Yesterday, I posted a proof on why the sigma sum for phi was true, but I kind of rushed it, so I will be going into details.

If you can do a little bit of “hard maths”, then it is easy to prove that:

is true.

That means that the sum is Φ^2 + Φ^3 + Φ^4 + Φ^5 + Φ^6 + …, but since Φ^2 = 1 – Φ, that means that Φ^3 = Φ^2 * Φ, or Φ-Φ^2. If we keep multiplying, we begin to see a pattern here, where the exponent keeps going up by one, and in the full sum, there is, for any positive integer k, a Φ^k and a -Φ^k, which implies that they can be cancelled out. But there is no term to cancel the one in Φ^2, so one is left, and Φ^∞, since Φ is smaller than one, the limit as x approaches infinity of Φ^x converges to 0. That means the final sum is 1-0, or just 1.

I will prove that the sigma sum is true given the value of phi, which is also the reciprocal of the golden ratio.

Remember how Φ^2 = 1 – Φ? The sigma sum states that 1 – Φ + Φ – Φ^2 + Φ^2 – Φ^3 + Φ^3 – Φ^4 + …, that means the powers of Φ keep cancelling themselves out, and if a number x is smaller than one, the limit as n goes to infinity of x^n approaches zero. And since Φ is less than one, Φ^∞ must converge to zero.

That means the final sum is 1 + Φ^∞ after all the cancellation, and since Φ^∞ converges to zero, the final sum is 1 + 0, or just one.

Before we start the boring post, there are already videos on the Conway’s Soldiers, so check them out if you don’t want to read this!

There is an infinite field, with infinite soldiers (as many as you need), but finite moves. Their job is to infiltrate over the line. It’s like checkers, except horizontally and vertically instead of diagonally. Only one jump may happen at a time. The jumped pieces are removed. The board:

With infinite soldiers and an infinite field, it seems like the soldiers can reach infinity. But no!

There is proof that level five is impossible. This was proved by the creator himself, John Horton Conway. The proof is on Wikipedia, but I’ll try to explain it myself:

There exists a number similar to phi where:

φ^2 = 1 – φ (I tried to copy a lowercase phi, but it all changes to the uppercase, so instead of the phi symbolised above, it will be φ).

In the other side to the soldiers, if we mark the target square (one square in level 5) with φ^0 (aka 1), and the surrounding squares with φ^n (where n is the Manhattan distance to the target square). There is a “score” of the configuration of the soldiers by totalling the value of the soldiers’ squares. So if there were two soldiers positioned to reach the target square on the next move, the configuration “score” would be φ^1 + φ^2.

There are three possible moves:

The jump towards the target square changes the score by 0.

The jump that changes nothing changes the score by -φ^(n+1).

The jump further away from the target changes the score by -2φ^(n+1).

Therefore, the score will never increment, because φ is positive.

The configuration score of a row of soldiers directly underneath the target square is φ + 2φ^2 + 2φ^3 + 2φ^4 + 2φ^5 + 2φ^6 + …, and the score for a row of soldiers n spaces below the target square is {φ^(n – 1)} * (φ + φ^2 + φ^3 + φ^4 + …).

Remember how φ^2 = 1 – φ? New fact, where:

So is the target square is in level one, the score with no moves is: S(1) = {φ + 2(φ^2 + φ^3 + φ^4 + …)} * (1 + φ + φ^2 + φ^3 + …) Side note: the number in the score “function” is usually subscripted (small and lower), but WordPress won’t allow it so :v

Using the sigma function from above, we get S(1) = (φ + 2) * (1 + φ + 1) = (φ + 2)^2 = 5 + 3φ.

Since the difference between two conjoined squares is φ, to get the value of S(n), it is φ * S(n-1).

S(2) = 3 + 2φ

S(3) = 2 + φ

S(4) = 1 + φ

S(5) = 1

When a soldier reaches the target square T, the ending score is E = φ^0 + ε, where φ^0 is the contribution of the soldier on the target square, and ε is the contribution of the soldiers behind the line (our region). ε is small but always positive.

