Irrationallem

James' Proof

Statement: Using the numbers $\sqrt{2}$ and $\sqrt{2}$$\sqrt{2} prove that there exist irrational numbers a,b such that ab is rational. Proof: We must first identify \sqrt{2} as an irrational number. We then consider two cases: 1) \sqrt{2}$$\sqrt{2}$ is rational, or 2) $\sqrt{2}$$\sqrt{2} is irrational. For the first case, let \sqrt{2}$$\sqrt{2}$ = cd when c = $\sqrt{2}$ and d = $\sqrt{2}$. Since $\sqrt{2}$ is an irrational number and $\sqrt{2}$$\sqrt{2} is rational, the statement is satisfied. Let us consider the second case. When a = \sqrt{2}$$\sqrt{2}$ and b = $\sqrt{2}$, ab = ($\sqrt{2}$$\sqrt{2})\sqrt{2} which is rational. Since \sqrt{2} is an irrational number and \sqrt{2}$$\sqrt{2}$ is an irrational number, the statement is also satisfied.

page revision: 2, last edited: 25 Jan 2018 19:47