A tail too small to be a whole number

The divisor search decides every root of every whole-number polynomial and it has nothing whatever to say about $e$. There is no polynomial with whole-number coefficients that $e$ satisfies, so there is no constant term to take divisors of, and the criterion does not fail — it does not apply.

What decides $e$ instead is a size argument, and it is short enough to fit on a postcard. Fourier wrote it down in 1815; it needs one series, one factorial and the observation that there is no whole number strictly between zero and one.

The series, and why the factorial is the right multiplier #

The number is defined by

$$e = \sum_{k \ge 0} \frac{1}{k!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \dots,$$

which is the exponential’s own series at $x = 1$.

Suppose $e = p/q$ with $p$ and $q$ whole numbers, $q \ge 2$. Multiply by $q!$:

$$q!\,e = q!\,\frac{p}{q} = (q-1)!\,p,$$

which is a whole number. That is the only use the assumption gets, and it is worth noticing how small a use it is: the argument does not care what $p$ is, only that $q!$ clears the denominator.

Now split the series at $k = q$:

$$q!\,e = \underbrace{\sum_{k = 0}^{q} \frac{q!}{k!}}_{\text{call it } A} + \underbrace{\sum_{k > q} \frac{q!}{k!}}_{\text{call it } T}.$$

Every term of $A$ is a whole number, because $k \le q$ means $k!$ divides $q!$. So $A$ is a whole number, and $q!e$ is a whole number, so $T$ is a whole number too.

The multiplier $q!$ was chosen for exactly that: it is the smallest thing that both clears the assumed denominator and makes the head of the series integral. Anything smaller leaves fractions in $A$; anything larger works but wastes the tightness of the bound below.

Squeezing the tail #

Write $T$ out:

$$T = \frac{1}{q+1} + \frac{1}{(q+1)(q+2)} + \frac{1}{(q+1)(q+2)(q+3)} + \dots$$

Every term is positive, so $T > 0$.

Every denominator is at least $(q+1)$ times the previous, so each term is at most $1/(q+1)$ times the one before. Comparing with the geometric series whose ratio is $1/(q+1)$,

$$T < \frac{1}{q+1}\cdot\frac{1}{1 - \frac{1}{q+1}} = \frac{1}{q}.$$

And $1/q \le 1/2 < 1$. So $0 < T < 1$, and $T$ is a whole number. There is no whole number strictly between zero and one, so the assumption is false.

That is the whole proof. The figure is a measurement of $T$ at each $q$, computed in exact rational arithmetic, with the band $(0, 1)$ drawn and the bound $1/q$ over it.

Which definition of e the proof needs #

The number arrives in several ways and only one of them makes this argument possible, which is worth saying because the choice looks like a convenience and is not.

As a limit of compound interest. $e = \lim (1 + 1/n)^n$, the amount a unit grows to under continuous compounding. This is the historical definition and the one with a story attached, and nothing whatever can be done with it here: the expression is a limit of numbers each of which is rational, and a limit of rationals says nothing about whether the limit is one.

As the base whose exponential is its own slope. This is the defining property and it is the one that makes $e$ matter. It also gives no purchase on irrationality; being its own derivative is a statement about a function, not about a decimal.

As the sum of the reciprocal factorials. This is the one the proof uses, and it is used for a single property: the terms fall faster than geometrically. Nothing else about the series matters — not that it converges, not what it converges to, not that the sum has any interpretation at all.

That the three definitions agree is a theorem and a good one. That only the third is usable here is a small lesson about proofs: the definition that explains what a number is for is rarely the definition that decides what kind of number it is, and choosing between them is most of the work in a great many arguments.

Why exact arithmetic, and not a calculation #

The figure computes $T$ in fractions of exact whole numbers rather than in decimals, and the reason is not fastidiousness.

The claim is that $T$ lies strictly between two numbers. At $q = 9$ the tail is about $10^{-1}$; at $q = 18$ it is around $5 \times 10^{-2}$ of a quantity whose head is $18!$, and a double-precision calculation of $q!e - A$ at that size is the difference of two numbers near $10^{16}$ — which is the size at which a double stops being able to tell consecutive whole numbers apart. The subtraction returns exactly zero, and zero is not strictly between zero and one.

