Golden Proof – Numberphile

Golden Proof – Numberphile

The Fibonacci numbers again start: 1, 1, 2, 3, 5 and the rule is: if each number here equals the one before it So, if this was the nth number and we’re gonna call that x n for the nth number it equals the one before it plus the one before that so that equals x n minus one plus x n minus two So that’s our finding thing and what we’re trying to find ultimately is what is the ratio between the two of the numbers and importantly, because we’re looking for the ratio it is approaching the limit it doesn’t matter which two terms it should be the same ratio, give or take and if you do infinitely many the same ratio for all of them but, rrrrhhhhh So. Uh. For any two of the terms so what we can say now is the ratio between x to the n and the one before that which was x to the n minus one. The one before. should equal the same as the ratio between x n minus one to the one before that which is x n minus two and we’re gonna call it What letter would you like Brady? awa aya a any any Greek letter? Do you have a favorite? Brady: uhhh…. delta? Guy: DELTA! ok, so lets do a capital delta because its quicker to draw so we’re trying [to come up] with this ratio, delta is. and we know its the ratio between any two consecutive terms now we know according to our relationship which generates this sequence that x n equals the two before it added together so I’m gonna put that in intstead of so all I’ve done is substitute that in but now I could start simplifying because I’ve got the same term there twice so, in fact, if I divide through that’s one plus x to the n minus two over x to the n minus one which still equals x to the n minus one over x to the n minus two which still equal to the ratio delta we’re looking for so now you’ll go hang on a second that is the same as this but upside down and in fact that equals that ratio delta so I can now change this to be one plus well that’s the inverse of this ratio so that’s one over delta equals and that there is the delta and so the relationship for the ratio of any sequence where you get the next term from adding the two previous ones one plus the inverse of that ratio equals the ratio Now what ratio could that possibly be? well let’s work it out! because we can turn this into a quadratic? If I multiply through by triangle it equals that plus one equals triangle squared and so delta squared minus delta minus one equals zero So, all I’ve done is rearrange that formula now we can solve this using our friend the quadratic equation cuz we know that if you have a times x squared plus b x plus c equals zero then x equals negative b plus or minus the square root of b squared minus four a c over two a so we can substitute in to use the quadratic formula So we now know delta equals negative b, which is the coefficient there which will be one plus or minus the square root of b squared –is going to be one minus four times one, times negative one, so it’s going to be plus four. Two times one is two. So that’s going to equal… Now I’m only going to use the positive for now. One plus root 5 divided by two, which is the golden ratio. If you actually work that out, that’s where 1.6180… comes from. And in fact, that is one of the definitions of the golden ratio And so that’s why, for any starting term, it always approaches the same sequence If you use the negative, you get the negative inverse of the golden ratio; another wonderful property of it. It I’d use the negative… If I’d used equals one minus root five over two and that equals -0.6180… So they’re the two solutions to that [particular] equation So, that’s why it works of any sequence where you add two terms to get the next one Brady: So the fact… The Fibonacci numbers meant nothing there? Guy: The Fibonacci numbers are one of [an] infinite family of such numbers because for any two whole numbers you want to start with, the exact same thing works out In fact, why use whole numbers? Go bananas! At no point did I specify whole numbers here. It’s just, whatever you start with, add two terms, you get the next one. You always get the golden ratio That’s not some special property Fibonacci numbers have. All of them have got that property. We want a better link. What we really want is a sequence of numbers which have a link the the golden ratio that other sequences don’t get. We want something above this. Brady: The Brady numbers! Guy: Well, the Brady numbers are wonderful, but I can go slightly better So, I’m going to show you the Lucas numbers. I, personally, am a massive golden ratio septic. I think it’s hugely overrated…


  1. But why does the ratio of two numbers from that sequence tend to the golden number ? Shouldn't it be equal to it ?

  2. i started dividing the numbers in fibonaccis sequence and their ratio wasnt exactly the golden ratio, instead it seemed to aproach more and more to that value as i divided numbers further into the sequence, but never quite reaching it, even though it should (or so i think) why does this happen despite the fact that this numbers are indeed the sum of its predecesors in the sequence? it really bugs me that their ratio is not exactly phi and not knowing why.

  3. Question :
    1 , 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 … What is the formula that will create this sequence beginning with the 3rd number ?It is not the fibonacci folrmula since the 3rd number is 1 not 2 .

  4. Actually he did NOT really prove that the Fibonacci ratio of two consecutive terms will tend towards the golden ratio.

    Look at 0:47. He assumes that consecutive terms will tend towards a fixed ratio.
    So what he actually proves here is "simply" that if a sequence tends towards a fixed ratio for two consecutive terms, then this ratio will be the golden ratio…

    But he did not prove here that the Fibonacci sequence will tend towards a fixed ratio. This is a missing element in his proof.

    Nice and instructive video anyway

  5. If I take the equation x_n/x_n-1= x_n-1/x_n-2 and multiply each side by (x_n-1)(x_n-2) I get
    (x_n)(x_n-2)=(x_n-1)^2 and that is not entirely true how does that work? Because for instance if we let x_n=5 then we get 5(2)=3^2 which is off by 1. If we let x_n=8 then 8(3)=5^2 which is also off by one. x_n=3 then 3(1)=2^2 and also off by one and so on. Could it be possible to demonstrate that by doing this we will always be off by one; meaning (x_n)(x_n-2)+1=(x_n-1)^2 and if so why does that happens, if the ratio is the same then so should be the equality without the +1.

  6. But the Fibbonacci no. make the best rational approximations of phi, as we get from its continued simple infinite fraction and that is <1;1;1;1;1;1;1;1;1;1;1;…>.Mathologer did a great video , search "the most irrational number".

