tisanurnr9c

2023-02-24

Prove that $3$ is an irrational number.

Jayden Landry

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

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