## How to prove square root of 2 is irrational (Constructive Approach)

In our previous post, we discussed how to prove that the square root of 2 is irrational, using a proof by contradiction.

There is a less well-known proof that is a direct constructive approach to proving that the square root of 2 is irrational!

We consider an arbitrary rational number $\displaystyle\frac{a}{b}$, and show that the difference between $\sqrt{2}$ and $\displaystyle\frac{a}{b}$ cannot be zero. Hence, the square root of 2 cannot be rational.

Firstly, we have:

$\displaystyle|\sqrt{2}-\frac{a}{b}|=|\frac{\sqrt{2}b-a}{b}|$

$\displaystyle =|\frac{\sqrt{2}b-a}{b}|\times \frac{\sqrt{2}b+a}{\sqrt{2}b+a}$     (Rationalizing the numerator)

$\displaystyle =\frac{|2b^2-a^2|}{\sqrt{2}b^2+ab}$

$\displaystyle =\boxed{\frac{|2b^2-a^2|}{b(\sqrt{2}b+a)}}$

Now, we analyse the numerator. We can write $a=2^\alpha\cdot x$,

$b=2^\beta\cdot y$, where $x,y$ are odd.

Then $2b^2=2^{2\beta +1}\cdot y^2$,

$a^2=2^{2\alpha}\cdot x^2$.

Since the largest power of two dividing $2b^2$ is an odd power, whilst for $a^2$ the largest power of two dividing it is an even power, $2b^2$ and $a^2$ cannot be the same number. Hence we have $|2b^2-a^2|\geq 1$.

Now, we analyse the denominator. Firstly, we can consider just the rationals $\displaystyle \frac{a}{b}\leq 3-\sqrt{2}\approx 1.59$. Because if $\frac{a}{b}>1.59$, it is clear that $\frac{a}{b}$ is not going to be $\sqrt{2}\approx 1.41$.

Rearranging, we have: $\displaystyle \sqrt{2}+\frac{a}{b}\leq 3$.

Multiplying throughout by $b$, $\sqrt{2}b+a\leq 3b$.

Going back to the original equation (boxed), we can conclude that:

$\displaystyle\boxed{\frac{|2b^2-a^2|}{b(\sqrt{2}b+a)}}\geq\frac{1}{b(3b)}=\frac{1}{3b^2}>0$.

We have shown constructively that $\sqrt{2}$ is not a rational number!

Reference: http://en.wikipedia.org/wiki/Square_root_of_2#Constructive_proof

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