  7. Great video.

    However, I think that it would be even better if you complete the video by also proving that the limit when n goes to infinity of x(n)/x(n-1) – x(n-1)/x(n-2) equals zero, therefore proving that indeed x(n)/x(n-1) tends to x(n-1)/x(n-2) when n goes to infinity.

    Thanks for your wonderful work.

  8. What about this sequence:

    the ratio doesn't tend to 1.618, but the terms satisfy the fibonacci property…

  9. Almost correct. If you begin exactly in the (1-sqrt(5))/2 subspace (starting with 1, (1-sqrt(5))/2, (3-sqrt(5))/2, …), then the ratio is the negative root of delta.

  10. "Add two terms, you get the next one . . . you always get the golden ratio"

    That's not entirely true. I can think of two sequences – one trivial and one not so trivial – that do not converge to Φ.

    There's the trivial sequence starting with 0,0 that doesn't work as others pointed out in the comments. The part that breaks down is assuming you can always take the ratio of consecutive terms since 0/0 is undefined.

    However, there is a more interesting sequence which doesn't converge to the golden ratio but does converge to something else. There are two solutions to the quadratic equation, namely Φ and -1/Φ. The second solution is always in the shadow of its big brother but a solution nonetheless.

    Take the sequence -1/Φ, 1/Φ^2, -1/Φ^3, 1/Φ^4, -1/Φ^5, . . ., (-1/Φ)^n , . . .
    which is approx. -0.618.., 0.382.., -0.236.., 0.146.., -0.090..,….
    The sum of two consecutive terms equals the term after that.
    But the ratio between two consecutive numbers is -1/Φ.
    You could also multiply the sequence by a constant and that too would have ratio -1/Φ.

  11. It seems that you are saying that any two numbers in a fib sequence are the golden ratio, which is kind of weird in itself, but why can't we apply this to geometry. A sequence of squares whose area is 1, 1, 2, 3, 5…. We can find the ratio of any two squares in that sequence by the golden ration and we may find something interesting since the golden ratio comes from geometry itself.

  12. There is a lot more to this. F(n)+hF(n+1)=F(n+2)   =>  F(n)=[((h+sqr(h^2+4))/2)^n – ((h-sqr(h^2+4)/2)^n]/(sqr(h^2+4)). Also see how to find square roots with Fibonacci. See Maths Montage. There is more to come including gh-sequences.

  13. Seriously, what is with those "x's"? He is not even simplifying it, his "x" is bigger and takes more effort than a standard x. Maddening, I could barely follow this because of his x's.

  14. i think it fails for 10,50,60,110,170,280,450 series ..
    and also it doesn't work if x1<<<<x2 , where x1 and x2 are the first two whole numbers in any series.
    Please Reply.

  15. No one will answer, but I try anyway. If I had discovered a new property of the Golden ratio, and new family of numbers with the same property, what I have to do? I'm not a mathematician professionally speaking, and I did not write an appropriate formal paper (because I don't know how), but the property is real and experimentaly proved! thank you.

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  17. The best Fibonacci-like sequence is:
    F(1) = 1
    F(2) = 0
    F(n) = F(n-1) + F(n-2)
    because every Fibonacci-like sequence:
    S(1) = x
    S(2) = y
    S(n)=S(n-1) + S(n-2)
    is equivalent to S(n)=F(n)x +F(n+1)y

  18. I'm confused. The proof given seems to imply that the ratio of any consecutive pair of numbers in a sequence, which obeys the property that the next number in the sequence is the sum of the previous two numbers, is equal to the golden ratio. However, this is not true; because only the limit of consecutive pairs in such a sequence, as n goes to infinity, is equal to the golden ratio. Why does the proof seem to imply the ratio of consecutive terms in the sequence is the golden ratio?

  19. 1:30 i thought you couldn't cancel out terms if they were added together, you could only do that if they are multiplied together. for example 4+3/4 wouldn't equal 3, but 4×3/4 would.

    EDIT: I just understood what he did by adding one. Ignore previous comment

  20. wow I had this interesting problem at my school which asked to calculate the number which added to its square is equal to its cube, and I found out it to be the golden ratio!!! came directly to this vid

  21. The other number that you get, 1-sqrt5/2 also satiasfies as a ratio for sequences like that. So wouldn't be fair to say that that number could also fit a sequence. Yes the Fibonacci sequence is one of them if you go to the left instead those numbers look like this
    …-8 5 -3 2 -1 1 0 1 1 2…. It's the same just back wards and every other number is negative. The ratios are -1/2,2/-3,-3/5,5/-8 which you can see is the ratios to right just flipped and negative, so it converges to -1/phi and 1/phi is phi -1 you can see this because phi^2 =phi +1 and divide by phi so it's -(phi-1) which is -phi +1 and you can do the math and see that that is 1-sqrt5/2

  22. Very, Very Nice Explanation!.  What I wonder though, is this.  You have proven that if a limit exists here, IT IS phi.  How would we prove that THE LIMIT EXISTS?  That's a lumpy stumper for me at this point.  Or have we in principle proven it here?

  23. It is interesting that the golden ratio squared is equal to the golden ratio + 1. This can be written as x^2=x+1. Solving for x, we get:
    Using the quadratic equation we get:
    x = (1+sqrt(5))/2 which is equal to golden ratio.

  24. As others have pointed out, there's errors in the way this way explained. Led to misunderstanding for my younger sister who's in calculus now. You need to talk about limits to do this right. For those with more experience, it's fine, but for those who don't get math as well, I think it causes more problems than it solves to leave things out.

  25. How do we know if we pick 2 arbitrary starting numbers, whether we will arrive with the golden ratio or its negative reciprocal?

  26. (Teacher Forming a question on board) : Gimme an example of whole number.

    Me : Why use whole numbers? Go Bananas!

    Teacher : Out!

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