2^n - 1 = Prime number - not divided by any number?
@Quantris9 минут бұрын
guarantee huh? b = 1
@douglasmagowan270910 минут бұрын
The proof that came to me while watching... if a = b-1, then a|b^n-1 is equivalent to a|(a+1)^n - 1. The binomial expansion of (a+1)^n has a lot of a^k terms and a constant term of 1.
@40watt5311 минут бұрын
You can also think of a small exponent like 2 or 3 and think of it geometrically!
@stevenlundy12 минут бұрын
I picked 17 randomly before Matt told me Nicole had also picked. I felt like Matt got meta-derricked by his recent video on random numbers kzread.info/dash/bejne/lmqds9ShYrexaZs.htmlsi=UKtMTcxogfTzMRtC&t=303
@taylorsmiles414312 минут бұрын
I started with 1. Divide by zero?
@tttITA1019 минут бұрын
Both of these proofs are so infuriatingly simple, I love it.
@shannonmarbut364821 минут бұрын
I had this realization during the summer I wasted thinking about the collatz conjecture.
@Smarwell12321 минут бұрын
This video brought a huge smile to my face, excellent proof
@4Kalmar22 минут бұрын
I solved it using polynomial division (which is pretty much equivalent)
@maxrs0724 минут бұрын
This is basically: x - integer b^n -1 = x*(b-1) mod b b^n -1 = x*(-1) mod b (ex 99 mod 100 = -1 mod 100) 3rd arm wavy way to prove this would be to use abstract algebra: using that b and b-1 are coprime and multiplicative order of b-1 in Z_b
@robinros259526 минут бұрын
Calculating mod b-1, b is equivalent to 1 and so b^n - 1 is equivalent to 1^n - 1 = 0, so b^n - 1 is divisible by b - 1. That's the easiest proof I could come up with
@robinros259521 минут бұрын
Similar, in base b-1, the number b is written as 11. Given that 11^n ends in 1, 11^n - 1 is divisible by a
@eli0damon27 минут бұрын
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) .
@kameronpeterson360129 минут бұрын
dont let the number be "x", let it be x+1 (x+1)^n, by the binomial theorem, equals x^n + x^(n-1) + x^(n-2) ... + x^3 + x^2 + x + 1 subtract 1 from the expansion, you get x^n + x^(n-1) + x^(n-2) ... + x^3 + x^2 + x (notice each term contains an x) subtract 1 from the original number x + 1 - 1 = x since each term contains an x, they're all divisible by x
@Krebzonide30 минут бұрын
You didn't specify b and n must be integers greater than 0.
@Stratelier31 минут бұрын
Fun fact: EVERY number base is "base 10" when written _in its own system._
@danielhale135 минут бұрын
I wanted to get spicy by picking _e_ as my base and _i_ (sqrt(-1)) as my exponent. Then I realized I'd have to do math with those, and these are the bully numbers that cause me the most emotional damage on a daily basis, so I chickened out. Best to not get cheeky with math when math is already not your friend. :P
@ompizz47 минут бұрын
For me it's an n-dimensional cube missing a corner unit cube. The edges next to the corner are b-1 long, faces next to those are (b-1)^2, cubes (b-1)^3.. All that's divisible by b-1.
@Akolyx49 минут бұрын
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.
@MrJaCraig50 минут бұрын
Small flaw with the beginning part: (12^-1) -1 = -0.9166666. Pretty sure n has to be >= 0.
@namkromh638152 минут бұрын
I love this. As soon as I heard base B, I knew where it was all going. Lovely
@MichaelPetito53 минут бұрын
Did not work for b=1 😢
@jurghaag276454 минут бұрын
Your first proof is a bit shakey. Just because b=1 is a root of a polynomial, this does not mean that the other factor is an integrer number for interer values of b. Using the same logic, we could say that 1/2(b-1) is always divisible by b-1, which it obviously is not. Of course it works for polynomials with integer coefficients, but the reason why, is not completely obvious unless you have a great intuition about long division with polynomials.
@aioia388556 минут бұрын
same as (b+1)^n - 1 = 0 (mod b) which is obvious from expanding (b+1)^n using the binomial theorem
@carlthepumpkinman57 минут бұрын
Uh b was 1 so what happens when I try to divide 0 by 0
@douglaswolfen782057 минут бұрын
I was lucky enough to jump to the "base" realisation in the first few seconds, so I could see it was true. But that didn't feel like a complete proof to me, so I actually went through a proof by induction for all integer values of n I found it was easier if I proved it for a+1 and a, instead of for b and b-1
@MichaelKatzmannСағат бұрын
The real magic using math is in encoding and decoding COFDM. Thousands of orthogonal carriers each with QAM generated with an IFFT! Just fantastic. See the BBC paper "The how and why of COFDM" by J.H. Stott
@TomahakaСағат бұрын
Base "b" this feels in the same vain as a recent demotro / Combo class video. Good stuff
@funtechuСағат бұрын
I picked b = 2 😊
@UlmaramluСағат бұрын
WTF did KZread change their layout? why? why would they do this?
