1. 9
lawrencecpaulson.github.io
1. 8

My favorite proof of irrationality looks like this:

• assume sqrt(2) = a/b, where a and b are natural numbers. Then 2b² = a².
• squaring doubles the prime factors of a number, so both a² and b² have an even number of prime factors
• but then 2b² has an odd number of prime factors.
• so there’s no way that equality could possibly hold.
• sqrt(2) is irrational.
2. 4

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.

3. 2

This is extremely confusing. Badly explained.

4. 2

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