2026-04-10 22:29:49
There’s a lot of mathematics just off the beaten path. You can spend a career in math and yet not know all there is to know about even the most basic areas of math. For example, this post will demonstrate something you may not have seen about decimal forms of fractions.
Let p > 5 be a prime number and 0 < k < p. Then the digits in k/p might be the same for all k, varying only by cyclic permutations. This is the case, for example, when p = 7 or p = 17. More on these kinds of fractions here.
The digits in k/p repeat for every k, but different values of k might have sequences of digits that vary by more than cyclic permutations. For example, let’s look at the values of k/13.
>>> for i in range(1, 13): ... print(i/13) ... 1 0.0769230769230769 2 0.1538461538461538 3 0.2307692307692307 4 0.3076923076923077 5 0.3846153846153846 6 0.4615384615384615 7 0.5384615384615384 8 0.6153846153846154 9 0.6923076923076923 10 0.7692307692307693 11 0.8461538461538461 12 0.9230769230769231
One cycle goes through the digits 076923. You’ll see this when k = 1, 3, 4, 9, 10, or 11. The other cycle goes through 153846 for the rest of the values of k. The cycles 076923 and 153846 are called the distinct repeating sets of 13 in [1].
If we look at fractions with denominator 41, thee are six distinct repeating sets.
02439 04878 07317 09756 12195 14634 26829 36585
You could find these by modifying the Python code above. However, in general you’ll need more than default precision to see the full periods. You might want to shift over to bc, for example.
When you look at all the distinct repeating sets of a prime number, all digits appear almost the same number of times. Some digits may appear one more time than others, but that’s as uneven as you can get. A corollary in [1] states that if p = 10q + r, with 0 < r < 10, then 11 − r digits appear q times, and r − 1 digits appear q times.
Looking back at the example with p = 13, we have q = 1 and r = 3. The corollary says we should expect 8 digits to appear once and 2 digits to appear twice. And that’s what we see: in the sets 076923 and 153846 we have 3 and 6 repeated twice and the remaining 8 digits appear once.
In the example with p = 41, we have q = 4 and r = 1. So we expect all 10 digits to appear 4 times, which is the case.
[1] James K. Schiller. A Theorem in the Decimal Representation of Rationals. The American Mathematical Monthly
Vol. 66, No. 9 (Nov., 1959), pp. 797-798
2026-04-10 01:54:21
Saw a post on X saying that the latitude of the Pyramid of Giza is the same as the speed of light.
I looked into this, expecting it to be approximately true. It’s exactly true in the sense that the speed of light in vacuum is 299,792,458 m/s and the line of latitude 29.9792458° N passes through the pyramid. The exact center of the pyramid is at 29.97917° N, 31.13417° E.
Of course this is a coincidence. Even if you believe that somehow the ancient Egyptians knew the speed of light, the meter was defined four millennia after the pyramid was built.
The post The Great Pyramid of Giza and the Speed of Light first appeared on John D. Cook.2026-04-10 01:25:08
I recently ran across a post on X describing a process for creating a random fractal. First, pick a random point c inside a hexagon.
Then at each subsequent step, pick a random side of the hexagon and create the triangle formed by that side and c. Update c to be the center of the new triangle and plot c.
Note that you only choose a random point inside the hexagon once. After that you randomly choose sides.
Now there are many ways to define the center of a triangle. I assumed the original meant barycenter (centroid) when it said “center”, and apparently that was correct. I was able to create a similar figure.
But if you define center differently, you get a different image. For example, here’s what you get when you use the incenter, the center of the largest circle inside the triangle.
2026-04-09 08:18:57
I recently found out about Andrica’s conjecture: the square roots of consecutive primes are less than 1 apart.
In symbols, Andrica’s conjecture says that if pn and pn+1 are consecutive prime numbers, then
√pn+1 − √pn < 1.
This has been empirically verified for primes up to 2 × 1019.
If the conjecture is true, it puts an upper bound on how long you’d have to search to find the next prime:
pn+1 < 1 + 2√pn + pn,
which would be an improvement on the Bertrand-Chebyshev theorem that says
pn+1 < 2pn.
The post Root prime gap first appeared on John D. Cook.
2026-04-09 07:30:15
Last week I wrote about the orbit of Artemis II. The orbit of Artemis I was much more interesting.
Because Artemis I was unmanned, it could spend a lot more time in orbit. The Artemis I mission took 25 days while Artemis II will take 10 days. Artemis I took an unusual path, orbiting the moon the opposite direction of the moon’s orbit around earth. This video by Primal Space demonstrates the orbit both from the perspective of earth and from the perspective of the Moon.
Another video from Primal Space describes the orbit of the third stage of Apollo 12. This stage was supposed to orbit around the sun in 1971, but an error sent it on a complicated unstable orbit of the earth, moon, and sun. It returned briefly to earth in 2002 and expected to return sometime in the 2040s.
2026-04-07 08:33:23
Landauer’s principle gives a lower bound on the amount of energy it takes to erase one bit of information:
E ≥ log(2) kBT
where kB is the Boltzmann constant and T is the ambient temperature in Kelvin. The lower bound applies no matter how the bit is physically stored. There is no theoretical lower limit on the energy required to carry out a reversible calculation.
In practice the energy required to erase a bit is around a billion times greater than Landauer’s lower bound. You might reasonably conclude that reversible computing isn’t practical since we’re nowhere near the Landauer limit. And yet in practice reversible circuits have been demonstrated to use less energy than conventional circuits. We’re far from the ultimate physical limit, but reversibility still provides practical efficiency gains today.
A Toffoli gate is a building block of reversible circuits. A Toffoli gate takes three bits as input and returns three bits as output:
T(a, b, c) = (a, b, c XOR (a AND b)).
In words, a Toffoli gate flips its third bit if and only if the first two bits are ones.
A Toffoli gate is its own inverse, and so it is reversible. This is easy to prove. If a = b = 1, then the third bit is flipped. Apply the Toffoli gate again flips the bit back to what it was. If ab = 0, i.e. at least one of the first two bits is zero, then the Toffoli gate doesn’t change anything.
There is a theorem that any Boolean function can be computed by a circuit made of only NAND gates. We’ll show that you can construct a NAND gate out of Toffoli gates, which shows any Boolean function can be computed by a circuit made of Toffoli gates, which shows any Boolean function can be computed reversibly.
To compute NAND, i.e. ¬ (a ∧ b), send (a, b, 1) to the Toffoli gate. The third bit of the output will contain the NAND of a and b.
T(a, b, 1) = (a, b, ¬ (a ∧ b))
A drawback of reversible computing is that you may have to send in more input than you’d like and get back more output than you’d like, as we can already see from the example above. NAND takes two input bits and returns one output bit. But the Toffoli gate simulating NAND takes three input bits and returns three output bits.