@jnsdroidСағат бұрын
I chose b = 164 and n = 42 ... and found out about a bug with the Mod function in excel
@zyxwvu3677Сағат бұрын
Would love to get my hands on a signed copy of the Love Triangle book! Unfortunately the website only allows me to pay through a credit card, not Paypal or iDeal/bank transaction. Still, thanks for the fun video and all of your work so far!
@JustinK0Сағат бұрын
i did 69^4= 22'667'121 (22'667'121-1)/(69-1)=333'340
@rikschaafСағат бұрын
The first method really reminded me of the pascal triangle. Could it be explained using that?
@Utesfan100Сағат бұрын
I picked b=i and n=8. Clearly 0 is divisible by anything.
@vonriel1822Сағат бұрын
Uh oh. I chose b=525600.
@TheDarkElderСағат бұрын
"Choose a number"... well, it should be an integer b >= 2 else it fails.
@qwaqwa1960Сағат бұрын
We musicians regularly use base-440....
@jell0goeswiggleСағат бұрын
I got ((2π)³-1)/(2π-1). Please help. Complex numbers with integer components fared surprisingly well after I stopped screwing up. Rational numbers not so well. B=2+(i/4) n=3, for example. (Matt didnt specify integers until @4:55!)
@RichardWinskillСағат бұрын
I ran into a bit of a problem by picking 1 at the beginning...
@WAMTATСағат бұрын
I chose 1, it didn't work.
@alexeynezhdanov2362Сағат бұрын
Didn't work for me. The end result wasn't divisible by n-1 (I chose 1 initially)
@DeJay7Сағат бұрын
Reading the comments this is what I realised about this phenomenal theorem: ONLY Matt finds this so unbelievably and utterly shocking.
@RandyKing314Сағат бұрын
Matt, you preempted everything I was gonna say about the problem so my comment is moot… but still thanks for the video!
@wyattstevens8574Сағат бұрын
When you said b=440 was where you started, I chose (for musical reasons- 440 Hz is almost always concert A, i.e. the one directly above middle C; sometimes tuned to 442 or 444 Hz instead) I divided by 2 a couple of times and settled on 110. Still A, but just over an octave below.
@wiggles7976Сағат бұрын
Maybe there is another mathematics lesson to be had here. We used the fact that in any base b, there is a unique representation of any integer as a linear combination of powers of the base where the scalars can only be anything from 0 to b - 1. (Why aaaaaaa? What about 123a125a9 or any other notation? You can't just assume b^n - 1 can be written as aaaaaaa.) We do take it for granted all the time in base ten, but now we see why we may want to look at the fundamentals and be grounded in rigor. When base systems were first devised, they didn't know every integer had a unique representation until they proved it.
@DeJay7Сағат бұрын
Not to be weird or whatever, but 999...99 being divisible by 9 as 9x111...11 is "obvious" because we know that for base ten ("base 10" just looks silly ...) But can you prove that for any base b and a = b - 1 that the number aaa...aa is equal to ax111...11? Do you even have to? You already answered that as no, but why is that the case?
@bruceleenstra6181Сағат бұрын
You said that the second option for a proof is to put it in an unusual context. So what's wrong with the usual geometric proof? The visual proof of b² - 1 = (b - 1)(b + 1) extended to bⁿ - 1. Geogebra and Desmos have joined the chat.
@salvaje1Сағат бұрын
The yellow dot in the corner is really bothering me
Пікірлер
2^n - 1 = Prime number - not divided by any number?
guarantee huh? b = 1
The proof that came to me while watching... if a = b-1, then a|b^n-1 is equivalent to a|(a+1)^n - 1. The binomial expansion of (a+1)^n has a lot of a^k terms and a constant term of 1.
You can also think of a small exponent like 2 or 3 and think of it geometrically!
I picked 17 randomly before Matt told me Nicole had also picked. I felt like Matt got meta-derricked by his recent video on random numbers kzread.info/dash/bejne/lmqds9ShYrexaZs.htmlsi=UKtMTcxogfTzMRtC&t=303
I started with 1. Divide by zero?
Both of these proofs are so infuriatingly simple, I love it.
I had this realization during the summer I wasted thinking about the collatz conjecture.
This video brought a huge smile to my face, excellent proof
I solved it using polynomial division (which is pretty much equivalent)
This is basically: x - integer b^n -1 = x*(b-1) mod b b^n -1 = x*(-1) mod b (ex 99 mod 100 = -1 mod 100) 3rd arm wavy way to prove this would be to use abstract algebra: using that b and b-1 are coprime and multiplicative order of b-1 in Z_b
Calculating mod b-1, b is equivalent to 1 and so b^n - 1 is equivalent to 1^n - 1 = 0, so b^n - 1 is divisible by b - 1. That's the easiest proof I could come up with
Similar, in base b-1, the number b is written as 11. Given that 11^n ends in 1, 11^n - 1 is divisible by a
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) .
dont let the number be "x", let it be x+1 (x+1)^n, by the binomial theorem, equals x^n + x^(n-1) + x^(n-2) ... + x^3 + x^2 + x + 1 subtract 1 from the expansion, you get x^n + x^(n-1) + x^(n-2) ... + x^3 + x^2 + x (notice each term contains an x) subtract 1 from the original number x + 1 - 1 = x since each term contains an x, they're all divisible by x
You didn't specify b and n must be integers greater than 0.
