# Prove the Square Root of 2 is irrational

by njkid
Tags: irrational, prove, root, square
 P: 666 You're off to a good start. Let "a" and "b" be natural numbers. $$\sqrt{2}=\frac{a}{b}\implies b\sqrt{2}=a$$ Now, how could "a" be a natural number? Well, "b" would be some multiple of $\sqrt{2}$. This in turn, would mean that "b" isn't a natural number. Can you see where this is going? You need to prove this by contradiction.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,345 Prove the Square Root of 2 is irrational "Now, how could "a" be a natural number? Well, "b" would be some multiple of $\sqrt{2}$. This in turn, would mean that "b" isn't a natural number." How does that follow? Saying b*(1/2) , for example, equals a natural number does not imply that b isn't a natural number! Of course 1/2 isn't an irrational number but the whole point here is to prove that $\sqrt{2}$ is irrational. Better to note that if $\frac{a}{b}= \sqrt{2}$ then, squaring both sides, $\frac{a^2}{b^2}= 2$ so that a2= 2b2 showing that a2 is even. Crucial point: the square of an odd integer is always odd: If p is an odd integer, then it can be written 2n+ 1 where n is any integer. p2= (2n+1)2= 4n2+ 4n+ 1= 2(2n2+2n)+1. Since 2n2+ 2n is an integer, p2 is of the form 2m+1 (m= 2n2+2n) and so is odd. Do you see why knowing that tells us that a must be even?
 Quote by HallsofIvy "Now, how could "a" be a natural number? Well, "b" would be some multiple of $\sqrt{2}$. This in turn, would mean that "b" isn't a natural number." How does that follow? Saying b*(1/2) , for example, equals a natural number does not imply that b isn't a natural number! Of course 1/2 isn't an irrational number but the whole point here is to prove that $\sqrt{2}$ is irrational. Better to note that if $\frac{a}{b}= \sqrt{2}$ then, squaring both sides, $\frac{a^2}{b^2}= 2$ so that a2= 2b2 showing that a2 is even. Crucial point: the square of an odd integer is always odd: If p is an odd integer, then it can be written 2n+ 1 where n is any integer. p2= (2n+1)2= 4n2+ 4n+ 1= 2(2n2+2n)+1. Since 2n2+ 2n is an integer, p2 is of the form 2m+1 (m= 2n2+2n) and so is odd. Do you see why knowing that tells us that a must be even?