Swedish Mathematics Olympiad | 2002 Question 4

[@patrickwienhoft7987]

8:35 underlining inequality signs is usually not the best way to emphasize them ;)

[@elefesel]

I think that's why he's using a different colour to do this:)

[@kurt.dresner]

I will draw a line through this equals sign to emphasize it ≠

[@RadicalCaveman]

The best way to emphasize a capital "I" is to draw a horizontal line on top of it. At least, T think so.

[@landsgevaer]

@@RadicalCaveman / think extreme italics work well.

[@MitanshilovecheesecakeMakwana]

😂😂​@@RadicalCaveman

[@arpansingh2116]

Can't believe I'm watching this for entertainment

[@thangphamngoc3897]

lool me too

[@dananajj]

I'm watching this to fall asleep

[@NickiRusin]

ikr, it's super fun

[@RosidinAli]

Lol me too

[@wassollderscheiss33]

That's because he is (somewhat) entertaining

[@davidbrisbane7206]

I have noticed that if there is an easy way to prove something, then Michael will point it out and then prove it a harder, but more entertaining and often a more general way :-).

[@falknfurter]

That's true. In mathematics important theorems are often proved in different ways using different techniques. Examples are e.g. theorem of pythagoras or the law of quadratic reciprocity. By this you can understand relations between different areas of mathematics.

[@wospy1091]

All he did in this proof was use Calculus instead of induction. You could even say he did it the easy way.

[@davidbrisbane7206]

@@wospy1091 Induction is easier.

[@gabriel7233]

@@davidbrisbane7206 not for everyone but it's clearly more elementary

[@iabervon]

@@gabriel7233 Additionally, if you know both techniques, the work for induction in this case is easier. 2^(n-7) > 128n > 128 implies 2^((n+1)-7)=2*2^(n-7)>128n+128=128(n+1).

[@pratikmaity4315]

He explained so many stuffs in a single video. This shows how different math problems are related to each other!!

[@ckeimel]

I think that this is one of the beauty of maths. There's so much ways to interpret something that you can always found a new one. For example, one of the most relevant theorems of math is The Fundamental Theorem of Algebra and one of the most important consequences is the fact that links algebra with the study of functions (calculus).

[@pratikmaity4315]

@@ckeimel True to say

[@vedhase2370]

May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.

[@ckeimel]

@@vedhase2370 wich part of the solution troubles you?

[@derhamcohomology]

What I've learnt from this channel is to start every math problem by considering case n=1.

[@Miju001]

I like the method that was chosen to prove the inequality - it proves it not only for natural numbers, but for all real numbers as well.

[@x_gosie]

Same!

[@SREproducciones]

This is a high school math problem, so derivatives are an overkill. There is a simpler way, and for a kid trying these problems that solution is more useful.

[@vedhase2370]

May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.

[@x_gosie]

@@vedhase2370 The first thing you need to know are the statement gaving, by the problem.

[@randomdude9135]

@@vedhase2370 cuz let's say 1=n^(1/n-7) Then taking log on bs we get 0=(1/n-7)xlog(n) Now (1/n-7) is non zero as n>=8 => logn =0 => n=1 Which is a contradiction. QED

[@takyc7883]

Fun fact, we know that 0

[@35571113]

Ha ha! How?

[@volodyanarchist]

@@35571113 You take 5th root of pi and 17 by hand, everybody is so impressed they ignore the fact that you didn't prove anything yet.

[@84y87]

@@volodyanarchist lmao

[@vedhase2370]

May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.

[@ASD128London]

@@vedhase2370 I was wondering that too.

[@tomaszgaazka6777]

Without induction, in my opinion the easiest way is to use binomial coefficients. n < 2^(n-7) for n >=11 is equivalent to n+7 < 2^n for n>=4. We check n=4 and n=5 by hand and for n>5 we have 2^n = (1+1)^n > (n choose 0) + (n choose 1) + (n choose n-1) + (n choose n) = 1 + n + n + 1 = 2 + 2n > n + 7.

[@robwsur]

beautiful idea

[@TheOneSevenNine]

great as always. love when you just underline something and mark it as "clear." that's how you know you're watching a real mathematician

[@xenofurmi]

Showing the "obvious" parts and "the long way" is so so good. It's how we all learn. Very good job!

[@Omar-of4tz]

set f(x) =x^(1/x-7) arguing the monotony of the function on [8,+∞) let a,b≥8 be real numbers s.t a1/b-7 ⇒ f(a)>f(b) ⇒the function is strictly decreasing on [8,+∞). therefore max(f(x))=f(8)=8. We have 0

[@ireallyhatemakingupnamesfo1758]

N^(1/n-7)>1 The proof of this is obvious and left as an exercise to the reader

[@AnindyaMahajan]

It is obvious. a^x is a strictly increasing function for a>1 and equals 1 at only x=0.

