When a complicated proof simplifies everything

Pre-order Love Triangle on Maths Gear now and get a signed copy! mathsgear.co.uk/products/love...
Also, a special book cover if you're fast enough. You'll be fast enough.
All UK options: www.penguin.co.uk/books/44315...
All USA options: bit.ly/3wCTesR
Video was inspired by this r/mathematics post and the discussion it lead to: / unjaymggts
Huge thanks to my Patreon supporters. I’m a huge exponent of all of
them, minus none.
/ standupmaths
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:
standupmaths.com/
New book!
mathsgear.co.uk/products/love...
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

[@zzzaphod8507]

That version of it may be obvious but it's a bit in-tens for me.

[@HunterJE]

Same, I saw the thumbnail and so knew I'd be raising it to an exponent, and so picked a number I knew would be easy to do that to without getting out a calculator. And yeah definitely gave away the trick on the second proof XD

[@johnnye87]

Once they've followed the step of converting to base b, technically everyone is using 10 as their starting number

[@malvoliosf]

I’m not convinced you are not.

[@gnaskar]

I picked 2. Which, yes, is divisible by 1.

[@karlmikko]

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

[@NonFatMead]

Now prove for case n+1 and you've got your inductive proof.

[@positivity3311]

me too

[@soberhippie]

I raised it to the power of 0, and it made no sense at all

[@gigantopithecus8254]

@@soberhippiei mean 0/(b-1) is an integrr for b!=1

[@lucas2nded461]

I chose 1 as my starting number "b", which I promptly regretted

[@mattgsm]

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

[@danielmcallister8902]

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.

[@ShongoStick]

ah yes, this complicated string of math that i don't understand....yes, i understood that perfectly

[@mathieuaurousseau100]

And now in the vein of the second proof, in (b^n)-1=(b-1)(b^n-1+b^n-2+...+b+1), if you write the second number in base b you get 111..1 (b ones)

[@saavyk1264]

@@mathieuaurousseau100 Whoa

[@nikos4677]

This was my first thought.

[@iang0th]

@@ShongoStick It's easier to understand (at least if you can write it out properly instead of trying to force it into a KZbin comment), if you look at like this: first set a=1, since we don't need the more general version, so we have (b-1)(b^n-1 + b^n-2 + ... b + 1) If you expand that out, you get (b^n + b^n-1 + ... b^2 + b) - (b^n-1 + b^n-2 + ... + b + 1). If you compare the positive terms with the ones being subtracted off, you'll notice that all of them cancel out except for the first and last, b^n-1.

[@hallojava2458]

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...

[@quentind1924]

I did that too!

[@kombat4135]

I thought of this as well

[@broskey4041]

was it specified that n is a set of natural numbers?

[@chaddaifouche536]

@@broskey4041 You mean *in* the set of natural numbers. And that's really a matter of conventions : if you use n as a value name for something else than a (natural) integer without precisions, you're an heretical monster in the mathematical community. Similarly, because we're discussing divisibility without more details, you can infer that b should be an integer too. Of course if you're writing a math paper or in your exam, you should **really** details all of this!

[@broskey4041]

@@chaddaifouche536 thank you for the correction and clarification. I didn't even know that divisibility implies working with integers that's kinda cool

[@henningsperr8063]

I thought of b=1

[@Elnadrius]

Me too!

[@JohnR31415]

Try b=pi, or root 2

[@revadiazairlangga5939]

same here

[@blackholedividedbyzero]

Zero divided by zero

[@EqSlay]

That's a bbbase case.

[@jihoonkim9766]

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 :)

[@samueldeandrade8535]

No sh1t, Sherlock.

[@novamc7945]

​@@samueldeandrade8535 You must be pretty smart wow

[@samueldeandrade8535]

@@novamc7945 thanks, but no. No one needs to be smart to NOT like obvious comments.

[@novamc7945]

​ @samueldeandrade8535 That's a stupid argument. According to your analogy, videos that explain a topic at a fundamental level should simply NOT exist. Afterall, it's obvious. Plus, it's certainly not like there are people out there that don't understand the subject you're so profoundly good at! If you'd already inferred what the comment was suggesting, you could've simply ignored it and moved on. There was simply no need to leave a discrediting reply. Nothing like a good old KZbin comment section argument.

[@samueldeandrade8535]

@@novamc7945 "According to your analogy ..." Wtf are talking about? What analogy? I made no analogy ...

