Prove that square root of 2 is irrational using the principle of Mathematical Induction How do I prove that the square root of 2 is irrational using the principle of mathematical induction?
The usual proof starts out something like "suppose for a contradiction $\sqrt 2$ were rational, and write it as $p/q$ in lowest terms ...".

The "in lowest terms" hides an instance of induction, which you can unfold to get a proof in the shape of


  Theorem. For all positive integers $p$ and $q$, it holds that $(p/q)^2\ne 2$.
  
  Proof. By long induction on $q$. If $p$ and $q$ have a common factor $n&gt;1$ then $(\frac pq)^2 = (\frac{p/n}{q/n})^2$, and since $q/n&lt;q$, the induction hypothesis guarantees that $(\frac pq)^2\ne 2$. Now we consider the case that $p$ and $q$ are relatively prime ...
This is as simple as I could make it in my mind, working on the simply basis of multiple factors (similar to the above) but less algebraic manipulation: