@@leistiEven coincidentally, you won’t believe the Greatest integer applied to the time (in sec) passed after you sent this message is to being an integer😅
This is from algebraic number theory. Essentially, both 2+sqrt(3) and 2-sqrt(3) are conjugates in the "ring of integers" Z[sqrt(3)] of Q(sqrt(3)) (basically numbers of the form a+b*sqrt(3), where a and b are integers in the former, and rationals in the latter). If you add up a number to all its conjugates (it's possible to have many in other rings of integers), you get an integer. But there's more to it than that: If you take any symmetric polynomial in these conjugates, then the result is an integer. So let's say x = 2+sqrt(3) and y = 2-sqrt(3). x+y is obviously an integer, but so is the polynomial x^n + y^n for any natural n. But here in our case, the conjugate y has absolute value less than 1, so y^n converges to 0, and this implies x^n gets closer and closer to being an integer. If you so happen to find conjugates such that one of them has absolute valu greater than 1 while all the other have absolute value less than 1, then you can do the same thing. w^n + x^n + y^n + z^n is an integer for the conjugates w,x,y,z, and so (WLOG) if |w| > 1 while the others have absolute value less than 1, w^n will get closer and closer to being an integer.
For me this was such a weird video. On the one hand, it had this interesting notion of integer rings and conjugates hidden behind the computation, but on the other hand, it seems that all the emphasis of the video was on writing more and more 9s after the decimal points. Most people are already familiar with complex conjugation and that the only numbers which equal their complex conjugation are the real numbers, which is true for any number plus its conjugate. A similar phenomenon happens in this Z[sqrt(3)] ring, and even without a formal proof, just mentioning this fact could have made this video much more informative: Not only we have this weird fact about these powers, they come from this interesting fact about adding conjugates, and actually this is not something brand new, but almost the same behavior that we know from complex numbers. Such a missed opportunity.
This is also the reason why we can easily calculate the Fibonacci numbers by an approximation Fn ≈ phi ^ n / √5 with phi = (1 + √5) / 2 as the error term is (1 - √5) / 2 to the power of n (divided by √5).
To find even better approximations just do m + sqrt(m^2 -1). The bigger the m you pick the better it will work. Like (100 + sqrt(99))^100 is really really close to an integer.
Very nice and interesting video! So the best combination, in general, would be: n + sqrt (n^2 - 1). But n + sqrt (n^2 + 1) would also work; the error term would alternate in sign, but rapidly decrease all the same.
This same expression can be used to get rational approximations of square roots: Expanding out gives (2 + sqrt(3))^10 = 262087 + 151316 sqrt(3) ≈ 524174. From this we can solve for sqrt(3) and get a good rational approximation: sqrt(3) ≈ 262087 / 151316, which is accurate to ten decimal places.
You can practically avoid calculating the 151316 coefficient by just doubling the 262087! That approximation is accurate as can be! Also, a good approximation for the error is 1 / 524174, because (2 + sqrt(3))^10 = 1 / (2 - sqrt(3))^10, with 1 being the result of 1 = 2^2 - sqrt(3)^2 = ((2 + sqrt(3)) * (2 - sqrt(3)))^k to any given power k. That also means, that using 524174 - 1 / 524174 as the result has an incredible precision! The absolute error is about 6.9434 * 10^-18, which is a relative error of circa 1.3246 * 10^-23. More than enough!
Appendix: As we know that (2 + sqrt(3))^10 + (2 - sqrt(3))^10 = 524174 and both values are reciprocal to each other, we can improve this insane precision even more by using x_{n+1} = 524174 - 1 / x_n... But who needs more precision than a relative error of 10^-23 for this?
sidenote: This approximation can actually be generated using the continued fraction expansion of irrational number. This is particularly useful/interesting for square root. One can look up Pell's equation and this algorithm is how you find a solution to it, which is closely related to units in quadratic fields (notice that (2 + sqrt(3)) is a unit).
This requires you to first be able to work out that 262087 + 151316 sqrt(3) ≈ 524174. I feel like this makes using this method to find rational approximations pretty pointless, as you will first need a very good approximation of sqrt(3) to calculate that 524174 figure.
@@alansmithee419 We don't need sqrt(3) at all (because we know that 151316 * sqrt(3) is almost exactly equal to the integer valu 262087, thus you can just double it to get the result)! The original poster just pointed out, that you could theoretically evaluate sqrt(3) as a fraction with this, but without pratical implications. How do we know, that both terms are almost equal? That's because the value and the conjugate value are reciprocal (at any power). That is that (2 - sqrt(3))^k = 1 / (2 + sqrt(3))^k = (a_k - b_k sqrt(3)) = 1 / (a + b_k sqrt(3)). As the sum approaches bigger values, the difference goes to 0. That means that a_k ≈ b_k sqrt(3) and a_k + b_k sqrt(3) ≈ 2 a_k. The rounding error for this is approximately 1 / (2 a_k).
When generalized to roots of arbitrary degree polynomials, numbers with this property are conveniently called Pisot-Vijayaraghavan numbers. The smallest such number is about 1.3247, a root of x³-x-1.
Interesting! This reminds me of Binet's formula for the Fibonacci sequence. It says that the n-th Fibonacci number is phi^n/sqrt(5) - (-1/phi)^n/sqrt(5), and if we apply the same reasoning as in this video we conclude that phi^n/sqrt(5) itself is very close to the n-th Fibonacci number (which is, of course, an integer).
That's so interesting! The "convergence" to an integer isn't something I'd have expected from raising sums with a square root to ever higher powers. Really was curious to see what's going on there.
What came to mind was the ratio of frequencies in the musical scale, for pairs of notes that are considered to be 'harmonies' in Western music. Of course, this is a bit of a circular argument, as the scale was constructed to make it come out this way. These ratios are close to being ratios of small integers, but not quite exactly. There was no guarantee that it would turn out this way, but we were lucky :)
@@brian8507 I agree. There's always someone somewhere who thinks that integers are taxable and punishable. So, it's important to not masquerade as an integer ... If on the other hand you're a near-integer, you should be able to avoid slavery. The only downside is that your conjugate soulmate would be extremely small and insignificant. All ass, there's more. For we live in greedy times. Even if you identify as an algebraic number far from any integer, then you're liable nowadays to be run and ruled by someone's AI Gore Rhythm ... And that's why it's so important to study maths. Because that's how you can learn to stay non-algebraic ✌️😎💯
Fascinating video! When I was at school we still used logarithm tables. I wander how we would have got on checking the answer. I can’t quite remember but I believe the values were 4 or 5 significant figures. 3.1732^10 is a lot further away from an integer than the examples here!
So basically you're looking for values with the format n+√(n²-1), and the larger the value of n, the more quickly powers of this value will approach integers.
Indeed, so really what this example is resulting from is the fact that sqrt(n^2-1) for integer n approaches an integer as n increases, because its very close to a square number without being one.
So it seems the easiest way to form better integer approximators is to take an integer, add the square root of its square minus 1, and raise to some number n. i.e. (m+sqrt(m^2-1))^n will approximate integers very effectively as m or n increases. This is because m-sqrt(m^2-1) rapidly approaches 0 as m approaches infinity.
a_{0} = 2 a_{1} = 4 a_{n}=4a_{n-1}-a_{n-2} As you can see all terms in this sequence are integers and as n get larger (2-sqrt(3))^n term is closer to zero so term (2+sqrt(3))^{n} is closer to an integer
So the pattern of improving the example is (x+sqrt(x^2±1))^n will get better results as x goes big. In fact, as x goes large, the 1 can be replaced with a larger values and you still get approximate integers.
I tried this with (2+sqrt(5))^n, and that alternates between being slightly more and slightly less than an integer! That's probably because the error term, 2-sqrt(5), is negative, so the signs of the powers alternate. (it still has to have an absolute value less than 1.)
Have to leave; will return to this intriguing question. I'm getting vibes from Pell's Equation here . . . Also, the integer it's so close to, is eerily near 2¹⁹ . . . 524174 = 2¹⁹ - 114 Fred
I was actually working on a problem, recreationally, where something like this came up. Essentially, I wanted to derive a similar equation to the closed form for Fibonacci numbers, but for the continued fraction up to n steps of sqrt(2), relying on the fact that the fibonacci numbers play a similar role in the nth-step continued fraction for phi, and the similarities of the continued fractions of phi and (1+sqrt(2)), and it was useful that the (1-sqrt(2))^n term tended off towards 0, so I could just round the (1+sqrt(2))^x term.
but to create Fibonacci numbers you should use phi^n = (1/2+sqrt(5)/2)^n. Because its conjugate is (1-phi)~-0.62 and phi^n+(1-phi)^n is an integer, in fact the n-th Fibonacci number.
@@deinauge7894 Yes, but as I said, I wasn't trying to create the Fibonacci numbers, but the numerator and denominator of the continued fraction representation of sqrt(2), and used the derivation of that phi^x+(-1/phi)^x from the continued fraction representation (seeing as (1+sqrt(2)) has a similar continued fraction representation as phi, except the 1s are replaced with 2s, the result is basically the same, though you get a different coefficient out front). That would get you the numerator (n) and denominator (n-1) for a terminated continued fraction representation of 1+sqrt(2), and for just sqrt(2) you subtract the denominator from the numerator (you can also use some algebraic manipulation to get a closed form for the resultant numerator)
So this essentially works for any pair (a,b) where the difference between a and √b is smaller than 1. I took √5 + 2 as an example and I found that it works. And it is an even better example than 2 + √3 because the difference between √5 and 2 is even smaller than that between 2 and √3
Similarly, because the nth Fibonacci number Fn = (φ^n + (1-φ)^n)/√5, and |1-φ| < 1, Fn is φ^n/√5 rounded to the nearest integer, and gets closer and closer to being exactly φ^n/√5 as n increases.
Another neat thing is that φ^n itself gets close to an integer as n increases (even though |1-√5| > 1), so [φ^n]/Fn gives rational approximations of √5. Eg, with n=10, we get √5≈122.99.../55≈123/55=2.23636...
The golden ratio also exhibits this property whose name I forgot Edit: These are called PV numbers, and the smallest one is the plastic ratio x³ = x + 1 The definition is similar to the one in the video, being about all "Galois conjugates" of a number being less than 1 in absolute value
Because as a mathematician, he knew that would work, because (2 - sqrt(3)) is the so-called _conjugate_ of (2 + sqrt(3)) . By the way, he is not presenting a roadmap to solve this question as if this were an exam/contest question; he's merely providing an explanation to why this phenomemon occurs. It's about creating understanding, not about presenting a rote algorithm that quickly creates a proof derivation.
Well, if you take (a+√(b))^n with integers a and b, you can take a binomial series expansion, where every term with an even power of √(b) would be an integer, and every odd power would be irrational (unless b is a perfect square). However, if you do the exact same thing, but with (a-√(b))^n, you get the exact same result, but every odd power is now negated. Thus, if you add them up, their odd terms cancel out, and you now have the sum of both their even terms, which is an integer. Thus, (a+√(b))^n+(a-√(b))^n is an integer. In our case, though, a=2, b=3, a-√(b) is between -1 and 1, so as the power n gets bigger, this number gets smaller. Thus, the difference between (2+√(3))^n and (2+√(3))^n+(2-√(3))^n gets smaller and smaller as n grows. Thus, the difference between (2+√(3))^n and an integer gets smaller and smaller. This occurs for all -1
I was quite shocked to see this because i've literally come across these numbers somewhere else. if we define a recursion s_{n} = (s_{n-1})**2 - 2 with s_{0} = 4 and solve this for a close form solution. we actually get: s(n) = (2 - sqrt(3))^(2^n) + (2 + sqrt(3))^(2^n) b.t.w that recursion is the one used in the Luas-Lehmer primality test
Another example, (7 +√47)¹⁰. There are eight 9's after the decimal in this one. I found it interesting because 7 and 47 are both prime numbers, as in your first example with 2 and 3.
Integers can't be measured directly because they may or may not have internal mechanitry, like being prime or composite, or being 0 versus 1. Instead their boundary measures consistently, in which 2+sqrt(3) seems to have interesting behavior. You need to take the nth root or higher of the result to find the likeliest outcome (a^n + b^n = c^[n or higher], both exactly and approximately), yet it seems the plain value is increasing accuracy in measuring an integer. A wonderful pattern, I wonder what science makes use of it.
By this logic, it seems that (2+sqrt3)^10/sqrt3 is also close to an integer. Because (2+sqrt3)^10 - (2-sqrt3)^10 is an integer multiple of multiple of sqrt3.
There are other examples of "almost integers" with different forms than the ones in this video. An example involves pi (3.14159...) and e, the base of natural logarithms (approximately 2.71828). From another source I found that e^(pi * square root(163)) = 262,537,412,640,768,743.999 999 999 999 250... (12 9s after the decimal point). There are numbers other than 163 you could use in the square root to get a number close to an integer (some give a result slightly larger than an integer instead of slightly smaller). My source for this said there is a way to determine such numbers but it did not elaborate so I am uncertain how to find them (other than by trial and error) or why they produce numbers close to an integer.
I'd be interested to know if there's a reason why this works with 163 in particular, and how we could generate more examples even closer to an integer. I've seen other forms of "almost integers" too, but some of them do just seem to be coincidences as far as I understand.
@DrBarker My source for my previous comment was a short article beginning on page 2475 in the book "CRC Concise Encyclopedia of Mathematics" (second edition) by Eric Weisstein. The information I present below is taken from that article. The number I referred to is called the Ramanujan constant (the reason is explained below). Numbers such as this can be found using the theory of modular functions and the J-function (unfortunately they are topics beyond my mathematical training). Although Ramanujan gave a few examples of "almost integers" (for instance the expression in my first comment but with 163 replaced by 58) he did not actually mention the expression I gave. The first to notice this property of 163 was Hermite (1859). The name Ramanujan constant appears to come from an April Fool's joke played by Martin Gardner on the readers of "Scientific American" (Apr. 1975). In his column Gardner claimed e^(pi * square root(163)) was exactly an integer and Ramanujan had conjectured this in his 1914 paper. Gardner admitted the hoax a few months later (July 1975). I searched for Ramanujan constant on RUclips and found an old video by Numberphile. But it doesn't provide much detail; initially the video discusses the apparently unrelated issue of factoring. It is only near the end of the video that the Ramanujan constant was mentioned. There is a deep relationship between the factoring problem and Ramanujan constant but Numberphile's video did not elaborate (probably because the mathematics required is far beyond the scope of the video). If you want to know more about the Ramanujan constant you could do an internet search. For now I will list numbers of the form e^(pi * square root(n)) (for n up to 1000) that differ from an integer by less than 0.01. This is also taken from Weisstein's book (under the entry "Almost Integer" beginning on page 57). Appropriate values of n are 6, 17, 18, 22, 25, 37, 43, 58, 59, 67, 74, 149, 163, 177, 232, 267, 326, 386, 522, 566, 638, 719, 790, 792, 928, and 986.
this number is a root of the polynomial f(x)=x^2-524174x+1. polynomials of the type x^2-nx+1 with large n have roots extremely close to integers as n grows, f(524174)=1 but f(524173)=-524172 so the root must be extremely close but not equal to 524174
So Doc, does this mean that if we considered a number (a + √b)^n (where the difference between a and √b is smaller than 1 and n is an integer) and then we considered the limit (a + √b)^n as n approaches infinity, then the limit (a + √b)^n would turn out to be an integer. Because when n is an integer, the number (a + √b)^n + (a - √b)^n is always an integer and when the difference between a and √b is smaller than 1, the limit (a - √b)^n approaches 0 as n approaches infinity.
From the way the numbers are written it appears, for example, that (2+ 3^.5)^11 is a terminating decimal and thus rational which is clearly not the case.
Going the other way it's more impressive. The principal tenth root of 524174 is 2 + sqrt(3) + appx 0.0000000000014. That's a difference of about 10^-12.
That's interesting. Occidental music theory is based on the fact 3^12 is near to 2^19 (1.3% error), because going up 12 fifth ( multiplying a notes frequency by (3/2)^12 ) is the same as going up 7 octaves ( multiplying by 2^7). is there a mathematical explanation for this? I mean for the approximation 3^12 = 2^19?
I'm familiar with what you're describing and I couldn't find a great reason for it, but you can create a great integer sequence of "best" approximators. I haven't quite been able to determine whether this article is quite quackery or not but it *seems* like the fundamental truth to these "good approximation" tuning systems is related to the Riemann Zeta function. en.xen.wiki/w/The_Riemann_zeta_function_and_tuning#Peak_EDOs
so when i saw that i was like difference of two squares, but then the ^n generalized it for all powers. i was like hmm. Next introduce to (8+sqrt(63))^n and was like hey this is generalizable with (a+sqrt(a^2-1))^n and we will result in integer - infinitesimal. This means as both a and n approach large numbers the error get infinitely smaller.
What I can't figure out is why this only seems to work for numbers of the form (a + b)^n where b is an irrational number only slightly larger than b. Why does this not work for a number where b is a rational number just slightly greater than a as well, like (2+ 401/200)^10. It seems like you could go through the whole same expansion and reasoning shown and then claim the "missing bit" is (2-401/200)^10; that (2-401/200)^10 is very small.
Interesting question! The reason it wouldn't work for e.g. 401/200 is that 401/200 raised to an even power doesn't give an integer, so (a+b)^n + (a-b)^n isn't an integer. In the examples with square roots, they always give an integer when raised to an even power, but this doesn't work for a rational number close to a, and it also wouldn't work e.g. for a cube root.
That's not really it. For example, (7 + sqrt(80))^n is also "just 16^n minus a little bit" (note that sqrt(81) = 9 ), but that doesn't mean that it gets arbitrarily close to integers with increasing n .
An ellipsis symbol (= the three dots) is always ambiguous, and hence is never really proper/formal notation. An ellipsis symbol basically means "and so on, as appropriate/clear from the context", and in this particular case is intended to mean "and the rest of the decimals of the value of this expression". In other words, it doesn't necessarily mean "and so on with repeating 9s". If he hadn't written the dots, then it would suggest than the outcome is exactly equal to (524173 + 99999/100000) . I guess in order to avoid ambiguity, he could/should have added at least the first non-9 decimal, and instead have written something like " = 524173.9999980... "
It depends on how the question is worded. The limit of (2 + sqrt(3))^n is infinite, but we could say that the decimal part converges to 1 as n tends to infinity, and make it more formal for example using the floor function: lim_{n -> infinity} [ floor( ( 2 + sqrt(3))^n ) - (2 + sqrt(3))^n ] = 1
Some weirder examples: 0.5 + sqrt(1.25) 1.5 + sqrt(3.25) 2.5 + sqrt(4.25) Interestingly for (1.5) + sqrt(2.75) we get closer and closer to an integer multiple of 0.5^(ceil(N/2) - 1) where N is the exponent of the original power. What's with the ceil(N/2). Whyyyy.
My guess is he just calculates (2 - sqrt(3))^n , and counts the 0s between the decimal point and the first significant digit. Because the sum of (2-sqrt(3))^n and (2+sqrt(3))^n is equal to the integer that you get from rounding (2+sqrt(3))^n . For example, if (2-sqrt(3))^n equals (6.386...) * 10^(-58) , then there are (58-1) = 57 instances of the decimal 0 between the decimal point and the first significant decimal (which is a 6), and hence (2+sqrt(3))^n will have 57 instances of the decimal 9 between the decimal point and the first non-9 decimal (which will be a (9-6) = 3).
This could be turned into a good Math Olympiad problem. Choose two consecutive integers a, b such that (a + sqrt(b)) or (b + sqrt(a)) raised to the exponent of [insert four-digit year here] has a maximum possible number of 9's after the decimal point. I think, if you insist on the numbers being consecutive, then (2 + sqrt (3)) is the answer regardless of what exponent you use.
No. For example, it wouldn't work for (a+sqrt(b)) = (8 + sqrt(23)) . But it does work for (a+sqrt(b)) = (8+sqrt(65)) . The correct requirement is that -1 < (a - sqrt(b)) < 1 . Or put differently: |a - sqrt(b)| < 1 (Or you could also rewrite it as sqrt(b)-1 < a < sqrt(b)+1 .)
isn't the underlying reason it works in both the cases of 2-sqrt(3) and 8-sqrt(63) because the term under the square root is exactly 1 lower than the square of the term outside of the square root?
Thats exactly what I was thinking and surprised how far down I had to scroll to find this in the comments. Its just n - sqrt(n^(2)-1). The difference will be smaller the larger n is. Example: 10,001−√(10,001^(2)−1) = 0.000049995
It would also work for 8-sqrt(62) or 8-sqrt(61) for example. It is only necessary that this yields something that has a magnitude less than 1. But the closer it is to zero, the quicker the powers converge to an integer. So using 8-sqrt(63) converges quicker than 8-sqrt(62) etc. …and 8-sqrt(50) would be MUCH slower.
@@Mike-oz3ox You mean "sqrt(64) - sqrt(63)" instead of "64 - sqrt(63)" (and "sqrt(64) - sqrt(62)" instead of "64 - sqrt(62)"), of course. By the way, good call on sqrt(64) - sqrt(50) ! If the first term is 8 (or sqrt(64) ), then the lowest integer n for the second term -sqrt(n) is n=50 ; because for lower iinteger values of n, the sequence of powers wouldn't be convergent anymore.
Well, them being infinite is not what makes some of them approximates of integers. The property you're looking for is density. By the way, this means that the same thing is true also for rationals.
according to the logic of the last bit of the video, can we say that we'll get closer and closer to integer for the expression (k + sqrt(k - 1))^n for very large k,n?
No. The reason we were able to get this result hinged on the fact that 2-sqrt(3) has absolute value less then 1. Raising it larger powers made it go to 0. The conjugate k - sqrt(k-1) (or even k - sqrt(k+1) if you wanted to follow the video's example) does not have absolute value less than 1 beyond small values of k. Instead, you should use k + sqrt(k^2-1) or k + sqrt(k^2+1), because their conjugates are close to 0.
@@f5673-t1h oh you're right. i just realised i typed the wrong expression. initially i wanna write k + sqrt(k^2 - 1) but my hands said no apparently lol
Its pretty clear no? He showed that it plus a related term is always an integer, and then showed that for large n the related term tends to 0. Is there a bigger picture “why“ that I’m not seeing?
I don't know why, but for some reason (((1+√5)÷2)^(2n)+3*(-1)^n+((1-√5)÷2)^(2n))÷(5n) turns out to be an integer for every prime number (except for n=5).
Even more amazing, you won't believe how close (2+sqrt(4))^n is to an integer!
Coincidentally, your number of likes is also currently an integer. I'll try to click the button and see whether it's still an integer.
@@leistiEven coincidentally, you won’t believe the Greatest integer applied to the time (in sec) passed after you sent this message is to being an integer😅
@@gamingwithtrikku2371Everything is secretly an integer. 🤫🧏♂️😂😂😂
where n>1
@@ok-gd8pn indeed, especially when n = 3.14159265
This is from algebraic number theory.
Essentially, both 2+sqrt(3) and 2-sqrt(3) are conjugates in the "ring of integers" Z[sqrt(3)] of Q(sqrt(3)) (basically numbers of the form a+b*sqrt(3), where a and b are integers in the former, and rationals in the latter).
If you add up a number to all its conjugates (it's possible to have many in other rings of integers), you get an integer.
But there's more to it than that: If you take any symmetric polynomial in these conjugates, then the result is an integer.
So let's say x = 2+sqrt(3) and y = 2-sqrt(3). x+y is obviously an integer, but so is the polynomial x^n + y^n for any natural n.
But here in our case, the conjugate y has absolute value less than 1, so y^n converges to 0, and this implies x^n gets closer and closer to being an integer.
If you so happen to find conjugates such that one of them has absolute valu greater than 1 while all the other have absolute value less than 1, then you can do the same thing.
w^n + x^n + y^n + z^n is an integer for the conjugates w,x,y,z, and so (WLOG) if |w| > 1 while the others have absolute value less than 1, w^n will get closer and closer to being an integer.
Damn, nice
For me this was such a weird video.
On the one hand, it had this interesting notion of integer rings and conjugates hidden behind the computation, but on the other hand, it seems that all the emphasis of the video was on writing more and more 9s after the decimal points.
Most people are already familiar with complex conjugation and that the only numbers which equal their complex conjugation are the real numbers, which is true for any number plus its conjugate. A similar phenomenon happens in this Z[sqrt(3)] ring, and even without a formal proof, just mentioning this fact could have made this video much more informative: Not only we have this weird fact about these powers, they come from this interesting fact about adding conjugates, and actually this is not something brand new, but almost the same behavior that we know from complex numbers. Such a missed opportunity.
What does it mean for more than 2 numbers to be conjugates?
This should come in handy when I need an integer but can't work it out exactly so I have to use an approximation.
💀💀💀💀
This is also the reason why we can easily calculate the Fibonacci numbers by an approximation Fn ≈ phi ^ n / √5 with phi = (1 + √5) / 2 as the error term is (1 - √5) / 2 to the power of n (divided by √5).
To find even better approximations just do m + sqrt(m^2 -1). The bigger the m you pick the better it will work. Like (100 + sqrt(99))^100 is really really close to an integer.
"finding the best approximation for an integer" sounds funny
Damn, I was going to comment that
I'm actually disappointed that he didn't mention this in the video
Also it should be (10 + √99) not (100 + √99)
I tried m = 1 and it worked perfectly 😂
Very nice and interesting video!
So the best combination, in general, would be:
n + sqrt (n^2 - 1).
But
n + sqrt (n^2 + 1)
would also work; the error term would alternate in sign, but rapidly decrease all the same.
Or if we allow a more general expression a+b.sqrt(c).. You can do even better. E. g. (2-sqrt(3))^2=7-4.sqrt(3),so 7+4.sqrt(3) must work "well".
@@tensor131 That's just (2+sqrt[3])^2. Taking (7+4.sqrt[3])^n=(2+sqrt[3])^{2n}
I thought of this too. It would’ve been nice to see 4+√17 talked about when 5+√17 was being talked about and comparing it to say 4+√15.
This same expression can be used to get rational approximations of square roots:
Expanding out gives (2 + sqrt(3))^10 = 262087 + 151316 sqrt(3) ≈ 524174. From this we can solve for sqrt(3) and get a good rational approximation: sqrt(3) ≈ 262087 / 151316, which is accurate to ten decimal places.
You can practically avoid calculating the 151316 coefficient by just doubling the 262087! That approximation is accurate as can be!
Also, a good approximation for the error is 1 / 524174, because (2 + sqrt(3))^10 = 1 / (2 - sqrt(3))^10, with 1 being the result of 1 = 2^2 - sqrt(3)^2 = ((2 + sqrt(3)) * (2 - sqrt(3)))^k to any given power k.
That also means, that using 524174 - 1 / 524174 as the result has an incredible precision!
The absolute error is about 6.9434 * 10^-18, which is a relative error of circa 1.3246 * 10^-23. More than enough!
Appendix: As we know that (2 + sqrt(3))^10 + (2 - sqrt(3))^10 = 524174 and both values are reciprocal to each other, we can improve this insane precision even more by using x_{n+1} = 524174 - 1 / x_n... But who needs more precision than a relative error of 10^-23 for this?
sidenote: This approximation can actually be generated using the continued fraction expansion of irrational number. This is particularly useful/interesting for square root. One can look up Pell's equation and this algorithm is how you find a solution to it, which is closely related to units in quadratic fields (notice that (2 + sqrt(3)) is a unit).
This requires you to first be able to work out that 262087 + 151316 sqrt(3) ≈ 524174.
I feel like this makes using this method to find rational approximations pretty pointless, as you will first need a very good approximation of sqrt(3) to calculate that 524174 figure.
@@alansmithee419 We don't need sqrt(3) at all (because we know that 151316 * sqrt(3) is almost exactly equal to the integer valu 262087, thus you can just double it to get the result)! The original poster just pointed out, that you could theoretically evaluate sqrt(3) as a fraction with this, but without pratical implications.
How do we know, that both terms are almost equal? That's because the value and the conjugate value are reciprocal (at any power). That is that (2 - sqrt(3))^k = 1 / (2 + sqrt(3))^k = (a_k - b_k sqrt(3)) = 1 / (a + b_k sqrt(3)). As the sum approaches bigger values, the difference goes to 0. That means that a_k ≈ b_k sqrt(3) and a_k + b_k sqrt(3) ≈ 2 a_k. The rounding error for this is approximately 1 / (2 a_k).
When generalized to roots of arbitrary degree polynomials, numbers with this property are conveniently called Pisot-Vijayaraghavan numbers. The smallest such number is about 1.3247, a root of x³-x-1.
Interesting! This reminds me of Binet's formula for the Fibonacci sequence. It says that the n-th Fibonacci number is phi^n/sqrt(5) - (-1/phi)^n/sqrt(5), and if we apply the same reasoning as in this video we conclude that phi^n/sqrt(5) itself is very close to the n-th Fibonacci number (which is, of course, an integer).
Same reasoning: phi - 1/phi = 1, phi^2 + 1/phi^2 = 3, ... etc. Here x + 1/x and all x^n + 1/x^n are integer...
BTW: (2 + sqrt(3)) is a root of x² - 4x + 1 = 0 and phi is a root of x² - x - 1 = 0. Similarities? Yes, a lot!
That's so interesting!
The "convergence" to an integer isn't something I'd have expected from raising sums with a square root to ever higher powers. Really was curious to see what's going on there.
What came to mind was the ratio of frequencies in the musical scale, for pairs of notes that are considered to be 'harmonies' in Western music. Of course, this is a bit of a circular argument, as the scale was constructed to make it come out this way. These ratios are close to being ratios of small integers, but not quite exactly. There was no guarantee that it would turn out this way, but we were lucky :)
Awesome stuff. I wish I was almost an integer, but I'm just a person =/
I believe in you! Your journey to becoming almost an integer has to start somewhere.
Trust me... you don't wanna be an integer... you will just be on a list somewhere
@@brian8507even as a person you're not free from lists.
I, for one, am on multiple lists that cause me nothing but problems.
@@brian8507
I agree. There's always someone somewhere who thinks that integers are taxable and punishable. So, it's important to not masquerade as an integer ... If on the other hand you're a near-integer, you should be able to avoid slavery. The only downside is that your conjugate soulmate would be extremely small and insignificant.
All ass, there's more. For we live in greedy times. Even if you identify as an algebraic number far from any integer, then you're liable nowadays to be run and ruled by someone's AI Gore Rhythm ... And that's why it's so important to study maths. Because that's how you can learn to stay non-algebraic ✌️😎💯
I wish I could be on a list. Except for indecency. it would be an honour to be on the number line itself.@@brian8507
Fascinating video! When I was at school we still used logarithm tables. I wander how we would have got on checking the answer. I can’t quite remember but I believe the values were 4 or 5 significant figures. 3.1732^10 is a lot further away from an integer than the examples here!
Your shirts are always nice, but this one is 🔥
So basically you're looking for values with the format n+√(n²-1), and the larger the value of n, the more quickly powers of this value will approach integers.
Indeed, so really what this example is resulting from is the fact that sqrt(n^2-1) for integer n approaches an integer as n increases, because its very close to a square number without being one.
So it seems the easiest way to form better integer approximators is to take an integer, add the square root of its square minus 1, and raise to some number n.
i.e. (m+sqrt(m^2-1))^n will approximate integers very effectively as m or n increases.
This is because m-sqrt(m^2-1) rapidly approaches 0 as m approaches infinity.
That’s so cool. Can’t believe I’ve never come across this concept before.
a_{0} = 2
a_{1} = 4
a_{n}=4a_{n-1}-a_{n-2}
As you can see all terms in this sequence are integers
and as n get larger (2-sqrt(3))^n term is closer to zero
so term (2+sqrt(3))^{n} is closer to an integer
So the pattern of improving the example is (x+sqrt(x^2±1))^n will get better results as x goes big. In fact, as x goes large, the 1 can be replaced with a larger values and you still get approximate integers.
The limit as x approaches infinity is (2x)^n, só yes, it does get closer to an integer the bigger x is
I tried this with (2+sqrt(5))^n, and that alternates between being slightly more and slightly less than an integer!
That's probably because the error term, 2-sqrt(5), is negative, so the signs of the powers alternate. (it still has to have an absolute value less than 1.)
I hate doing math, but I love watching other people do math. Y'all feel me?
YouFeeAreIrr
Here's is the same in generalized form
For any integer, n, k and m, the value (n+ (n^(2)-k)^(1/2))^m approximates to an integer, where 0
basically (A+√B)^n is almost an integer so long as 0
Long life Dr Barker 🎉❤😊
Very well explained and a very interesting result.
FWIW, 3-sqrt(8) seems to have the smallest nonzero absolute value among all a-sqrt(b) for single digit integers a and b.
Hi Dr. Barker!
The best integer-approximator of the form (a + sqrt(a^2 - 1))^n will always be (1 + sqrt(0))^n. :)
Great explanation! You have the darkest eyes I've ever seen. Profound understanding must be hidden behind.
Have to leave; will return to this intriguing question. I'm getting vibes from Pell's Equation here . . .
Also, the integer it's so close to, is eerily near 2¹⁹ . . . 524174 = 2¹⁹ - 114
Fred
I was actually working on a problem, recreationally, where something like this came up. Essentially, I wanted to derive a similar equation to the closed form for Fibonacci numbers, but for the continued fraction up to n steps of sqrt(2), relying on the fact that the fibonacci numbers play a similar role in the nth-step continued fraction for phi, and the similarities of the continued fractions of phi and (1+sqrt(2)), and it was useful that the (1-sqrt(2))^n term tended off towards 0, so I could just round the (1+sqrt(2))^x term.
but to create Fibonacci numbers you should use phi^n = (1/2+sqrt(5)/2)^n.
Because its conjugate is (1-phi)~-0.62 and phi^n+(1-phi)^n is an integer, in fact the n-th Fibonacci number.
@@deinauge7894 Yes, but as I said, I wasn't trying to create the Fibonacci numbers, but the numerator and denominator of the continued fraction representation of sqrt(2), and used the derivation of that phi^x+(-1/phi)^x from the continued fraction representation (seeing as (1+sqrt(2)) has a similar continued fraction representation as phi, except the 1s are replaced with 2s, the result is basically the same, though you get a different coefficient out front). That would get you the numerator (n) and denominator (n-1) for a terminated continued fraction representation of 1+sqrt(2), and for just sqrt(2) you subtract the denominator from the numerator (you can also use some algebraic manipulation to get a closed form for the resultant numerator)
This actually works in general as long as n is sufficiently large.
(a+√b)^n goes to an integer as n goes to infinity as long as a-√b < 1
(a+√(a²-1))
So therefore...
10^9 + √(10^18-1) is quite close
Very impressive comment ❤️
I did a presentation about this in my senior dissertation class. However we went in a broader case with all but 1 root within (-1,1)
It’s in my head.
So this essentially works for any pair (a,b) where the difference between a and √b is smaller than 1.
I took √5 + 2 as an example and I found that it works. And it is an even better example than 2 + √3 because the difference between √5 and 2 is even smaller than that between 2 and √3
Can we call these Parker Integers? They are close, but not quite right.
Barker integers, they’re almost Parker integers lol.
Matt Parker actually did a video himself about this with regards to the powers of the golden ratio!
@@joe_z I've got to check that out! Thank you for the recommendation. :)
Similarly, because the nth Fibonacci number Fn = (φ^n + (1-φ)^n)/√5, and |1-φ| < 1, Fn is φ^n/√5 rounded to the nearest integer, and gets closer and closer to being exactly φ^n/√5 as n increases.
Another neat thing is that φ^n itself gets close to an integer as n increases (even though |1-√5| > 1), so [φ^n]/Fn gives rational approximations of √5. Eg, with n=10, we get √5≈122.99.../55≈123/55=2.23636...
The golden ratio also exhibits this property whose name I forgot
Edit: These are called PV numbers, and the smallest one is the plastic ratio x³ = x + 1
The definition is similar to the one in the video, being about all "Galois conjugates" of a number being less than 1 in absolute value
1:17 I’m not a mathematician. Why did he do this operation? Seems like he magically pulled (2-root3)^n out of no where.
Because as a mathematician, he knew that would work, because (2 - sqrt(3)) is the so-called _conjugate_ of (2 + sqrt(3)) .
By the way, he is not presenting a roadmap to solve this question as if this were an exam/contest question; he's merely providing an explanation to why this phenomemon occurs. It's about creating understanding, not about presenting a rote algorithm that quickly creates a proof derivation.
Look at 8 mins in and there's more examples on the board
Well, if you take (a+√(b))^n with integers a and b, you can take a binomial series expansion, where every term with an even power of √(b) would be an integer, and every odd power would be irrational (unless b is a perfect square).
However, if you do the exact same thing, but with (a-√(b))^n, you get the exact same result, but every odd power is now negated.
Thus, if you add them up, their odd terms cancel out, and you now have the sum of both their even terms, which is an integer. Thus, (a+√(b))^n+(a-√(b))^n is an integer. In our case, though, a=2, b=3, a-√(b) is between -1 and 1, so as the power n gets bigger, this number gets smaller.
Thus, the difference between (2+√(3))^n and (2+√(3))^n+(2-√(3))^n gets smaller and smaller as n grows. Thus, the difference between (2+√(3))^n and an integer gets smaller and smaller.
This occurs for all -1
I was quite shocked to see this because i've literally come across these numbers somewhere else. if we define
a recursion s_{n} = (s_{n-1})**2 - 2 with s_{0} = 4 and solve this for a close form solution. we actually get:
s(n) = (2 - sqrt(3))^(2^n) + (2 + sqrt(3))^(2^n)
b.t.w that recursion is the one used in the Luas-Lehmer primality test
This also works for 2 + root 2 and 2 + root 5 as well.
Another example, (7 +√47)¹⁰. There are eight 9's after the decimal in this one. I found it interesting because 7 and 47 are both prime numbers, as in your first example with 2 and 3.
Another great video! 😊
Thank you!
So (a+sqrt(b))^n comes closer and closer to a whole number as n goes to infinity if a-1
Integers can't be measured directly because they may or may not have internal mechanitry, like being prime or composite, or being 0 versus 1. Instead their boundary measures consistently, in which 2+sqrt(3) seems to have interesting behavior. You need to take the nth root or higher of the result to find the likeliest outcome (a^n + b^n = c^[n or higher], both exactly and approximately), yet it seems the plain value is increasing accuracy in measuring an integer. A wonderful pattern, I wonder what science makes use of it.
1:31
This is obviously an integer because it's an algebraic integer, and it's rational because 2+sqrt3 and 2-sqrt3 are conjugate over Q
I would love to be an ancient Greek who knew this and proposed a visual "proof" to prank everyone
(2+Sqrt[3])^10+(2-Sqrt[3])^10=524174
정말 멋진 영상이네요
Also interesting is (1/sqrt[5]) ((1 + sqrt[5])/2)^n, as you no doubt know.
By this logic, it seems that (2+sqrt3)^10/sqrt3 is also close to an integer.
Because (2+sqrt3)^10 - (2-sqrt3)^10 is an integer multiple of multiple of sqrt3.
(2+Sqrt[3])^n+(2-Sqrt[3])^n n is an integer equals an integer.
There are other examples of "almost integers" with different forms than the ones in this video. An example involves pi (3.14159...) and e, the base of natural logarithms (approximately 2.71828). From another source I found that e^(pi * square root(163)) = 262,537,412,640,768,743.999 999 999 999 250... (12 9s after the decimal point). There are numbers other than 163 you could use in the square root to get a number close to an integer (some give a result slightly larger than an integer instead of slightly smaller). My source for this said there is a way to determine such numbers but it did not elaborate so I am uncertain how to find them (other than by trial and error) or why they produce numbers close to an integer.
I'd be interested to know if there's a reason why this works with 163 in particular, and how we could generate more examples even closer to an integer. I've seen other forms of "almost integers" too, but some of them do just seem to be coincidences as far as I understand.
@DrBarker My source for my previous comment was a short article beginning on page 2475 in the book "CRC Concise Encyclopedia of Mathematics" (second edition) by Eric Weisstein. The information I present below is taken from that article. The number I referred to is called the Ramanujan constant (the reason is explained below). Numbers such as this can be found using the theory of modular functions and the J-function (unfortunately they are topics beyond my mathematical training). Although Ramanujan gave a few examples of "almost integers" (for instance the expression in my first comment but with 163 replaced by 58) he did not actually mention the expression I gave. The first to notice this property of 163 was Hermite (1859). The name Ramanujan constant appears to come from an April Fool's joke played by Martin Gardner on the readers of "Scientific American" (Apr. 1975). In his column Gardner claimed e^(pi * square root(163)) was exactly an integer and Ramanujan had conjectured this in his 1914 paper. Gardner admitted the hoax a few months later (July 1975).
I searched for Ramanujan constant on RUclips and found an old video by Numberphile. But it doesn't provide much detail; initially the video discusses the apparently unrelated issue of factoring. It is only near the end of the video that the Ramanujan constant was mentioned. There is a deep relationship between the factoring problem and Ramanujan constant but Numberphile's video did not elaborate (probably because the mathematics required is far beyond the scope of the video). If you want to know more about the Ramanujan constant you could do an internet search. For now I will list numbers of the form e^(pi * square root(n)) (for n up to 1000) that differ from an integer by less than 0.01. This is also taken from Weisstein's book (under the entry "Almost Integer" beginning on page 57). Appropriate values of n are 6, 17, 18, 22, 25, 37, 43, 58, 59, 67, 74, 149, 163, 177, 232, 267, 326, 386, 522, 566, 638, 719, 790, 792, 928, and 986.
this number is a root of the polynomial f(x)=x^2-524174x+1. polynomials of the type x^2-nx+1 with large n have roots extremely close to integers as n grows, f(524174)=1 but f(524173)=-524172 so the root must be extremely close but not equal to 524174
So Doc, does this mean that if we considered a number (a + √b)^n (where the difference between a and √b is smaller than 1 and n is an integer) and then we considered the limit (a + √b)^n as n approaches infinity, then the limit (a + √b)^n would turn out to be an integer.
Because when n is an integer, the number (a + √b)^n + (a - √b)^n is always an integer and when the difference between a and √b is smaller than 1, the limit (a - √b)^n approaches 0 as n approaches infinity.
Then it will be same for 3 + sqrt(8) or 4 + sqrt(15)
Or anything where N1 - sqrt(N2) is
From the way the numbers are written it appears, for example, that (2+ 3^.5)^11 is a terminating decimal and thus rational which is clearly not the case.
Amazing stuff! Replace 3 with pi.
Going the other way it's more impressive. The principal tenth root of 524174 is 2 + sqrt(3) + appx 0.0000000000014. That's a difference of about 10^-12.
That's interesting. Occidental music theory is based on the fact 3^12 is near to 2^19 (1.3% error), because going up 12 fifth ( multiplying a notes frequency by (3/2)^12 ) is the same as going up 7 octaves ( multiplying by 2^7). is there a mathematical explanation for this? I mean for the approximation 3^12 = 2^19?
I'm familiar with what you're describing and I couldn't find a great reason for it, but you can create a great integer sequence of "best" approximators. I haven't quite been able to determine whether this article is quite quackery or not but it *seems* like the fundamental truth to these "good approximation" tuning systems is related to the Riemann Zeta function.
en.xen.wiki/w/The_Riemann_zeta_function_and_tuning#Peak_EDOs
19/12 is a good rational approximation to ln 3/ln 2.
@@rosiefay7283 i guess but that leaves the question as to why that is.
@@maxime_weill there have to be *some* rational approximations, and that's one of them. i don't think there's anything deeper than that here
so when i saw that i was like difference of two squares, but then the ^n generalized it for all powers. i was like hmm. Next introduce to (8+sqrt(63))^n and was like hey this is generalizable with (a+sqrt(a^2-1))^n and we will result in integer - infinitesimal. This means as both a and n approach large numbers the error get infinitely smaller.
This is really interesting
0.999... is also almost an integer.
What I can't figure out is why this only seems to work for numbers of the form (a + b)^n where b is an irrational number only slightly larger than b. Why does this not work for a number where b is a rational number just slightly greater than a as well, like (2+ 401/200)^10. It seems like you could go through the whole same expansion and reasoning shown and then claim the "missing bit" is (2-401/200)^10; that (2-401/200)^10 is very small.
Interesting question! The reason it wouldn't work for e.g. 401/200 is that 401/200 raised to an even power doesn't give an integer, so (a+b)^n + (a-b)^n isn't an integer. In the examples with square roots, they always give an integer when raised to an even power, but this doesn't work for a rational number close to a, and it also wouldn't work e.g. for a cube root.
I have been thinking of this method..what about coefficients of Paschal Triangles
Well mod(2+sqrt 3)^n) converges to 1 (which is 0 (mod 1))
It’s not….after the sixth decimal place, it retains its irrational behavior.
So how is (5+sqrt(17)) special? Wouldn't (5+sqrt(24)) give a much better result as the difference between 5 and sqrt(24) is tiny?
makes sense to me once you go with the 8 example,
sqrt63 is very close to sqrt64 which is 8, so the sum is just 16^n minus a little bit
That's not really it. The difference 16^n-(8+sqrt(63))^n gets arbitrarily large with increasing n. But its fractional part still approaches 0.
That's not really it. For example, (7 + sqrt(80))^n is also "just 16^n minus a little bit" (note that sqrt(81) = 9 ), but that doesn't mean that it gets arbitrarily close to integers with increasing n .
So.... That to the power of infinity IS an interger? It certainly approaches it. ;)
If therr is five nines after the decimal place, why do you have 3 dots after it? That signifies that it continues on forever.
An ellipsis symbol (= the three dots) is always ambiguous, and hence is never really proper/formal notation. An ellipsis symbol basically means "and so on, as appropriate/clear from the context", and in this particular case is intended to mean "and the rest of the decimals of the value of this expression". In other words, it doesn't necessarily mean "and so on with repeating 9s".
If he hadn't written the dots, then it would suggest than the outcome is exactly equal to (524173 + 99999/100000) .
I guess in order to avoid ambiguity, he could/should have added at least the first non-9 decimal, and instead have written something like
" = 524173.9999980... "
Cool
Stupid question, but would it be correct to say this kind of expressions is equal to integer when n-》inf?
No, because it doesn't converge.
Though you can say something like (2+sqrt(3))^n/[(2+sqrt(3))^n+(2-sqrt(3))^n] tends to 1 and n tends to infinity.
It depends on how the question is worded. The limit of (2 + sqrt(3))^n is infinite, but we could say that the decimal part converges to 1 as n tends to infinity, and make it more formal for example using the floor function:
lim_{n -> infinity} [ floor( ( 2 + sqrt(3))^n ) - (2 + sqrt(3))^n ] = 1
So sin((2+sqrt(3))^n pi)-->0 when n-->+\infty, right?
@@DrBarker I think you mean the fractional part, not the floor.
@@f5673-t1h Good point! I forgot to subtract the original (2 + sqrt(3))^n to turn it into the fractional part.
Some weirder examples:
0.5 + sqrt(1.25)
1.5 + sqrt(3.25)
2.5 + sqrt(4.25)
Interestingly for (1.5) + sqrt(2.75) we get closer and closer to an integer multiple of 0.5^(ceil(N/2) - 1) where N is the exponent of the original power.
What's with the ceil(N/2). Whyyyy.
How do you calculate the amount of consecutive decimal 9s in the numbers? Is it just counting them from wolframalpha?
My guess is he just calculates (2 - sqrt(3))^n , and counts the 0s between the decimal point and the first significant digit. Because the sum of (2-sqrt(3))^n and (2+sqrt(3))^n is equal to the integer that you get from rounding (2+sqrt(3))^n .
For example, if (2-sqrt(3))^n equals (6.386...) * 10^(-58) , then there are (58-1) = 57 instances of the decimal 0 between the decimal point and the first significant decimal (which is a 6), and hence (2+sqrt(3))^n will have 57 instances of the decimal 9 between the decimal point and the first non-9 decimal (which will be a (9-6) = 3).
This could be turned into a good Math Olympiad problem. Choose two consecutive integers a, b such that (a + sqrt(b)) or (b + sqrt(a)) raised to the exponent of [insert four-digit year here] has a maximum possible number of 9's after the decimal point. I think, if you insist on the numbers being consecutive, then (2 + sqrt (3)) is the answer regardless of what exponent you use.
Why would you use 5+/17 instead of 5+/24? Surely that's more comparable to 2+/3
How did you come up with this vid idea
5000+sqrt(5000^2-1) is 9999.9999
Isn’t it the case, then, that (a+sqrt(b))^n forms such an integer approximation whenever a > sqrt(b)?
No. For example, it wouldn't work for (a+sqrt(b)) = (8 + sqrt(23)) . But it does work for (a+sqrt(b)) = (8+sqrt(65)) .
The correct requirement is that -1 < (a - sqrt(b)) < 1 . Or put differently:
|a - sqrt(b)| < 1
(Or you could also rewrite it as sqrt(b)-1 < a < sqrt(b)+1 .)
Root(10) + pi
(2+√3)^10+(2-√3)^10 is integer, 0 < 2-√3 < 0,3, 0 < (2-√3)^10 < 0,3^10 = 0,0000059049 so (2-√3)^10 is almost 0, and (2+√3)^10 is almost integer.
isn't the underlying reason it works in both the cases of 2-sqrt(3) and 8-sqrt(63) because the term under the square root is exactly 1 lower than the square of the term outside of the square root?
Thats exactly what I was thinking and surprised how far down I had to scroll to find this in the comments.
Its just n - sqrt(n^(2)-1). The difference will be smaller the larger n is.
Example: 10,001−√(10,001^(2)−1) = 0.000049995
It would also work for 8-sqrt(62) or 8-sqrt(61) for example. It is only necessary that this yields something that has a magnitude less than 1. But the closer it is to zero, the quicker the powers converge to an integer. So using 8-sqrt(63) converges quicker than 8-sqrt(62) etc. …and 8-sqrt(50) would be MUCH slower.
@@Mike-oz3ox You mean "sqrt(64) - sqrt(63)" instead of "64 - sqrt(63)" (and "sqrt(64) - sqrt(62)" instead of "64 - sqrt(62)"), of course.
By the way, good call on sqrt(64) - sqrt(50) ! If the first term is 8 (or sqrt(64) ), then the lowest integer n for the second term -sqrt(n) is n=50 ; because for lower iinteger values of n, the sequence of powers wouldn't be convergent anymore.
@@yurenchu Thank you for pointing out the typo. Now corrected.
Why? Because there are an infinite number of irrational numbers, so if you look at enough of them, you'll find some that are near integers.
Well, them being infinite is not what makes some of them approximates of integers. The property you're looking for is density.
By the way, this means that the same thing is true also for rationals.
Popular I+f problem
احسنت
according to the logic of the last bit of the video, can we say that we'll get closer and closer to integer for the expression (k + sqrt(k - 1))^n for very large k,n?
No.
The reason we were able to get this result hinged on the fact that 2-sqrt(3) has absolute value less then 1. Raising it larger powers made it go to 0.
The conjugate k - sqrt(k-1) (or even k - sqrt(k+1) if you wanted to follow the video's example) does not have absolute value less than 1 beyond small values of k.
Instead, you should use k + sqrt(k^2-1) or k + sqrt(k^2+1), because their conjugates are close to 0.
@@f5673-t1h oh you're right. i just realised i typed the wrong expression. initially i wanna write k + sqrt(k^2 - 1) but my hands said no apparently lol
More generally (a+sqrt(b)) where a is either the floor rounding or ceil rounding of sqrt(b) like (6+sqrt(43)) or (8+sqrt(53))
(2+sqrt(3))^n =x
As n approaches infinity, x approaches an integer?
Excacly, cuz (2 - sqrt(3))^n approaches 0
But the approaching of n must be discrete
Not really. As n approaches infinity, x approaches infinity, which is not an integer.
You didn't show WHY it is nearly an integer. You showed that IT IS nearly an integer. Why ? No one ever found that maths has a motive.
Its pretty clear no? He showed that it plus a related term is always an integer, and then showed that for large n the related term tends to 0. Is there a bigger picture “why“ that I’m not seeing?
A distinction without a difference
This was fun!
The history of maths failures is littered with “almost”. 0.999 etc to 10 trillion places to a pure mathematician, is nowhere near close.
I was trying to figure out why the number was rational. Then I realized that after the five nines came infinite garbage.
May as well ask why [sqrt(2)] ^2 is an integer.
Sorry to say you are wrong, but you really are. The infinite repeating decimal 0.999… is equal to the whole number 1
Bro there r numbers with infinite point nines
Are you filming from jail?😳 LOL JK
Get a proper mic, dude. They're not expensive.
It is : Continued fraction: 524173 + 1/(1 + 1/(524172 + 1/(1 + 1/(524172 + 1/(1 + 1/(524172 + 1/(1 + 1/(524172 + 1/(1 + 1/...)))))))))
I don't know why, but for some reason
(((1+√5)÷2)^(2n)+3*(-1)^n+((1-√5)÷2)^(2n))÷(5n)
turns out to be an integer for every prime number (except for n=5).