[@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.

[@vsm1456]

I like your idea! When a problem is visualised it makes it much more simple for me.

[@jonathanallan5007]

This. I visualised exactly the same almost straight away.

[@jonathanallan5007]

...and going up in dimensions by one always adds (b-1) copies of the currently existing cubelets prior to the removal from the very first strip.

[@amethystklintberg7436]

This is beautiful!

[@Naryoril]

This should get more upvotes

[@dtfd_]

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

[@Mitchpott]

Does 0/0 = 1 or undefined or infinity

[@kutsen39]

​@@Mitchpott0/0 is considered undefined. Imagine a function (or graph it in Desmos) where it's some number divided by x, like y=1/x. If you start at x=1 and move towards x=0, you're dividing by a smaller number which means you're actually multiplying by a larger number, i.e. 1÷½=1×2. So as x approaches 0 from the positive side, the number races off to infinity. Now what happens if you start from x=-1? Well, the exact same thing, but the number is now negative. As x gets infinitely close to 0, the 'limit' (basically what we've just done) is different depending on how you look at it. So the limit is said to not exist. Lastly, imagine the age-old 0.99999... = 1. You might know it works because it's an infinitely long string of O's, but if you try to decide what its limit is (the number it gets infinitely close to), it approaches 1. There are some great proofs here for you to find. But in this respect of limits, we arrive at the conclusion that they are actually equivalent. So, because the limit does not exist, it is impossible to divide by 0. In actual calculus, it is said that if your result is 0/0, you've taken the wrong approach.

[@danceswithowls]

Factors aren't defined by division but by multiplication into a product. Given h × k = j, then h and k are factors of j. Likewise, since 3 × 0 = 0, both 3 and 0 are factors of 0. Nothing in there about division.

[@nickpro8116]

From the point of view of the definition of divisibility, 0 is actually divisible by 0, since there exists a number (any number in fact) that you can multiply by zero to get zero. The funny thing is, this still doesn't imply you can divide by zero, it's just the one single case where the terms "divisibility" and "division" collide and contradict each other. But for any non-zero number, divisibility for sure implies that you can divide by that number and get a single unambiguous result.

[@19divide53]

@@nickpro8116 The terms "divisibility" and "division" do not contradict each other. Division is defined as multiplication (one of the defining operation in a ring) by the multiplicative inverse (which exists for any nonzero element in a field). By definition, "division" excludes dividing by zero already.

[@mocliamtoh573]

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

[@kullen2042]

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).

[@jameshart2622]

Nice one. Really clean. It does have the downside of not knowing the resulting alternate factor, though. Still a good proof, though.

[@preetichandra7113]

Just got that answer too🤪🤪🤪

[@nura8578]

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

[@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

[@Subatomicfish]

I believe that coincides with how synthetic division works, which was how I went about thinking through the problem

[@softy8088]

I really like this one.

[@Penrose707]

This is how I went about it as well. Let F(b, n) = (b^(n) - 1) / (b - 1). Now evaluate at F(b + 1, n) = ((1 + b)^(n) - 1) / (b). Now expand the binomial and note that C(n ,n) b^0 = 1, which cancels the one in the numerator. Now we have a polynomial expression in terms of b multiplied by 1/b, which we can negate by subtracting one in all of our terms. Now evaluate this expression at b - 1 to show that F(b, n) = (b^(n) - 1) / (b - 1) = sum(k = 0, n - 1, C(n, k) * (b - 1)^(n-k-1))

[@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.

[@akaHarvesteR]

Base B is just mod B with a carry 😊

[@gideonk123]

But then when you need to divide by b-1 mod b-1, it would be like dividing by zero, or perhaps I didn’t understand you?

[@JdeBP]

The base b approach is natural when one is in school and one hasn't done modular arithmetic yet, but one _has_ encountered things like how to test in base 10 for divisibility by 9.

[@Pieter31]

@@gideonk123 if a number is 0 mod b-1, that means its divisible by b-1

[@quentind1924]

A way to find it would’ve been if you tried b=10 at some point in your tests

[@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.

[@AkiSan0]

well. technically he didnt say to pick a "fully random" number, but just a number we like, thus reducing the numbers chosen by a LOT.

[@Lombravia]

17 is just a decently likable number.

