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 =|\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:


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



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Why is e irrational?

Anyone who has taken high school math is familiar with the constant \boxed{e=2.718281828\cdots}.


Today we are going to prove that e is in fact irrational! We will go through Joseph Fourier‘s famous proof by contradiction. The maths background we need is to know the power series expansion: \displaystyle \boxed{e=\sum_{n=0}^{\infty}\frac{1}{n!}}. The proof is slightly tricky so stay focussed!


Suppose to the contrary that e is a rational number, so \displaystyle e=\frac{a}{b}.

Using the power series formula mentioned above, we have \displaystyle\sum_{n=0}^\infty \frac{1}{n!}=\frac{a}{b}

Multiply both sides by b!, \displaystyle \sum_{n=0}^{\infty}\frac{b!}{n!}=\frac{ab!}{b}=a(b-1)!

Now, we split the sum into two parts:

\displaystyle \sum_{n=0}^b \frac{b!}{n!}+\sum_{n=b+1}^\infty \frac{b!}{n!}=a(b-1)!


\displaystyle \sum_{n=b+1}^\infty \frac{b!}{n!}=a(b-1)!-\sum_{n=0}^b \frac{b!}{n!}

Now, denote \displaystyle x=\sum_{n=b+1}^\infty \frac{b!}{n!}>0. x is an integer since both \displaystyle a(b-1)! and \displaystyle\sum_{n=0}^b \frac{b!}{n!} are integers and their difference (which is x) will be an integer.

We now prove that x<1. For all terms with n\geq b+1 we have the upper estimate

\displaystyle\begin{array}{rcl}  \frac{b!}{n!}&=&\frac{1\times 2\times \cdots \times b}{1\times 2\times \cdots \times b \times (b+1) \times \cdots \times n}\\  &=&\frac{1}{(b+1)(b+2)\cdots (b+(n-b))}\\  &\leq& \frac{1}{(b+1)^{n-b}}  \end{array}

This inequality is strict for every n\geq b+2. Changing the index of summation to k=n-b and using the formula for the infinite geometric progression S_\infty = \frac{a}{1-r}, we obtain:

\displaystyle x=\sum_{n=b+1}^\infty \frac{b!}{n!} < \sum_{n=b+1}^\infty \frac{1}{(b+1)^{n-b}}=\sum_{k=1}^\infty \frac{1}{(b+1)^k}=\frac{\frac{1}{b+1}}{1-\frac{1}{b+1}}=\frac{1}{b}\leq 1

We have that x is an integer but 0<x<1. This is a contradiction (since there is no integer strictly between 0 and 1), and so e must be irrational. (QED)

Interesting? 🙂

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Watch this video for another proof that e is irrational!

How to prove square root of 2 is irrational?

A rational number is a number that can be expressed in a fraction with integers as numerators and denominators.

Some examples of rational numbers are 1/3, 0, -1/2, etc. Now, we know that \sqrt{2}\approx 1.41421\cdots.

Is the square root of 2 rational? Or is it irrational (the opposite of rational)? How do we prove it? It turns out we can prove that the square root of two is irrational using a technique called proof by contradiction. (One of the earlier posts on this blog also used proof by contradiction to show that there are infinitely many prime numbers.)

First, we suppose that \displaystyle\sqrt{2}=\frac{p}{q}, where \displaystyle\frac{p}{q} is a fraction in its lowest terms.

Next, we square both sides to get \displaystyle 2=\frac{p^2}{q^2}.

Hence, 2q^2=p^2. We can conclude that p^2 is even since it is a multiple of 2. Thus, p itself is also even. (the square of an odd number is odd).

Thus, we can write p=2k for some integer k. Substituting this back into 2q^2=p^2, we get 2q^2=4k^2, which can be simplified to q^2=2k^2.

Hence, q^2 is also even, and hence q is also even!

But if both p and q are even, then \displaystyle\frac{p}{q} is not in the lowest terms! (we could divide them by two). This contradicts our initial hypothesis!

Thus, the only possible conclusion is that the square root of two is not a rational number to begin with!


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