So a figure drawn from floating-point arithmetic would show the tail arriving at the boundary of the band and would be showing the arithmetic’s failure rather than the number’s behaviour. Every value in the drawing is a ratio of exact big integers, and the comparisons that decide the two bounds are comparisons of whole numbers.

This is the same discipline that the exhaustive searches elsewhere on this site are written to, and it bites in the same place: a claim about a strict inequality cannot be checked by an arithmetic that has already rounded one side of it.

What the proof does and does not give #

It gives irrationality and nothing more, and the distinction matters because the next rung of this ladder is about a stronger property.

It does not show $e$ is transcendental. For all this argument says, $e$ could be a root of $x^2 - 3x + 1$ or of some quintic. Transcendence — being a root of no whole-number polynomial at all — was proved by Hermite in 1873, sixty years later, and the proof is several pages rather than several lines.

It does not measure how irrational $e$ is. The argument shows $T$ is between zero and one; it does not extract from that any statement about how closely fractions can approach $e$. That question has an answer and it is a rather good one, and it is the last rung that asks it.

It does not generalise to $\pi$. There is no series for $\pi$ whose terms fall fast enough for the same squeeze; the closest classical series, $\pi/4 = 1 - 1/3 + 1/5 - \dots$, has terms falling like $1/n$, which is nowhere near. What replaces the series is an integral, and the head that has to be whole is built out of derivatives.

A stronger version, almost free #

The same argument, pushed a little, gives something the bare statement does not.

Multiply the strict inequality by nothing and read it as an approximation statement instead. For any $q$, the partial sum $A/q!$ is a fraction with denominator $q!$, and it differs from $e$ by $T/q!$, which is less than $1/(q \cdot q!)$. So

$$\left| e - \frac{A}{q!} \right| < \frac{1}{q \cdot q!}.$$

That is a very good approximation for the size of the denominator — much better than the $1/q^2$ that a generic irrational is guaranteed. The factorials in the denominators are producing fractions that come unusually close for their size.

Two consequences follow, and they point in opposite directions.

It makes $e$ easy to compute. Ten terms give seven decimal places; twenty give eighteen. Nothing about the series is delicate and no cancellation occurs, because every term is positive.

It makes $e$ nearly the kind of number the last rung of this ladder rules transcendental. A number approachable much faster than its degree allows is not algebraic, and the natural question is whether $1/(q \cdot q!)$ is fast enough. It is not — the denominators being used are $q!$ rather than $q$, so the exponent gained is only about one — and closing that gap is where the transcendence proof’s extra pages go.

The continued fraction, which sees the same thing #

There is a second window on $e$’s irrationality that gives the answer instantly and is much harder to justify.

$$e = [2; 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, \dots]$$

The pattern is visible and it never terminates. A continued fraction terminates exactly when the number is a fraction, so a non-terminating expansion is irrationality — provided the pattern really does continue, which is a theorem of Euler’s and is genuinely harder than the series argument.

The two routes are worth comparing because they show what irrationality proofs cost. The series argument is elementary and finishes in a paragraph. The continued fraction is more informative — it says how well $e$ is approximated as well as that it is irrational — and the price of the extra information is a much harder theorem underneath.

The same argument on a number that is a fraction #

A proof should be tested on a case where it must fail, and the test is instructive here because the failure is visible rather than mysterious.

Run the argument on $3/2$ instead of $e$. The number has an expansion — $3/2 = 1 + 1/2$, or as a series with all further terms zero — and the head-and-tail split works exactly as before: multiply by $2!$ and the whole thing is $3$, the head is $3$, and the tail is $0$.

Zero is a whole number, and it is not strictly between zero and one. So the contradiction does not arrive, and the argument correctly declines to conclude anything. The step that fails is the one that fails, and it fails for the right reason: the tail is positive for $e$ because every term of the series is positive and there are infinitely many of them, and that is exactly what a terminating expansion does not have.