[@X1ma_]

I picked 17, but only because it used to be my favourite football player's number when I was a kid, and now it's my favourite number xD

[@sol_in.victus]

I picked 7 and i think there's something to it why is 7 a number people often pick

[@JdeBP]

I have changed my strategy for picking random numbers under 100 since I watched that. 30 and 90 are preferred random numbers, now. But, of course, there's the question of how many other people in the world have thought the same. (-:

[@Nolys-bk4kd]

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

[@prof.dr.jorgmeuthenne765]

same

[@samueldeandrade8535]

That's probably the original proof of this fact.

[@Schpeeedy]

well you can do your way without proving that for any a satisfying a polynomial f(x), (x-a) divides f(x). So I think its nicer

[@TechToppers]

I mean you would have to show after dividing it out, the polynomial has integer coeffs? If rationals then, we can't guarantee?

[@samueldeandrade8535]

@@Schpeeedy exactly.

[@ShinySwalot]

instructions unclear, I picked "b" to be the big famous constant e and it didn't work :(

[@Cruzz999]

Similarly, I picked pi. Fairly sure that's not working either.

[@spaceshipable]

I think 1 has to divide into your "b". I imagine if you picked b = ½, you could say b ^ n - ¼ is divisible by b - ¼. (I picked ¼ because it divides into ½). Therefore it makes sense that b has to be an integer for the video's equation to work. p.s. this could all be wrong, I've not actually checked the maths

[@adityaflashraj]

No it definitely works

[@oinkoink3669]

it only applies to integers. He should have said that

[@katherinescheper1951]

I picked -1/4. Can you believe that didn't work?

[@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

[@BigDBrian]

I'm sure the 440Hz playing bassist also had some sheets of paper: A0 = 1 m² A1 = 1/2 m² A2 = 1/4 m² A3 = 1/8 m² A4 = 1/16 m² ...

[@musicat3330]

@@BigDBrian Then if you multiply the pitch by the paper size, you get a very strange way to express kinematic viscosity: A4 × A4 = 440 Hz × 1/16 m² = 27.5 m²/s = 275 kilostokes (kSt)

[@AlRoderick]

A440 isn't a bass note, it's the very definition of mid-range.

[@efhiii]

@@AlRoderick it's not a bass note, but that doesn't mean it can't be played on a bass.

[@helsing7423]

My b was 2. Yep, i was underwhelmed

[@charstringetje]

That's odd

[@Centauris-ty8wn]

I think it’s prime time to retry with another number.

[@nathangamble125]

@@charstringetje No, 2 is definitely even.

[@helsing7423]

@@Centauris-ty8wn 2 is my favourite number. I would never forsake it.

[@kilroy1964]

​@@helsing7423Two forever!

[@Cmanorange]

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

[@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.

[@estherstreet4582]

Yeah, I found it fairly easy to visualise a proof for n=2 and n=3 just by chopping up shapes, but I wasn't sure how to visualise n=4 lmao

[@jacovisscher]

My thoughts exactly! Sad it wasn't original, happy people think in hypercubes!

[@UnCavi]

Yes, this is my favorite proof. You can also mix it with a bit of induction to make it simpler: b^n -1 is the volume of a n-cube with side lenght b, which has a small (n-cube)-shaped hole with side length 1 at one of its corners. Take a slice off of it from the top: you’re left with a n-parallelepid with one side lenght (b-1) and all other sides b, and the extra slice you cut off. You can “flat” this extra slice by looking at it from the front, suppressing the dimension along its side of lenght 1. Then, you’ve got the same shape as you started with, but which lives in 1 dimension lower: a (n-1)-cube with a unit corner missing. Repeat n times and you get all shapes with at least a side lenght of (b-1)

[@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

[@threesixtydegreeorbits2047]

The base change is sooo poggers

[@notnilc2107]

very skibidi indeed.

[@movieidiots5542]

Spoiler alert!

[@123MondayTuesday]

The naughty orange dot in the top right corner bugged me

[@MrCheeze]

Not only is the second proof easy to follow, it also immediately tells you what the other factor is and why (1, b+1, b^2+b+1, etc, depending on n)

[@raresaturn]

What is the second proof?

[@jacksonpercy8044]

The base B proof. Also I wasn't expecting to see MrCheese while scrolling through the comments. Neat.

