# Add Odd Numbers. A DIY Proof of the Irrationality of $\sqrt{2}$

### Samuel G. Moreno, Esther M. García-Caballero

Gentle reader: it is hardly a case of being in favor of, or against, the irrationality of $\sqrt{2}$. At this point in time, and after so many different proofs, the only thing to do is to try to contribute to THE MUSEUM OF THE IRRATIONALITY OF $\sqrt{2}$ with a new artifact. In the main gallery of this museum you can find one of the central pieces of the collection, which reads: "If $\sqrt{2}$ is rational, then $\sqrt{2}=a/b$ for some $a,b$, positive and relative prime integers; since $a^2=2b^2$, then $a^2$ is even and so is $a$. From $a=2c$ we get $b^2=2c^2$, so $b$ is also even. Contradiction".

Our aim is to prove that in the irrationality of $\sqrt{2}$, ODDNESS ALSO MATTERS. More than an exquisite cabinet, we offer a bricolage piece, in which the old formula

(1)

$n^2=1+3+5+\ldots +(2n-1),$ $n=1,2,3,\ldots$

has been recycled to give an irrationality result. Before stating our main theorem, let us recall that the discovery of the irrationality of $\sqrt{2}$ is sometimes credited to Hippasus of Metapontum (5th century BC), a Greek Pythagorean philosopher who was finally drowned at sea as a punishment from his peers, the Pythagoreans, for divulging the existence of the irrational numbers (the Pythagoreans advocate the idea that all numbers SHOULD be the ratio of two integers).

### Theorem

Dear Professor Pythagoras:

We are sorry. Not only $\sqrt{2}$ is not rational, but this can be proved with just the unworthy odd numbers.

### A Do-It-Yourself-Proof for bricoleurs.

Assume that $\sqrt{2}$ is rational, so that it equals $p/q$ for $p,q$ positive integers with $q$ the smallest possible. Use formula (1) to replace $p^2$ and $q^2$ in $2=(\sqrt{2})^2=p^2/q^2$. Now you must first simplify, next simplify once again, ..., and continue simplifying until concluding that $\sqrt{2}$ equals a fraction with smaller denominator than the original one. After this you must decide if your proof deserves to be exhibited in THE MUSEUM.

### The same Proof in a more formal way.

Suppose $\sqrt{2} = p/q$, where $p$ and $q$ are positive integers and $q$ is the smallest possible. Since $1\lt\sqrt{2}\lt 2$, then $q\lt p\lt 2q.$ Squaring $\sqrt{2} = p/q$ and using (1) we have

$\begin{align}\displaystyle 2=\frac{p^2}{q^2}&=\frac{1+3+\cdots +(2q-1)+(2q+1)+\cdots +(2p-1)}{1+3+\cdots +(2q-1)}\\ &=1+\frac{(2q+1)+\cdots +(2p-1)}{1+\cdots +(2q-1)}, \end{align}$

and thus

$\begin{align} 1+3+\cdots +(2q-1)&=(2q+1)+(2q+3)+\cdots +(2p-1)\nonumber\\ &=(2q+1)+(2q+3)+\cdots +\Big(2q+\big(2(p-q)-1\big)\Big). \end{align}$

The bound $p\lt 2q$, which is equivalent to $p-q\lt q$, makes it possible to rewrite the previous equation in the form

$\begin{multline*} 1+3+\cdots +\big(2(p-q)-1\big)+\big(2(p-q)+1\big)+\cdots+(2q-1) \\=(1+2q)+(3+2q)+\cdots +\Big(\big(2(p-q)-1\big)+2q\Big), \end{multline*}$

which, after canceling the terms $1,3,\ldots,\big(2(p-q)-1\big)$, simplifies to

(2)

$\big(2(p-q)+1\big)+\big(2(p-q)+3\big)+\cdots+(2q-1)=2q(p-q).$

Now note that the left hand side of (2) can be rewritten as

$\begin{align} \big(2(p-q)+1\big)+&\big(2(p-q)+3\big)+\cdots+\Big(2(p-q)+\big(2(2q-p)-1\big)\Big)\\ &=2(p-q)(2q-p)+\Big(1+3+\cdots+\big(2(2q-p)-1\big)\Big)\\ &=2(p-q)(2q-p)+(2q-p)^2\\ &=p(2q-p), \end{align}$

so that equation (2) transforms to $p(2q-p)=2q(p-q)$, which in turn yields to

$\displaystyle\frac{2q-p}{p-q}=2\frac{1}{p/q}=2\frac{1}{\sqrt{2}}=\sqrt{2},$

but this contradicts the minimality of $q$.

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