Prove that $\sqrt{5}$ is irrational.

(N/A) Assume that $\sqrt{5}$ is a rational number.
Therefore,we can find two integers $a$ and $b$ $(b \neq 0)$ such that $\sqrt{5} = \frac{a}{b}$,where $a$ and $b$ are co-prime (i.e.,their only common factor is $1$).
Squaring both sides,we get $a^2 = 5b^2$.
This implies that $a^2$ is divisible by $5$,and by the Fundamental Theorem of Arithmetic,$a$ is also divisible by $5$.
Let $a = 5k$ for some integer $k$.
Substituting this into the equation $a^2 = 5b^2$,we get $(5k)^2 = 5b^2$,which simplifies to $25k^2 = 5b^2$,or $b^2 = 5k^2$.
This implies that $b^2$ is divisible by $5$,and consequently,$b$ is also divisible by $5$.
Since both $a$ and $b$ are divisible by $5$,they have a common factor of $5$,which contradicts our initial assumption that $a$ and $b$ are co-prime.
Therefore,our assumption that $\sqrt{5}$ is rational is false,and we conclude that $\sqrt{5}$ is irrational.