[@Leandro-vy7nj]

I am also very stunned by the implications this could have on quickly determining if a number is divisible by another specific number. Powerful stuff

[@skyforger8102]

I Love the shade thrown at Amazon!

[@hoblesy]

I think a pretty intuitive way of thinking about it is geometrically, If you make a b X b square (or cube or whatever) it can be seen to be a load of sets of (b - 1). In the case of a square b rows of b - 1 and one extra column of b - 1

[@sumner1107]

This is how I visualized it as well

[@JavedAlam-ce4mu]

Doesn't the extra column have b elements? E.g if b = 4 and we square it: Say these asterisks represent b, so here is a 4x4 grid: * * * * * * * * * * * * * * * * I see b (4) rows of b-1(3) * * * * * * * * * * * * But this remaining extra column you refer to has b, not b-1 elements * * * * I just realised why this makes sense - because we are dividing by (b^n) - 1, so if you subtract this extra one off, then yes it is divisible by (b - 1) * * * But the way you described it was incorrect and confusing for me lol.

[@hoblesy]

@@JavedAlam-ce4mu sorry if I was unclear but yes we ended up at the same solution, if b = 4, you end up with: b rows of b - 1 (4 rows of 3) and one extra column of b - 1 (3) ###| ###|# ###|# ###|#

[@sumner1107]

@@JavedAlam-ce4mu if you remove the corner square then yeah, youll have a square of (b-1)^2 plus 2(b-1) sides

[@Patagonicus42]

I picked b=1. I feel like "a number" was not specific enough. Or is 0 divisible by 0? 🤔

[@galoomba5559]

Sure 0 is divisible by 0, 0 = k*0 for some k

[@chemicalbrother5743]

@@galoomba5559 That would make 0 / 0 = k. Which is not a unique value, so u can't divide 0 by 0.

[@mudkip_074]

Zero is indeed divisible by zero. Since there exists a whole number k such that kx0=0 (this is true for all k, but we can take k=0 as an example, 0x0=0). Generally speaking "a is divisible by b" and "a is a multiple of b" are the same statement. This trips a lot of people up because there's no answer to "zero divided by zero", which sounds like it should mean the same thing, but it doesn't.

[@Patagonicus42]

@@mudkip_074 Ah, right, hadn't considered that you can define divisibility by multiplication

[@Mmmm1ch43l]

@@chemicalbrother5743 unfortunately, the mathematical definition of "a is divisible by b" is in general not the same as "you can divide b by a". this is because the concept of divisibility is very useful even in situations where you don't have a concept of division at all. in general, "a divides b" is defined as "there's some k such that k*a=b", which is obviously satisfied for 0 and 0, because 0*k=0 for all k.

[@Huntracony]

My mind was blown and I immediately understood when you said to use the base of the number, wonderful proof. I love the bigger videos with locations and large production values, but I love these simple "Here's a cool math(s) thing!" videos too, and I'm glad you're doing both.

[@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.

[@kpaasial]

I like big exponents and I can not lie.

[@cosumel]

I was obsessed with the Mersenne series in middle school. I discovered early that 2^(2n)-1 was divisible by 2^2-1 and 2^(3n)-1 was divisible by 2^3-1. That’s how i checked my work.

[@Takame9]

I asked myself this exact questions a few years ago during a calm shift, and came up with 3 solutions, two of them being those you present (but I cannot remember the third)! My favourite one was using base b. Love those kind of reflections, great video

[@kenhaley4]

Very nice! I love these kinds of insights. I'll never forget this little gem now.

[@bighammer3464]

Pick any number but irrationals don’t work. He really Mat Parkered those instructions.

[@QuantumCurvature]

Actually, irrationals DO work, depending on your definition of divisibility. No matter the number, b^n - 1 can always be factorized as (b^(n-1) + b^(n-2) + ... + b + 1) * (b - 1). If you divide by b-1, you're left with just the b^(n-1) + ... + 1 term, whose representation in base b is always just 111...111. Thus, in base b, the resulting number is always an "integer" even if b itself is irrational.

[@twojuiceman]

If I remember correctly, putting maths in an unusual context because it tells you something about how the formula was derived, was exactly the sort of thing you were arguing _against_ in your tau vs pi smackdown with Steve Mould lol. I'd love to see a tau vs pi rematch with the two of you

[@hiccupwarrior89]

W pfp

[@samueldeandrade8535]

Oh my Euler, you are ins4ne.

