1

So I find solving proving irrationals with even numbers inside the square root like $\sqrt{2}$ easy which gives us even q and even p but when it comes to odd square roots or just straight out unknowns like $\sqrt{pq}$(yes they are 2 distinct prime numbers) i follow the same procedure of $\frac{q^2}{p^2}$ then i stop there and i dont see anything to follow up with I want to know if there is a general rule of thumb I can use to prove irrationality by contradiction

oma
  • 81
  • There isn't a "general rule of thumb" for proving irrationality. It is difficult. However, square roots are a special case, and they are easier (the classic $\sqrt2$ proof works almost without changes, using $p\mid n^2\implies p\mid n$). For your specific proof, if you provide us some more details, it will be much easier for us to understand where you are stuck and what you should do next. – Arthur May 09 '19 at 04:44
  • Proving irrationality is notoriously difficult. For example here is a link with several proofs of the irrationality of $\pi$ https://en.wikipedia.org/wiki/Proof_that_%CF%80_is_irrational

    How many can you follow? Yes, there are some easy ones such as $\sqrt 2$ but these are the exceptions.

     – user317176 May 09 '19 at 05:41

1 Answers1

1

An alternative method to showing irrationality of roots, assuming you have proved this already, is to consider the Rational Roots Theorem

For example, let $a$ be a non-perfect square natural number. Then $\sqrt{a}$ is definitely a root of the polynomial $x^2 - a = 0$

Let's assume that $\sqrt{a}$ is rational.

The Rational Roots Theorem states that any possible rational roots of this polynomial must be a divisor of $a$ (positive or negative). Once you show that none of these options are roots, we must have that $\sqrt{a}$ isn't a rational root, a contradiction

WaveX
  • 5,440