Fun fact: EVERY number base is "base 10" when written _in its own system._
I wanted to get spicy by picking _e_ as my base and _i_ (sqrt(-1)) as my exponent. Then I realized I'd have to do math with those, and these are the bully numbers that cause me the most emotional damage on a daily basis, so I chickened out. Best to not get cheeky with math when math is already not your friend. :P
For me it's an n-dimensional cube missing a corner unit cube. The edges next to the corner are b-1 long, faces next to those are (b-1)^2, cubes (b-1)^3.. All that's divisible by b-1.
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.
Small flaw with the beginning part: (12^-1) -1 = -0.9166666. Pretty sure n has to be >= 0.
I love this. As soon as I heard base B, I knew where it was all going. Lovely
Did not work for b=1 😢
Your first proof is a bit shakey. Just because b=1 is a root of a polynomial, this does not mean that the other factor is an integrer number for interer values of b. Using the same logic, we could say that 1/2(b-1) is always divisible by b-1, which it obviously is not. Of course it works for polynomials with integer coefficients, but the reason why, is not completely obvious unless you have a great intuition about long division with polynomials.
same as (b+1)^n - 1 = 0 (mod b) which is obvious from expanding (b+1)^n using the binomial theorem
Uh b was 1 so what happens when I try to divide 0 by 0
I was lucky enough to jump to the "base" realisation in the first few seconds, so I could see it was true. But that didn't feel like a complete proof to me, so I actually went through a proof by induction for all integer values of n I found it was easier if I proved it for a+1 and a, instead of for b and b-1
The real magic using math is in encoding and decoding COFDM. Thousands of orthogonal carriers each with QAM generated with an IFFT! Just fantastic. See the BBC paper "The how and why of COFDM" by J.H. Stott
Base "b" this feels in the same vain as a recent demotro / Combo class video. Good stuff
I picked b = 2 😊
WTF did KZread change their layout? why? why would they do this?
I chose b = 164 and n = 42 ... and found out about a bug with the Mod function in excel
Would love to get my hands on a signed copy of the Love Triangle book! Unfortunately the website only allows me to pay through a credit card, not Paypal or iDeal/bank transaction. Still, thanks for the fun video and all of your work so far!
i did 69^4= 22'667'121 (22'667'121-1)/(69-1)=333'340
The first method really reminded me of the pascal triangle. Could it be explained using that?
I picked b=i and n=8. Clearly 0 is divisible by anything.
Uh oh. I chose b=525600.
"Choose a number"... well, it should be an integer b >= 2 else it fails.
We musicians regularly use base-440....
I got ((2π)³-1)/(2π-1). Please help. Complex numbers with integer components fared surprisingly well after I stopped screwing up. Rational numbers not so well. B=2+(i/4) n=3, for example. (Matt didnt specify integers until @4:55!)
I ran into a bit of a problem by picking 1 at the beginning...
I chose 1, it didn't work.
Didn't work for me. The end result wasn't divisible by n-1 (I chose 1 initially)
Reading the comments this is what I realised about this phenomenal theorem: ONLY Matt finds this so unbelievably and utterly shocking.
Matt, you preempted everything I was gonna say about the problem so my comment is moot… but still thanks for the video!
When you said b=440 was where you started, I chose (for musical reasons- 440 Hz is almost always concert A, i.e. the one directly above middle C; sometimes tuned to 442 or 444 Hz instead) I divided by 2 a couple of times and settled on 110. Still A, but just over an octave below.
Maybe there is another mathematics lesson to be had here. We used the fact that in any base b, there is a unique representation of any integer as a linear combination of powers of the base where the scalars can only be anything from 0 to b - 1. (Why aaaaaaa? What about 123a125a9 or any other notation? You can't just assume b^n - 1 can be written as aaaaaaa.) We do take it for granted all the time in base ten, but now we see why we may want to look at the fundamentals and be grounded in rigor. When base systems were first devised, they didn't know every integer had a unique representation until they proved it.
Not to be weird or whatever, but 999...99 being divisible by 9 as 9x111...11 is "obvious" because we know that for base ten ("base 10" just looks silly ...) But can you prove that for any base b and a = b - 1 that the number aaa...aa is equal to ax111...11? Do you even have to? You already answered that as no, but why is that the case?
You said that the second option for a proof is to put it in an unusual context. So what's wrong with the usual geometric proof? The visual proof of b² - 1 = (b - 1)(b + 1) extended to bⁿ - 1. Geogebra and Desmos have joined the chat.
The yellow dot in the corner is really bothering me
based