[@Anonymous-df8it]

It's on Numberphile

[@robinsparrow1618]

love the jab at amazon right at the end!

[@heighRick]

Matt releasing the video I didn't think I needed today. Thanks, helps a lot!

[@ottertvmtg6229]

i picked b=900.1 it wasnt an integer, but it still worked

[@enderyu]

No it doesn't work?

[@kilroy1964]

Nope, it doesn't. I think b has to be a natural number greater than 1.

[@nathangamble125]

No, it does not still work. (900.1^2-1)/899.1 = 901.1. If it works the end result will be an integer.

[@QuantumCurvature]

@@nathangamble125 But it IS an integer, so long as you are working in base 900.1, in which case the answer is 11.

[@nattyzepko167]

That is GENIOUS! The base transformation is just so simple, I love it so much! I'm going to show everyone I know

[@_notch]

Oh wow, that second proof is incredible

[@dannysharpe6119]

Great work Jess! I’ve just run my first marathon and looking for inspiration for what to do next. I think you’ve inspired me to have another go at running a sub 45 minute 10k. I was 12 seconds short at last year’s Run Norwich… so have my goal for this September! 🤞

[@vincentlevarrick6557]

This. *This* is my favourite comment on a maths video. 😁

[@frenchertoast]

I like it when Matt does a video on a topic I fully understand, it makes me feel much smarter than I actually am.

[@mthielssalvo]

Brilliant video - the first proof came to mind immediately but the second one blew me away! Just pre-ordered the book because trig is my favorite, and also because the UK cover is much better than the US cover. (That being said I do hope there's significant respect paid to our friend the unit circle!)

[@MartyR-nm8nf]

3:45 s'all about dat base 'bout dat base no treble

[@nickkirkpatrick396]

This my friends, is a classic old school standup maths. Well done, Matt!

[@adityavardhanjain]

This video is so cool. Need more of such short fun videos.

[@artemisSystem]

In the introductory logic course at my university, an exercise in the book is to prove 6^n-1 is divisible by 5, by induction on n. I discussed it with some other students and we realized it works for any base

[@robertdarcy6210]

Even better, (x^n - y^n) is divisible by x-y

[@bigjukebox3370]

what a neat way to look at the problem!

[@9darkspells]

the revelatory feeling of understanding I got out of this was so incredibly unique. Exactly the feeling that I always am seeking when working in math or programming.

[@LeeSmith-cf1vo]

I love how intuitive the 2nd proof is (as long as you understand bases)

[@keyem4504]

That's a great technique. Maybe it'll come handy for other stuff as well. Loving it.

[@thentoxd]

I literally tried this problem yesterday from the Stanford Maths Problem book, and the next day BANG a video from the man himself.

[@platinummyrr]

Matt Parker: "Now, I don't think anyone's ever used base 440 before" Domotro: *exists*

[@phiefer3]

As others have pointed out, there are several other ways to prove this. One that I thought of is using modular arithmetic. If a=b-1, then b=a+1, which means that b is congruent to 1(mod a), and because you can multiply remainders this means that b^n is congruent to 1^n (mod a), or in other words b^n is 1 more than a multiple of a.

[@RexxSchneider]

Surely if you consider b^(n-1) + b^(n-2) + ... + b^2 + b + 1, and multiply it by (b-1), you get (b^n + b^(n-1) + ... + b^3 + b^2 + b) - (b^(n-1) + b^(n-2) + ... + b^2 + b + 1). It should be pretty clear that the expression collapses to b^n - 1 since all the other terms are both added and subtracted. So (b-1)(b^(n-1) + b^(n-2) + ... + b^2 + b + 1) = b^n - 1. A rather trivial result, IMHO.

[@calvinrollins4957]

Haven't watched the video yet, but since I use different bases often enough I notice that in base n any exponent of n would always be a 1 followed by a given number of zeros and subtracting 1 from any of those numbers would give you a series of (n-1) and thus be factored easily into (n-1)(1111...) so yeah that makes a lot of sense that (b^n)-1 would be divisible by b-1

[@amethystklintberg7436]

YEEEESSSS!!! Did you overhear my meeting with my academic supervisor a few months ago? This is so validating! I told him it’s so easy to see divisibility when you write the Mersenne number (2^mn)-1 in binary and then just look at the number! You can see so much by just looking at it! (The string of mn 1s is divisible by a string of n 1s, so the original is divisible by the smaller Mersenne number 2^n-1, and this is why a Mersenne number with a composite power can never be prime.) His reaction was that the difference of powers formula is an easier proof, and my immediate reaction was just 👀 Big lesson for me: translate concepts into what’s most familiar to my audience! And, in the academic community, that means translating my visuals into familiar formulas!

