let a = sqrt(2) and b = sqrt(2) - 1. Note that b < 1, so multiplying by b makes a number smaller.

put K = { k in N | k a in N }. So this set contains all the natural numbers that you can multiply by sqrt(2) and get a natural number. If sqrt(2) is rational, i.e. a = p/q then q is in K. So K is nonempty. So it is valid to take the minimum of the set.

Let k be the minimum.

k a is in N

k b a in N by the line of algebra.

The implication now is that there we can find a smaller element of K. The number is x = k b. This is a natural number by the same line of algebra.

So the fact that x is in K and x < k completes the proof by contradiction.

Not keen on the way the proof was written on the blackboard. Makes it much harder to understand that necessary.

My favorite proof of irrationality looks like this:

oddnumber of prime factors.let a = sqrt(2) and b = sqrt(2) - 1. Note that b < 1, so multiplying by b makes a number smaller.

put K = { k in N | k a in N }. So this set contains all the natural numbers that you can multiply by sqrt(2) and get a natural number. If sqrt(2) is rational, i.e. a = p/q then q is in K. So K is nonempty. So it is valid to take the minimum of the set.

Let k be the minimum.

The implication now is that there we can find a smaller element of K. The number is x = k b. This is a natural number by the same line of algebra.

So the fact that x is in K and x < k completes the proof by contradiction.

Not keen on the way the proof was written on the blackboard. Makes it much harder to understand that necessary.

This is extremely confusing. Badly explained.

You can find proofs of irrationality of sqrt2 (or sqrtp) for various formal provers here https://www.cs.ru.nl/~freek/comparison/comparison.pdf