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!

irrational

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


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