When a complicated proof simplifies everything

Also, a special book cover if you're fast enough. You'll be fast enough.

Huge thanks to my Patreon supporters. I’m a huge exponent of all of
them, minus none.

CORRECTIONS
- None yet, let me know if you spot anything!

Filming by Alex Genn-Bash
Editing by Gus Melton
Graphics by Sam Hartburn and Matt Parker
Written and performed by Matt Parker
Produced by Nicole Jacobus
Music by Howard Carter
Design by Simon Wright and Adam Robinson

MATT PARKER: Stand-up Mathematician
Website:
New book!
parker-signed

[KhanStopMe]

i thought this was blindingly obvious until i realised the reason i thought that is because i had already picked 10 as my starting number so the proof was immediately intuitive

for a brief moment, i thought i was a genius

[karlmikko]

I raised to the power of 1. Made a load of sense.

[danielmcallister]

Love the second proof! In the vein of the first proof, there's the identity b^n - a^n = (b-a)*(b^n-1 + b^n-2 * a + .... + b * a^n-2 + a^n-1) that makes the divisibility quite clear.

[hallojava]

Nice proof by induction I made:

b^1 - 1 is obviously divisible by b - 1, as they are the same
If b^n - 1 is divisible by b - 1, then so is b^(n+1) - 1, as b^(n+1) - 1 = b^(n+1) - b + b - 1
= b(b^n - 1) + (b - 1)
First term divides by (b - 1), as b^n - 1 divides by (b - 1)
Second term (b - 1) divides by (b - 1)
Induction: True of first case, and second, and third, and so on...

[mattgsm]

"Wait it's all 9s"
"Always has been"

[jihoonkim]

The two proofs are actually very closely related!
In the first proof, you basically factor b^n - 1 into (b - 1) * (b^(n-1) + b^(n-2) + b^(n-3) + ... + b^0).
And in the second proof, the reason aaa...a in base b is divisible by a is because aaa...a (base b) = a * b^(n-1) + a * b^(n-2) + a * b^(n-3) + ... + a * b^0 = a * (b^(n-1) + b^(n-2) + b^(n-3) + ... + b^0).
Since a = b - 1, they are the same equation :)

[dtfd_]

0:05 b = 1
0:30 n = 0
0:41 oh...
So it turns out you can divide by 0 after all

[amorphant]

MATT! I think there's a lovely geometric proof as well! Visualize 5^2 as a 5x5 grid of squares. Remove the southeast corner. You now have a row of 4 (n-1) to the west of the missing square. Remove it. You also have a column of 4 to the north of the missing square. Remove it. You're left with a 4x4 grid, which is just more rows of 4. This works for any n^2.

It extrapolates to higher dimensions as well. If you make a 5x5x5 cube, then take out one corner, you can remove one entire face using the method above. Then, remove the strip of 4 that's aligned with the missing corner on the Z-axis -- just like you did with the "row" and "column" in the x^2 example -- which leaves you with a bunch of identical slices, each of which is identical to the 2-dimensional example after removing one corner piece.

[mocliamtoh]

I like how "complicating" the initial problem leads to a much more intuitive understanding of the proof. (And the dig at Amazon)

[kullen]

My first intuition when I heard "divisible by something" was to consider the whole situation modulo this something. There it also becomes quickly apparent, because modulo (b-1) we have:
b ≡ 1 (mod b-1) and therefore
b^n ≡ 1^n = 1 (mod b-1), hence
b^n - 1 ≡ 0 (mod b-1),
but the last equation is exactly synonymous to "(b^n - 1) is divisible by (b - 1)".

Of course this than relates to the fact, that 1 is a zero of the polynomial (x^n - 1), because the calculation above is just evaluating this polynomial in the according ring of residual integers (Z/(b-1)Z).

[yoavshati]

If you use b-1=a you can rewrite the problem as (a+1)^n - 1 being divisible by a, and if you were to expand it you would get a lot of terms with some power of a, and then +1-1, which just cancels

[Cmanorange]

i love the orange circle in the top right, brings the whole thing together

[nura]

I REALLY LOVE WHEN WE SWITCH NUMBER BASES AND IT BECOMES OBVIOUS
IT IS MY FAVORITE GENRE OF MATH

[efhiii]

"I don't think anybody's ever used base 440 before"
Hmm... 440 Hz is what A440 (Stuttgart pitch) tuning is based on.

I suppose it's more of a unit conversion than base though.

I'm sure someone's played 440 Hz on a bass before though, and maybe that would be bass 440.

A0 = 27.5 Hz
A1 = 55 Hz
A2 = 110 Hz
A3 = 220 Hz
*A4 = 440 Hz*
A5 = 880 Hz
A6 = 1760 Hz
A7 = 3520 Hz
A8 = 7040 Hz

[polygrum]

The "base b" insight is wonderful, but it really doesn't answer how you would decide to do that. For me it's much more natural to switch to mod b-1, where b^n - 1 = 1^n - 1 = 0 mod b-1, since when you have to prove that x is divisible by y it's often useful to switch to proving that x = 0 mod y.

[Nolys-bkkd]

To be honest, the first thing I thought of were geometric sums ( [b^n-1]/[b-1] = 1 + b + b² + ... + b^[n-1] ). But that's an admittedly more convoluted way of proving that [b^n-1]/[b-1] is an integer than proving that 1 is a root of x^n-1

[StarchedPie]

There is also a simple way to prove this by induction, for increasing n. (It's going to sound complicated in text but with pictures it's obvious)
For n = 2, imagine the it as a square grid, now take one off the corner, now you have a rectangle of b*(b-1) plus a line of b-1, which are both divisible by b-1.
For n = 3, imagine a cube, taking one off the corner leaves you with the same n = 2 square case on one face, plus a block of (b^2)*(b-1), this block is also divisible by b-1.
This can be extended to arbitrary n, where the 'block' will always be (b^(n-1))*(b-1), divisible by b-1.

[xHyperElectric]

I paused the video immediately after you said you picked 440, thought for a moment, decided on 17. I unpaused the video and the next sentence you said was someone else picked 17. I was reminded of Veritasium’s recent video on the human inability to create randomness.

[pitust]

I tried proving this myself without watching the rest of the video, and I made something like:
1. Base case: for n=1, b^n-1 is clearly divisible by b-1 (because they are equal)
2. Inductive case: Let's say b^n-1 is some x(b-1) + 1. Then: b^(n+1)-1 = b (b^n) - 1 = b(x(b-1)+1)-1 = bx(b-1) + b - 1 = bx(b-1) + 1(b-1) = (bx+1)(b-1) which is also divisible by b-1. QED

[Zerotan]

I thought of the cubes from grade school.
Your n-hypercube can be decomposed into a bunch of sticks and slabs and cubes and hypercubes that have (b-1) side lengths.... and.... 1, but you have conveniently subtracted that out.