That is the shape of a well-behaved impossibility proof: it should be possible to point at the line that stops working when the hypothesis is dropped. Here the line is every term is positive, and the site’s habit of feeding a piece of machinery something it must refuse is the same discipline applied to figures rather than to proofs.

The shape, stated generally #

Strip the setting away and the argument is a template that has been used many times.

Assume the number is $p/q$. Find a multiplier that turns the assumption into “some explicitly computable quantity is a whole number”. Then show that quantity lies strictly between zero and one.

The multiplier is $q!$ here. In the proof that $\pi$ is irrational it is a polynomial’s integral, and the whole number is produced by derivatives at the endpoints. In the proof that $\zeta(3)$ is irrational — Apéry’s, from 1978, and the most celebrated recent instance — it is a pair of sequences satisfying a recurrence, and demonstrating that the resulting quantity is small was the hard part for over two centuries.

In every case the difficulty sits in the same place. Producing a quantity that would have to be whole is usually straightforward; making it small is the work. For $e$ the factorials do it unaided, which is why this is the easy case and the one to learn first.

What a decimal expansion cannot settle #

$e$ is known to tens of trillions of decimal places, and none of them is evidence for anything this essay proves.

That is worth stating flatly because the intuition runs the other way. A decimal expansion that shows no repeating block looks like irrationality, and it is not: a fraction with denominator $q$ repeats with a period of at most $q - 1$, so an expansion with no repeat in its first trillion places rules out every denominator below a trillion and no more. There are larger denominators.

The direction of the failure is worth being exact about. Computation can refute irrationality — find a repeat, and the number is a fraction — and cannot establish it. So the trillions of digits are a very strong test that the argument above has passed, and are not a substitute for it.

The same asymmetry runs through this whole subject and it is why the proofs look the way they do. The primes on a spiral show a pattern nobody has explained; the pattern is evidence and the explanation is missing, and no amount of further computation converts one into the other. Here the explanation exists and takes a paragraph, and the digits are decoration.

There is one place where the digits earn their keep. The continued fraction pattern $[2; 1, 2, 1, 1, 4, \dots]$ was found by computation, and Euler then proved it; a proof of something nobody had noticed is much rarer than a proof of something a calculation suggested. Computation is very good at generating the statement and useless at closing it.

What the picture cannot show #

The figure shows the tail at finitely many denominators, and the proof is about all of them. A reader could reasonably ask whether the tail might climb above one somewhere past the right edge, and the picture cannot answer — the bound $1/q$ drawn over it is the answer, and it is prose.

It also cannot show the head. $A$ is a whole number of size roughly $q! \cdot e$, which at $q = 9$ is near a million and at $q = 14$ is past $10^{11}$; there is no axis on which the head and the tail both appear. The picture is of the remainder after the interesting quantity has been subtracted, and the interesting quantity is off the page by many orders of magnitude.

And the band between zero and one is drawn as a region, which invites the reading that the tail is near a whole number and misses. It is not near anything. The statement is that a whole number would have to be there and none is, and the emptiness of the band is the content — which is a hard thing to draw, since an empty region looks the same whatever is absent from it.

Where the ladder goes next #

Above: $\pi$, by the same template with a much more elaborate multiplier. And then the question this ladder has been circling — whether a number satisfies any whole-number polynomial at all — which needs an argument about how fast fractions can approach rather than about whether one hits.

One debt. Euler’s continued fraction for $e$ is quoted here and not proved, and the proof is a genuinely nice piece of work with a Riccati equation in it. It belongs on this ladder and is not on it.

What the factorial was for #

Multiply by enough to make the assumption say that something is a whole number, then show that something is smaller than one.

The factorial is not magic and it is not about $e$. It is the smallest multiplier that clears both denominators at once — the assumed one and the series’ — and having cleared them it leaves behind a tail whose own factorials make it too small to be anything. The proof is one well-chosen multiplication and one comparison, and everything harder in this subject is an attempt to find the multiplication for a number where it is not obvious.