Editing Square root of 2 (section)

===Proof using reciprocals===
Assume by way of contradiction that <math>\sqrt 2</math> were rational. Then we may write <math>\sqrt 2 + 1 = \frac{q}{p}</math> as an irreducible fraction in lowest terms, with coprime positive integers <math>q>p</math>. Since <math>(\sqrt 2-1)(\sqrt 2+1)=2-1^2=1</math>, it follows that <math>\sqrt 2-1</math> can be expressed as the irreducible fraction <math>\frac{p}{q}</math>. However, since <math>\sqrt 2-1</math> and <math>\sqrt 2+1</math> differ by an integer, it follows that the denominators of their irreducible fraction representations must be the same, i.e. <math>q=p</math>. This gives the desired contradiction.