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Prove that for any prime positive integer $p$, $\sqrt{p}$ is an irrational number.
Given: A positive integer $p$.
To prove: Here we have to prove that for any prime positive integer $p$, $\sqrt{p}$ is an irrational number.
Solution:
Let us assume, to the contrary, that $\sqrt{p}$ is rational.
So, we can find integers $a$ and $b$ ($≠$ 0) such that $\sqrt{p}\ =\ \frac{a}{b}$.
Where $a$ and $b$ are co-prime.
Now,
$\sqrt{p}\ =\ \frac{a}{b}$
Squaring both sides:
$(\sqrt{p})^2\ =\ (\frac{a}{b})^2$
$p\ =\ \frac{a^2}{b^2}$
$pb^2\ =\ a^2$ ...(1)
Therefore, $p$ divides $a^2$. Which implies that $p$ divides $a$ also. So, we can write $a\ =\ pc$ for some integer $c$.
$a\ =\ pc$
Squaring both sides:
$a^2\ =\ p^2c^2$
Putting the value of $a^2$ from eq (1):
$pb^2\ =\ p^2c^2$
$b^2\ =\ pc^2$
Therefore, $p$ divides $b^2$. Which implies that $p$ divides $b$ also.
So, $a$ and $b$ have at least $p$ 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 $\sqrt{p}$ is rational.
So, we can conclude that $\sqrt{p}$ is irrational.
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