Mathematics

More properly, \( e \) is not the root of a quadratic equation with integer coefficients, which is the same as saying rational coefficients, because denominators can be cleared. This is sometimes stated as \( e \) is not a quadratic irrationality. Liouville proved the theorem in his journal in 1840.^[1] It was a step towards proving that \( e \) is transcendental, meaning that it is not the root of any polynomial equation with integer coefficients. Apparently Liouville regarded it as such^[2], but couldn't push through the general proof. Hermite was the first to prove that \( e \) is transcendental in 1873.^[3] Hilbert simplified the proof^[4], among others. But let's get back to the quadratic case and Liouville's proof. Here it is in its entirety, translated into English:

Janot de Stainville ascribes this proof to Joseph Fourier.^[1] Start with:

The reason for the inequality is that each factorial is greater than a product of as many twos as there are factors in the factorial, discounting the one; for example, \( 4! = 1 \cdot 2 \cdot 3 \cdot 4 \gt 1 \cdot 2 \cdot 2 \cdot 2 = 2^3. \) The final step results from summing the geometric series. It follows that \( 2 \lt e \lt 3 \) and in particular, that \( e \) is not a whole number.

Eratosthenes Measures the Earth

Like Aristotle^[1] before him and in keeping with virtually all educated opinion in ancient Greece^[2], Eratosthenes (c. 276 BC - c. 194 BC) assumed the earth was spherical. He set out to measure its size using a sound method that has stood the test of time. His results were good (about \( 5.4\% \) too high) and no one did better for over a thousand years. His near correct result was ignored well into the modern era, including by Columbus. Snellius and Picard finally nailed down the earth's circumference in the seventeenth century using Eratosthenes' experimental design and taking advantage of advances in mathematics and instrumentation that had accrued over the intervening \( 1900 \) years.

A system of three congruences is shown on the right, but start with the simpler system:
\[ \begin{align*}
x &\equiv 1 \hspace{-.6em} {\pmod{2}}\\
x &\equiv 2 \hspace{-.6em} {\pmod{3}}.
\end{align*} \]
Values congruent \( \text{mod} \; 6 \) are certainly congruent \( \text{mod} \; 2 \) and \( \text{mod} \; 3, \) so in looking for an \( x \) solving both congruences simultaneously, it suffices to consider congruence classes \( \text{mod} \; 6 \) and in particular their smallest positive residues, namely \( 0, 1, 2, 3, 4, 5. \) We're seeking an odd number among those \( 6 \) since \( x \equiv 1{\pmod{2}}, \) one that is also congruent to \( 2 \; \text{mod} \; 3. \) \( x = 1 \) won't do, since \( 1 \equiv 1{\pmod{3}} \) neither will \( x = 3, \) since \( 3 \equiv 0{\pmod{3}}. \) \( x = 5 \) is the solution, since it satisfies both congruences, and it is the only solution \( \text{mod} \; 6. \)

The theorem in question is:

If \( p \) is an odd prime with \( p \equiv 1 \; (\text{mod} \; 4), \) then \( p \) is the sum of two squares.\( (1) \)

Only if is easy, because for all natural numbers \( n, \; n^2 \equiv 0, 1 \; (\text{mod } 4), \) so \( n^2 + m^2 \equiv 0, 1, 2 \; (\text{mod} \; 4) \) and a sum of two squares cannot be congruent to \( 3 \; (\text{mod} \; 4). \) Obviously \( 2 = 1^2 + 1^2 \) as well. The Brahmagupta-Fibonacci identity assures that a product of sums of two squares is itself a sum of two squares:

\[ \begin{equation}{(a^{2}+b^{2})(c^{2}+d^{2})=(ac+bd)^{2}+(ad-bc)^{2}.}\tag{2} \end{equation} \]

Integers under \( 40 \) that are the sum of two squares. \( \color{red}{25} \) is the first that is the sum of two squares in two ways.

\( 5 = 1^2 + 2^2 \) is the sum of two squares, \( 3 \) is not. Dealing with whole numbers only, including \( 0, \) it's a bit of a riddle coming up with the criterion distinguishing the two situations. Based on empirical investigations, mathematicians in the \( 17^\text{th} \) century found the key. According to Leonard Dickson^[1]:

A. Girard (Dec 9, 1632) had already made a determination of the numbers expressible as a sum of two integral squares: every square, every prime \( 4n + 1, \) a product formed of such numbers, and the double of one of the foregoing.

