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Prove that $\sqrt5$ is irrational.
To find:
We have to prove that $\sqrt5$ is irrational.
Solution:
We know that,
If $p$ is a prime number and if $p$ divides $a^2$, then $p$ divides $a$, where $a$ is a positive integer.
Now,
Let us assume, to the contrary, that $\sqrt5$ is rational.
So, we can find integers $a$ and $b (≠ 0)$ such that $\sqrt5= \frac{a}{b}$.
Where $a$ and $b$ are co-prime.
⇒ $(\sqrt5)^2= (\frac{a}{b})^2$
⇒ $5 = \frac{a^{2}}{b^{2}}$
⇒ $5b^2 =a^2$
Therefore, $5$ divides $a^2$
This implies,
$5$ divides $a$.
So, we can write $a = 5c$ for some integer $c$.
⇒ $a^2 = 25c^2$
⇒ $5b^2 = 25c^2$ (Using, $5b^2= a^2$)
⇒ $b^2 = 5c^2$
Therefore, 5 divides $b^2$.
This implies,
$5$ divides $b$.
Therefore, $a$ and $b$ have at least $5$ as a common factor.
But this contradicts the fact that $a$ and $b$ have no common factors other than 1.
This contradiction has arisen because of our incorrect assumption that $\sqrt5$ is rational.
So, we conclude that $\sqrt5$ is irrational.
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