[@Packbat]

I had to pause at 3:47 to process just how patently obvious the result just became. That's so good.

[@LukeNAndo]

Wow I love that! The initial proof confirms to me that it works, but the second proof shows me how it works!

[@palpatinewasright]

I feel so pleased I worked this out while the video was running! I paused the video as requested, and tried this with b=10, saw lots of 99999s and had to leap from the bathtub and run down the street shouting "BASES! BASES! BASES!!"

[@EmilJ-gx3et]

By induction: 1. for n=1 the equasion holds true as b-1=0(mod b-1). 2.We need to show that If b^n -1 = 0 (mod b-1) then b^(n+1)=0 (mod b-1) Here is the proof: b^n-1=0(mod b-1)b^(n+1)-b=0(mod b-1)b^(n+1)-1=0(mod b-1)

[@arandomdiamond2]

I went ahead and tried it myself geometrically before watching this video. Obviously I can't imagine more than 3 dimensions but I can take a length of boxes and make a square or cube out of them. Then I reframe b as a=b-1 and then the question becomes "Why does squaring or cubing a+1 then subtracting 1 leave a mass that is divisible by a?" That way, I can see the original square from a and then the extra bits at the ends from the a+1 and of course there is always an extra box in a corner that doesn't fit in the other sections. And with that understanding, I can easily prove this for higher dimensions too.

[@arandomdiamond2]

Actually same idea as the second proof, we even both reframed as a too.

[@soyalguien335yt4]

Induction: b-1 divides (b^1) -1 = b-1 Let's assume b-1 divides (b^n)-1 = k b^(n+1) = b^n * b = (k+1) b = kb + b We need to prove b-1 divides b^(n+1)-1 = kb + b -1 And since b-1 divides k, b-1 divides kb, and b-1 divides b-1, we proved b-1 divides kb+ b+1 which equals to b^(n+1)-1

[@hyperbaroque]

Discrete Math ftw.

[@adamluhring2482]

The n=2 & n=3 cases have a very satisfying visual proof for this.

[@saavyk1264]

That second method was just beautiful. Beautiful. Wow.

[@namkromh6381]

I love this. As soon as I heard base B, I knew where it was all going. Lovely

[@Akolyx]

OK, I was surprised by the title, but the proof! That's very cool! I don't know any other example of using every number as a base for solving a problem, so that makes it much cooler. Might want to try giving this as an exercise in understanding the bases.

[@tttITA10]

Both of these proofs are so infuriatingly simple, I love it.

[@marcvanleeuwen5986]

The most standard way to test divisibility by m it to compute modulo m. And indeed working modulo b-1 we just get 1^n-1=0, which is even more trivial than your base b approach. Also you might recall the formula for the geometric series 1+b+b^2+...+b^{n-1} = (b^n-1)/(b-1) which implies this divisibility.

[@andrewdenne6943]

another interseting way you can prove it (It might be the same method and I just got confused) is to start with a^0+a^1+a^2+...+a^b=a^0+a^1+a^2+...+a^b then change the start to a 1 1+a^1+a^2+...+a^b=a^0+a^1+a^2+...+a^b then move it over and factorise (a^0+a^1+a^2+...+a^(b-1))a=a^0+a^1+a^2+...+a^b-1 (a^0+a^1+a^2+...+a^(b-1))a-1(a^0+a^1+a^2+...+a^(b-1))=a^b-1 (a^0+a^1+a^2+...+a^(b-1))(a-1)=a^b-1 a^0+a^1+a^2+...+a^(b-1)=(a^b-1)/(a-1) you get the original equation the cool thing about doing it this way is you can also do if you start with a^(0c)+a^c+a^(2c)+...+a^(bc)=a^(0c)+a^(c)+a^(2c)+...+a^(bc) and do the same process 1+a^c+a^(2c)+...+a^(bc)=a^(0c)+a^c+a^(2c)+...+a^(bc) (a^(0c)+a^c+a^(2c)+...+a^(bc-c))a^c=a^(0c)+a^c+a^(2c)+...+a^(bc)-1 (a^(0c)+a^c+a^(2c)+...+a^(bc-c))a^c-(a^(0c)+a^c+a^(2c)+...+abc-c)=a^(bc)-1 (a^(0c)+a^(c)+a^(2c)+...+a^(bc-c))(a^(c)-1)=a^(bc)-1 a^(0c)+a^(c)+a^(2c)+...+a^(bc-c)=(a^(bc)-1)/(a^(c)-1)

[@felixmerz6229]

Woah, that's such a cool proof. So simple and approachable.

