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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