorrect.

Therefore, $\sqrt{3}$ is an irrational number.

Then there are positive integers a and b such that $\sqrt{3}=\frac{a}{b}$, where a and b are co-prime, meaning their HCF is 1.

$3=\frac{a}{b}\Rightarrow 3=\frac{{a}^{2}}{{b}^{2}}\Rightarrow 3{b}^{2}={a}^{2}\Rightarrow 3divides{a}^{2}[\because 3divides3{b}^{2}]\Rightarrow 3dividesa.....................\left(i\right)\Rightarrow a=3cforsomeintegerc\Rightarrow {a}^{2}=9{c}^{2}\Rightarrow 3{b}^{2}=9{c}^{2}[\because {a}^{2}=3{b}^{2}]\Rightarrow {b}^{2}=3{c}^{2}\Rightarrow 3divides{b}^{2}[\because 3divides3{c}^{2}]\Rightarrow 3dividesb..............................\left(ii\right)$

We can see that a and b share at least 3 as a common factor from (1) and (2).

The fact that a and b are co-prime, on the other hand, contradicts this and indicates that our hypothesis is incorrect.

Therefore, $\sqrt{3}$ is an irrational number.