The starting configuration score is S(5) = 1, and the ending configuration score is E = 1 + ε, and this leads to a contradiction, since S(5) < E because ε is always positive, and the score never increases, so S(5) ≥ E, but a number cannot be smaller than another number that is smaller than or equal to that number.

But

hold

UP!!!!!!!!!!!!!!!!!!!!!!

Simon Tatham and Gareth Taylor proved that the impossible level 5 in Conway’s Soldiers is actually possible! Now if your first thought is, “how can that be true when Conway proved that it is impossible?” And you’re right. It uses up every soldier on the infinite board. We just have to tweak the rules and manipulate time a little bit.

If this doesn’t make sense, go to the link at the bottom, with all the links I used.

I’m not taking any credit for this solution, the credit goes to Simon Tatham and Gareth Taylor.

Think of making a large amount of moves in parallel, so using a full row of soldiers to get every one of them into level one. Not helpful, but it springs a new concept:

Doing a similar concept, but along one row. If a finite number of soldiers were in one row without any gap in between, the first one (far right) can be jumped by the second, the third by the fourth, the fifth by the sixth, and so one. First tweak! If an infinite number of moves can be made in finite time, that is legal. So the first jump in the second concept can be done at time t=0, the second jump at t=1/2, the third at t=3/4, the fourth at t=7/8, and so on, and at t=1, all jumps have been made, with time to spare. We made infinite moves in one second!

Now, we can only move one soldier at a time, so the idea of making large amounts of moves at once is scrapped, but if we take the “infinite moves, finite time” approach, we could do it in 1 second. We don’t know if t’s units of measurement is seconds, minutes or days, let’s call it arbunit (Arbitary Time Unit).

Another useful move that can be used is a reversed infinite sequence, or an ecneuqes etinifni. A clear end, but an uncertain beginning, and this strategy is very useful. We can visualise one soldier after consecutive jumps as we did in the second approach to the infinite sequence jump, but reversed. The two people created an awesome sequence which Gareth Taylor called a “whoosh”, most likely because it does look like a whoosh. In one row, the second jumps the first, fourth jumps third, and so on, until there are two soldiers at the left “end”, then the second part of the whoosh begins, where the leftmost soldier jumps over every soldier in the line until it finishes the rightmost jump, and is left alone after the whoosh.

The finite time? The first half of the whoosh can be done in one second, then the second half in another second, in the “keep halving and add, get 1” way, but in the second half, multiplying by two until reaching one arbunit.

Now, to advance further, Gareth Taylor and Simon Tatham created a “megawhoosh”, where it could literally turn a “quarter-plane” of soldiers into one soldier using a series of whooshes. The problem was that the megawhoosh only took the soldier to level two, so the two people started with a soldier at level three and worked their way backwards, doing one reverse jump, then reverse-whooshing both soldiers. If visualising it is hard, the visualisation is in the website linked down below. Move the left soldier left by reverse-jumping, then reverse-whoosh again.

Repeat infinitely, with the “halve every time” time management tool to fill the staircase of soldiers.

After the infinite whooshing, take every other soldier and whoosh them to the left, and then reverse-jump the soldiers to fill the gaps, which only takes a finite amount of steps in each column.

When that is done, the “quarter-plane” will be filled, and that is the reverse-megawhoosh. Now, to use the megawhoosh, do the exact same thing explained above, but in reverse.

Now, for the grand solution that Simon Tatham and Gareth Taylor devised!

Draw a line in the middle, where the final soldier will end up on the fifth level.

Megawhoosh the left half (not including the middle line) up to level three.

In the column that the line was drawn on, whoosh the column to level two, then jump the level three soldier to get to level four.

Megawhoosh the right half to level three, like the left half, then jump the soldier directly above the result of the second megawhoosh to reach level five.

