{\huge Descending Proofs Into Algorithms}

A way to make indirect reasoning more palpable \vspace{.25in}

Nicholas Saunderson was the fourth Lucasian Professor at Cambridge, two after Isaac Newton. He promoted Newton's Principia Mathematica in the Cambridge curriculum but channeled his original work into lecture notes and treatises rather than published papers. After his death, most of his work was collected into the book, The Elements of Algebra in Ten Books, whose title recalls Euclid's Elements. It includes what is often credited as the first ``extended'' version of Euclid's algorithm.

Today we raise the idea of using algorithms such as this as the basis for proofs.

Saunderson was blind from age one. He built a machine for doing what he called ``Palpable Arithmetic'' by touch. As described in the same book, it was an enhanced abacus---not a machine for automated calculation of the kind a later Lucasian professor, Charles Babbage, attempted to build.

We take the ``palpable'' idea metaphorically. Not only beginning students but we ourselves still find proofs by contradiction or ``infinite descent'' hard to pick up at first reading. We wonder how far mathematics can be developed so that the hard nubs of proofs are sheathed in assertions about the availability and correctness of algorithms. The algorithm's proof may still involve contradiction, but there's a difference: You can interact with an algorithm. It is hard to interact with a contradiction.

An Example

It was known long before Euclid that the square root of 2 is irrational. In the terms Saunderson used, the diagonal of a square is ``incommensurable'' with its side.

Alexander Bogomolny's great educational website Cut the Knot has an entire section on proofs. Its coverage of the irrationality of $latex {\sqrt{2}}&fg=000000$ itemizes twenty-eight proofs. All seem to rely on some type of infinite descent: if there is a rational solution then there is a smaller one and so on. Or they involve a contradiction of a supposition whose introduction seems perfunctory rather than concrete. We gave a proof by induction in a post some years ago, where we also noted a MathOverflow thread and a discussion by Tim Gowers about this example.

We suspect that one reason the proof of this simple fact is considered hard for a newcomer is just that it uses these kinds of descent and suppositions. Certainly the fact itself was considered veiled in antiquity. According to legend the followers of Pythagoras treated it as an official secret and murdered Hippasus of Metapontum for the crime of divulging it. To state it truly without fear today, we still want a clear view of why the square root of 2 is irrational.

Our suggestion below is to avoid the descent completely. Of course it is used somewhere, but it is encapsulated in another result. The result is that for any co-prime integers $latex {r}&fg=000000$ and $latex {s}&fg=000000$ there are integers $latex {x}&fg=000000$ and $latex {y}&fg=000000$ such that

$latex \displaystyle rx + sy = 1. &fg=000000$

An Attempt

Here is the closest classical proof we could find to our aims. We quote the source verbatim and will reveal it at the end.

Proposition 15. If there be any whole number, as $latex {n}&fg=000000$, whose square root cannot be expressed by any other whole number; I say then that neither can it be expressed by any fraction whatever.

For if possible, let the square root of $latex {n}&fg=000000$ be expressed by a fraction which when reduced to it's least integral terms is $latex {\frac{a}{b}}&fg=000000$, that is, let $latex {\frac{a}{b} = \sqrt{n}}&fg=000000$, then we shall have $latex {\frac{a}{b}\frac{a}{b} = \frac{n}{1}\;}&fg=000000$; but the fraction $latex {\frac{a}{b}\frac{a}{b}}&fg=000000$ is in it's least terms, by the third corollary to the twelfth proposition, because the fraction $latex {\frac{a}{b}}&fg=000000$ was so; and the fraction $latex {\frac{n}{1}}&fg=000000$ is in it's least terms, because $latex {1}&fg=000000$ cannot be further reduced; therefore we have two equal fractions $latex {\frac{a}{b}\frac{a}{b}}&fg=000000$ and $latex {\frac{n}{1}}&fg=000000$, both in their least terms; therefore by the tenth proposition, these two fractions must not only be equal in their values, but in their terms also, that is, $latex {aa}&fg=000000$ must be equal to $latex {n}&fg=000000$, and $latex {bb}&fg=000000$ to $latex {1:}&fg=000000$ but $latex {aa}&fg=000000$ cannot be equal to $latex {n}&fg=000000$, because $latex {a}&fg=000000$ is a whole number by the supposition, and $latex {n}&fg=000000$ is supposed to admit of no whole number for its root; therefore the square root of $latex {n}&fg=000000$ cannot possibly be expressed by any fraction whatever. Q.E.D.

\bigskip The quoted propositions are that two fractions in lowest whole-number terms must be identical and that if $latex {a}&fg=000000$ and $latex {b}&fg=000000$ are co-prime to $latex {c}&fg=000000$ then so is $latex {ab}&fg=000000$. The proof of the latter starts with the for-contradiction words ``if this be denied,'' so the absence of such language above gets only part credit. This all does not come trippingly off the tongue; rather it stays trippingly in the throat. Let's try again.

A Proof

In fact, we don't need the concepts of ``lowest terms'' or co-primality or the full statement of the identity named for \'Etienne Bézout. It suffices to assert that for any whole numbers $latex {r}&fg=000000$ and $latex {s}&fg=000000$, there are integers $latex {x}&fg=000000$ and $latex {y}&fg=000000$ such that the number

$latex \displaystyle d = rx + sy &fg=000000$

Then for the proof, suppose that $latex {\sqrt{n} = \frac{r}{s}}&fg=000000$ for integers $latex {r}&fg=000000$ and $latex {s}&fg=000000$. We take $latex {x}&fg=000000$ and $latex {y}&fg=000000$, and let $latex {r' = r/d}&fg=000000$, $latex {s' = s/d}&fg=000000$ be the resulting integers. Now let's do some simple algebra:

$latex \displaystyle r'x + s' y = 1,\quad\text{so}\quad r' = \frac{1 -s' y}{x},\quad\text{so}\quad x^2r'^2 = (1 - s' y)^2. &fg=000000$

$latex \displaystyle n = \frac{r^2}{s^2} = \frac{r'^2}{s'^2},\quad\text{so}\quad r'^2 = s'^2 n. &fg=000000$

$latex \displaystyle x^2 s'^2 n = (1 - s' y)^2 = 1 - 2s' y + s'^2 y^2. &fg=000000$

$latex \displaystyle \frac{1}{s'} = 2y + s' x^2 - s' y^2 = \text{some integer}. &fg=000000$

This is a contradiction. But it is a palpable contradiction. For instance, of course we can see that $latex {\sqrt{2}}&fg=000000$ isn't an integer. Thus we claim that the effect of this proof is more concrete.

Open Problems

Is this a new proof? We doubt it. But the proof is nice in that it avoids any recursion or induction. The essential point---the divisibility of $latex {d}&fg=000000$ into $latex {r}&fg=000000$ and $latex {s}&fg=000000$---is coded into the Euclidean algorithm.

Is ours at least smoother than the classical proof we quoted? The latter is from Saunderson's book, on pages 304--305 which come soon after his presentation of the algorithm on pages 295--298.

What other proofs can benefit from similar treatment by ``reduction to algorithms''?