<s>[INST] Prove that $\sqrt 5$ is irrational I have to  prove that $\sqrt 5$ is irrational. 
Proceeding as in the proof of $\sqrt 2$, let us assume that $\sqrt 5$ is rational. This means for some distinct integers $p$ and $q$ having no common factor other than 1, 
$$\frac{p}{q} = \sqrt5$$
$$\Rightarrow \frac{p^2}{q^2} = 5$$
$$\Rightarrow p^2 = 5 q^2$$
This means that 5 divides $p^2$. This means that 5 divides $p$ (because every factor must appear twice for the square to exist). So we have, $p = 5 r$ for some integer $r$. Extending the argument to $q$, we discover that they have a common factor of 5, which is a contradiction.
Is this proof correct?
 [/INST] Let us assume $√5$ is rational.
$$√5=\frac{x}{y}$$
Square both sides of the equation above
$$5 =\frac{x^2}{y^2}$$
Multiply both sides by $y^2$
$$5  y^2 =\frac{x^2 }{y^2}$$
We get $5  y^2 = x^2$
Another important concept before we finish our proof: Prime factorization.
Key question: is the number of prime factors for a number raised to the second power an even or odd number?
For example, $6^2$, $12^2$, and $15^2$
$6^2 = 6 × 6 = 2 × 3 × 2 × 3$ ($4$ prime factors, so even number)
$12^2 = 12 × 12 = 4 × 3 × 4 × 3 = 2 × 2 × 3 × 2 × 2 × 3$ ($6$ prime factors, so even number)
$15^2 = 15 × 15 = 3 × 5 × 3 × 5$ ($4$ prime factors, so even number)
There is a solid pattern here to conclude that any number squared will have an even number of prime factors
In order words, $x^2$ has an even number of prime factors.
Let's finish the proof then!
$5  y^2 = x^2$
Since $5  y^2$ is equal to $x^2$, $5  y^2$ and $x^2$ must have the same number of prime factors.
We just showed that
$x^2$ has an even number of prime factors, $y^2$ has also an even number of prime factors.
$5  y^2$ will then have an odd number of prime factors.
The number $5$ counts as $1$ prime factor, so $1$ + an even number of prime factors is an odd number of prime factors.
$5  y^2$ is the same number as $x^2$. However, $5  y^2$ gives an odd number of prime factor while $x^2$ gives an even number of prime factors.
This is a contradiction since a number cannot have an odd number of prime factors and an even number of prime factors at the same time.
The assumption that square root of $5$ is rational is wrong. Therefore, square of $5$ is irrational.
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