[@christopherwork9676]

My approach was to rewrite b^n as ((b-1)+1)^n and look at binomial expansion using Pascal's Triangle. Treat (b-1) and 1 as the two terms in a binomial. b^n = (b-1)^n*1^0 + ....[several terms with some power of (b-1) in them]....+(b-1)^0*1^n So the +1 at the end is the only term in the binomial expansion without (b-1) as a factor. Subtracting 1 leaves only numbers divisible by (b-1). From here you can generalize it to (b^n-a^n)/(b-a) is a whole number. I love Pascal's Triangle for approaching problems where you want a generalization for any power. It can be used to prove the power rule for derivatives (for positive powers at least). The derivative of x^n can be found in the second number in the nth row of Pascal's Triangle if you do a binomial expansion of the numerator of lim as h->-0 of [f(x+h)-f(x)]/h for f(x)=x^n and then cancel every remaining h.

[@highlandyeoman]

Awesome video Matt! I found it was easiest to build intuition for this problem by visualizing the n = 2 and n = 3 cases. If we draw a b x b grid and remove the top right corner, it's pretty easy to see that the remaining cells are a b-1 x b-1 square starting in the bottom left, a b-1 row at the top, and a b-1 column on the right. And the 3-d version just scales each of these pieces up - there's a b-1 x b-1 x b-1 cube, 3 b-1 x b-1 squares, and some b-1 columns and rows. (that said, I'm pretty sure this is entirely analogous to your base-b approach)

[@androidlogin3065]

Very beautifull proof, easy to do and obvious proof, just what i like the most. But not so easy to think on it, without any external help.

[@tszhin814]

I am so glad I stayed through this video. I was like yeah yeah yeah roots and stuff, then BAM! That was MAGNIFICENT❤

[@eli0damon]

If you look at b^n-1 as an integer polynomial in b, it's factors have a really interesting structure. Each positive integer has a corresponding polynomial (call it p_i), and b^n-1 is the product of p_i(b) for each factor i of n (and a minus sign). For example, b^4-1=-(1-b)(1+b)(1+b^2)=-p_1(b)*p_2(b)*p_4(b) and b^6-1=-(1-b)(1+b)(1+b+b^2)(1-b+b^2)=-p_1(b)*p_2(b)*p_3(b)*p_6(b) .

[@bobthegiraffemonkey]

As a speedcuber and having solved 4d and 5d cubes, i very quickly understood it geometrically. Very closely related to the second proof. Helped that i picked 6 which is small and easy to visualise.

[@ThatShushi17-mc7ct]

I've always thought about the factors of consecutive numbers, but only now after you said it did it become obvious to me why they wouldn't have any.

[@quintopia]

The second way is the way I've always thought about it, but I liked the simplicity of the first proof.

[@MitchBurns]

I literally figured this out very recently when trying to fall asleep at night. Figuring out things like this is my version of counting sheep.

[@ktkrelaxedscience]

Awesome. 😃 Sharing it at once. 👍

[@dbrunnermusic]

There's a geometric proof too. It's hard to describe in words because it's geometric, but I'll give it a go. If it sounds complex in the general case, picture it with n=3. Let's start by defining a "nibbled hypercube". Start with an n-dimensional hyper-cube with side length b. Take a little bite out of one corner - a hyper-cube with side length 1. The volume of this is just the volume of the original cube, minus the volume of the nibble, i.e. b^n - 1. Now, on a face which touches the nibbled bit, shave off a width 1 slice of the hyper-cube. Since it has width 1, the volume of this slice is the same as the (n-1)-dimensional volume of the "face" of the slice. And that face is an (n-1) dimensional nibbled hypercube. Its volume is b^(n-1) - 1, and by our inductive hypothesis, that is divisible by b-1. Having shaved a width-1 slice off our original nibbled cube, what we're left with is a hyper-rectangle. The side perpendicular to the slice has length b-1, so the volume of this is also divisible by b-1. Since we've shown that an n-dimensional nibbled hypercube can be broken into a two shapes whose volumes are both divisible by 1, we know that the it's total volume, i.e. b^n 1, is divisible by n-1.