This was pain to write, but definitely not as much as Simon Tatham and Gareth Taylor when they had to find the manipulated solution to Conway’s Soldiers.

QED.

P.S. I want to post the gif for the solution provided by the two people who found the solution, but for legal reasons, I won’t. Check out the website for the animation, or this link.

Also, none of this information originates from me.

When I say uninteresting, I don’t mean that this paradox is uninteresting. It means “relating to the interesting/uninteresting numbers” and this is a paradox that is about uninteresting numbers.

When we say interesting number, they’re numbers with special features. Like 1 is the first positive integer, 2 is the only even prime, 3 is the first odd prime, 4 is the first composite, and so on. But what about uninteresting numbers?

If the list of uninteresting numbers went 7, 13, 35, 57, 89, 103, 205, 784, 1083, 1849, 3456, 8465, 9172, 10928, 40923, 98402, … then obviously 7 is interesting because it’s the first uninteresting number. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. But wait. For 7 to be interesting, it has to be an uninteresting number, which means that 7 is uninteresting. But wait. There’s no number below 7 that’s uninteresting. So 7 is interesting. (And on and on and on and on and on and on and on and on and on and on and on and on…)

Sorry about that, I just wanted to do it. So what it suggests is that, if the list is as above, then 7 can neither be interesting or the opposite (or it could be both, but what?) because to be interesting, it has to be the smallest uninteresting number (which 7 is), but that also implies that 7 is uninteresting, so it’s not one or the other, but either both or none.

I mean, yeah. But in a prime/composite definition? It’s neither, the one exception, the golden, whatever.

Why is it not prime? Let’s have a look.

Take 25321, a prime number. 1 is also prime, right? So the factorization of 25321 would be 25321 and 25321*1 at the same time. But we have broken two golden rules:

Every number, even prime numbers, have exactly one prime factorization

A number cannot be both prime and composite at the same time.

Also, a proof that came in my head while writing:

You know Euclid’s Theorem that states ‘there is no prime because of this, that and stuff’? Full Proof:

Say the list of primes were finite, like 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, and that’s it. Multiply them all together, add one, it’s 6469693231 (2*3*5*7*11*13*17*19*23*29+1). Since this number is bigger than the “last” prime, it must be composite (some cases are when it actually is prime). But if you divide it by any “Prime” number, it will result in a remainder 1. So there must be some other “prime” out there that is prime.

Curtains, lights, action!

Say that one was prime, then Euclid’s Theorem to primes to 30 multiplication would still be 6469693231. But since 1 can divide 6469693231 evenly, and one is “prime”, Euclid’s Theorem has this weird exception, and that contradicts the fact that there are infinitely many primes.

1 isn’t composite either, since it doesn’t have any prime numbers that can divide it evenly.

Which is it? Well, it’s not 3.14 because pi is not rational, and that is the same process for 4, right?

Well, if you thought so, Congratulations! You are

WRONG!

It’s actually 4.

Take a square with perimeter 4 (side length 1) and a circle exactly in the square with perimeter pi (diameter 1). Cut the corners with squares the largest it can go. Repeat as much as possible, again and again, …

But for a better proof, it’s going to be awesome. First, draw lines, which will now be:

Now, since BF = BA and BC = BD, triangles BFC and ABD are congruent. Also, since A, K and L are collinear (all three points can form a straight line), and AL is parallel to BD, triangle ABD is half of rectangle BDLK. In the same way, BAGF is twice of FBC.

Since ABD and FBC are congruent, BDLK must be equal in area to BAGF. That is AB^2. WHEW! Also, draw lines AE and IB to find that KCEL = AHIC. Also known as AC^2!

Because the square BC^2 is made up of BDLK (aka. AB^2) and KCEL (aka. AC^2), we can now say that AB^2 + AC^2 = BC^2.

And that, everyone, is the Pythagorean Theorem (aka. Pythagoras’s Theorem) and I hope you enjoyed proving it.