Taken from pg-53 of Roger Penrose's road to reality,
Suppose $\sqrt{2}$ is rational, then:
$$ \sqrt{2} = \frac{p}{q}$$
Where $p$ and $q$ are some integers with $ q \neq 0$:
Squaring and rearrangnig:
$$ 2 q^2 = p^2$$
This means $p$ is even, and hence, $p=2p_1$, meaning that:
$$ q^2 = 2p_1^2$$
This means that $q$ is even, we can put $q=2q_1$, where $q_1$ is some integer, meaning that:
$$ 2q_1^2 = p_1^2$$
See that this looks same as the equation we begun with $2q^2 =p^2$, repeating the procedure we done it again we can further find that $p_1$ and $q_1$ is again even writing $p_1 = 2p_2$ and $q_1=2q_2$, after simplifications we will find again:
$$ 2q_2^2 = p_2^2$$
Repeating the procedure many times, we could repeat this as many times as we want , we get a recurrence for $q_i$s as:
$$q_i = 2q_{i+1}$$
Or, $$\frac{q_i}{2} = q_{i+1}$$
Obeying:
$$q>q_1>q_2...$$
Penrose writes:
"..All of these integers being positive. But any decreasing sequence of positive integers must come to an end, contradicting the fact that this sequence is unending. This provides us with a contradiction to what has been supposed , namely that there is a rational number which squares to $2$"
Firstly where was it assumed that the sequence is 'unending' and what exactly does it mean for a sequence to end? I thought sequence could go on forever..
P.S: I understand the square root of two's irrationality proof already, the point is not to prove the statement in itself but understand this particular proof.