The part about primes \( p \equiv 1 \; (\text{mod} \; 4) \) is central, because a product of two numbers each of which is the sum of two squares is itself the sum of two squares. Since \( 5 = 1^2 + 2^2 \) and \( 13 = 2^2 + 3^2, \) for example, \( 65 = 5 \cdot 13 \) is also the sum of two squares: \( 65 = 4^2 + 7^2. \) In fact there is a second representation: \( 65 = 1^2 + 8^2, \) and the number of representations is of interest too (this exact example is from Diophantus).

Progressive differences of the first few cubes.

Write down the first few cubes, then put their differences \( \Delta \) in the second column, the differences of those differences \( \Delta^2 \) in the third column, and so on. Remarkably, \( \Delta^3 = 6 \), and that is true for any contiguous sequence of cubes (obviously \( \Delta^4 = 0 \)). Do that with the fourth powers and you find that \( \Delta^4 = 24, \) and in general for contiguous \( n^{th} \) powers, \( \Delta^n = n!. \) The key to unlocking this mystery is the Calculus of Finite Differences, out of vogue now apparently, but with a hallowed history going back to Newton and before and studied in depth by George Boole in 1860.^[1] His book can still be read with profit, as can C. H Richardson's little text from 1954. ^[2]

Euler used \( \Delta^n x^n = n! \) in 1755 to prove the two squares theorem. Boole and those following him employed the term "calculus" advisedly, many theorems in the finite case matching similar ones in the familiar infinitesimal calculus. Which stands to reason, considering all there is in common, it's just that now \( \Delta x = 1. \)

All Abelian groups of order 72.

The Fundamental Theorem of Finite Abelian Groups decisively characterizes the Abelian finite groups of a given order. Its remote origins go back to Gauss in the Disquisitiones Arithmeticae in 1801 and it was nailed down by Schering (1869) and by Frobenius and Stickelberger (1879)^[1]:

Fundamental Theorem of Finite Abelian Groups

Let \( G \) be a finite Abelian Group of order \( n. \) Then: \[ \begin{equation}{G \cong \mathbb{Z}_{n_1} \times \mathbb{Z}_{n_2} \times \cdots \times \mathbb{Z}_{n_s},} \tag{1} \end{equation} \] where \( s \) and the \( n_i \) are the unique integers satisfying \( s \geq 1, n_i \geq 2 \) for all \( i, \) and \( n_{i+1} \; | \; n_i \) for \( 1 \leq i \leq s - 1. \) And also: \[ \begin{equation}{G \cong \mathbb{Z}_{p^{\beta_1}} \times \cdots \times \mathbb{Z}_{p^{\beta_t}} \times \cdots \times \mathbb{Z}_{q^{\gamma_1}} \times \cdots \times \mathbb{Z}_{q^{\gamma_u}},} \tag{2} \end{equation} \] for \( p \) and \( q \) and all the other primes dividing \( n, \) again in a unique way, where \( \sum \beta_i \) is the exponent of the greatest power of \( p \) dividing \( n, \) \( \sum \gamma_i \) is the exponent of the greatest power of \( q \) dividing \( n, \) and so on for all the other primes dividing \( n. \)

Click image for original.

When a problem contains several unknowns whose relationships are so complicated that one is obliged to form several equations; then, to discover the values of the unknowns, one makes all of them vanish, except one, which combined only with known quantities, gives, if the problem is determined, a final Equation, whose resolution reveals this first unknown, and then by this means all the others.

Here are the midterm and final exams for Math 542, Modern Algebra, at the University of Wisconsin-Madison in the spring semester 2015-2016, Professor Paul Terwilliger officiating. It is an undergraduate class, junior or senior level, for (mostly) math majors. The class takes up basic ring theory, following on Math 541, which is mostly group theory. The text, followed pretty closely, is Abstract Algebra, 3rd ed., by David S. Dummit and Richard M. Foote (Wiley, 2004) — D&F — widely used it appears. It certainly is comprehensive, with many examples and a great set of exercises, and is an impressive work in its own right, well-organized, demanding, and thorough. Typo-free as well, I haven't found a single one.