[@dbrunnermusic]

PS I didn't start by thinking in n-dimensions, I'm not a genius! I started in 2-d, then thought about 3-d, and fortunately that was enough to see how the generalisation would work :)

[@pranaypallavtripathi2460]

Loved the second proof 👍

[@seizan88]

I love this 😂❤ it makes me feel smart despite not doing anything, really

[@BrettDalton]

Love it... That is beyond elegant

[@cdsmithus]

Two more proofs: 1. Work in mod (b-1). b is congruent to 1, so b^n is congruent to 1^n, which is also 1. Since b^n is congruent to 1, b^n - 1 is congruent to 0, so it's divisible by b-1. 2. By induction on n. If n = 0, then b^0 - 1 = 0, which is divisibly by absolutely anything. But if b^(n-1) - 1 is divisible by b-1, then there's some a where b^(n-1) - 1 = (b-1) a, so b^n - 1 = (b - 1) b^(n-1) + (b^(n-1) - 1) = (b-1) b^(n-1) + (b-1) a = (b - 1) (b^(n-1) + a).

[@electricportals3644]

b-1=a b^n-1=(a+1)^n-1 All terms in the polynomial have a as a factor therefore the number is divisible by a. I usually think of these problems using convenient bases but I found my method simpler

[@alansmithee419]

My thinking: You have a number b. This is of course 1 more than a multiple of b-1 (specifically 1 more than b-1 itself), which I will call A. If you multiply a number that is 1 more than a multiple of A by m, you get a number that is m more than a multiple of A, which if m is greater than A is also m mod A greater than a multiple of A. So if you multiply b by itself, you get a number b bigger than a multiple of A, so the resulting number must be b mod b-1 = 1 more than a multiple of A. If you multiple this by b you can go through the same logic as in the last step, and so on. So b^n is always 1 more than a multiple of b-1, or b^n-1 is divisible by b-1.

[@divVerent]

My favorite way is still rephrasing with a substitution a=b-1: b^n - 1 = 0 (mod b-1) (a+1)^n - 1 = 0 (mod a) At this point you either know modular arithmetics, replace by 1^n - 1 = 0 (mod a) - or you don't, and you expand that power into a^n + whatever a^(n-1) + ... + whatever a + 1 where all the whatevers are binomial coefficients, and you are done too.

[@ai22sbro]

You can also think of it geometrically as a square/cube of side b where you remove one corner. You're left with b rows of b-1, and a remaining column of b-1 (in the cube instance this also leaves b squared "depths" of height b-1).

[@Adamdun11]

A neat mental visualisation for this I always come back to for this (which is basically just the (b - 1)(b + 1) solution) is I imagine a grid of n x n dots (I’ve used the neodymium magnetic balls), remove the top row and append it as a new column. You should have a grid of (n - 1) x (n + 1) and one ball overhanging. This shows that for any n^2 - 1 the value is technically divisible by both (n - 1) as Matt showed, but ALSO (n + 1). Again, this is just the original proof but I always think of it as a nice, practical representation of it :)

[@NeoJackBauer]

Love this!

[@flinkbeder]

Surprised you didn't mention that x^n-1 can be factored into (x-1)(x^(n-1)+x^(n-2)+....1). Is that the same as the polynomial root fact you mentioned?

[@joeeeee8738]

Finally background noise while talking and not uncomfortable silence Matt !!

[@ThaOnlyHatman]

I thought about it before you started explaining and came up with this: a = b-1 b^n = a*b^(n-1) + b^(n-1) b^5 = a*b^4 + b^4 b^4 = a*b^3 + b^3 b^3 = a*b^2 + b^2 b^2 = a*b + b b^1 = a*b^0 + b^0 = 1 We see that the rest we get when dividing b^n with b-1 is ultimately 1. Therefore removing 1 from the original b^n will give us a number divisible by b-1. Hope this makes sense!

[@sergiohernandezdiaz6032]

Ordered my copy of the book!

[@azrobbins01]

I like it! Makes total sense when you see what is going on!