[@richardconrardy9673]

*n^(1/(n-7))>1

[@josem138]

I don't quite get the fact he is doing too much to prove that sentence. There is no number greater than 8 multiplied by itself n-7 times going to give a number lesser or equal than 1, enough proof.

[@jamesjohnson4695]

@@josem138 Not multiplied by itself (n-7) times, but 1/(n-7) times. But yes the claim is still trivial.

[@rhaq426]

@@AnindyaMahajan I don't see why. a^y means a is a real constant and y is the input. Here a = x (so a is not a constant) and y = 1/(x-7). After x>=8, the function (x^(1/(x-7))) is strictly decreasing. Apart from looking at the graph what other proof is there? Edit: Actually after giving it some thought I see it is obvious since x^(1/(x-7)) = 1 when 1/(x-7) = 0 which leads to: 1 = 0 So x^(1/(x-7)) is never 1 and since the function is strictly decreasing it will never be less than 1 either.

[@hrishikeshkulkarni2320]

I love this channel, awesome the way you spent a several seconds and a couple of sentences in proving that 10 is not a perfect cube.

[@MK-13337]

7:10 is not equivalent, but you only need the reverse implication here. f(x) >0 everywhere => f(n) > 0 for natural numbers

[@stevenwilson5556]

This channel is such a gem, more people should be watching this stuff.

[@shurik3nz346]

No, a person.can watch what they want on youtube.

[@shurik3nz346]

Abacus, should and must are different

[@shurik3nz346]

Abacus , ahh I see. Thanks for enlightening me.

[@stevenwilson5556]

@@shurik3nz346 I wasn't advocating strapping people to chairs and prying their eyes open Clock of Orange style; I meant that people would benefit but most people are ignorant of this channel's existence.

[@shurik3nz346]

@@stevenwilson5556 haha alright

[@dariensisko0]

Although it was excluded right at the beginning, n=1 would be a solution as 1^x=1\in N for any x.

[@red0guy]

Man who I wish I would have had such videos in high-school almost 20 years ago.

[@jole0]

Love your videos, participating in my first olympiad in a month or two

[@rishabhsinha4765]

Good luck!

[@James_Moton]

What olympiad are you attending?

[@jole0]

@@James_Moton Norwegian one

[@asdfghjkl6506]

All the best✌️👍👍

[@35571113]

Good luck!

[@demenion3521]

when i first looked at the problem, i found it pretty clear that either the exponents needs to be 1 (which gives 8 as a solution) or n needs to be a perfect power (because otherwise we would never get an integer by taking an (n-7)th root). the rest would be quite similar to what you did, just showing that f(x)=x^(1/(x-7)) is decreasing for x>=8.

[@tomatrix7525]

This was a really lovely problem. Well presented Michael, loving your stuff

[@basmamjouel2893]

In order for it to be an element of N then n should be written as n=m^(n-7), with m>=3>e cuz n>8>e (the case of n=8) is trivial) also m=n^m when we play around with the inequality and by replacing values we get n

[@lawandeconomics1]

I probably would have turned back after taking the derivative. Thanks for an unusually clear and direct work-through!

[@harish6787]

I felt this much easier than any other question

[@roboto12345]

For example question 6 of IMO 2020

[@pedroafonso8384]

Yes

[@xz1891]

Bcz it is

[@xz1891]

@@roboto12345 I can only solve Q1 of imo2020

[@paulchapman8023]

Finding the solutions is easy; the hard part is proving that there are no other solutions. It may be easy enough to grasp intuitively, but math requires more proof than just intuition.

[@dumbledort9755]

the equivalence at 7:30 is false. But the implication b=>a is enough to prove the result (b result with x in R).

[@Nicolas-zf3pv]

Why the equivalence is false? I had seen other comments claiming the same thing, one of them said that f(x) is negative for some real numbers, but how can this be true for x>= 11?

[@dyidyirr]

@@Nicolas-zf3pv It's not clear that if f(n)>0 for n>=11, n integer then f(x)>0 for x>11, x real. In this case the equivalency holds because f is continuous and monotonous. But it doesn't hold for any f. He proves later that it's monotonous, so with that bit from the proof (and the fact that the function is clearly continuous) he could prove that the equivalency holds as well. He doesn't need to, because the other direction is sufficient. But the way he writes it, it's not sufficient.

[@manucitomx]

A backflip would have been great. I love these videos. It makes me kinda sad I didn’t go for pure instead of applied math.

[@iabervon]

He can't do backflips in videos any more, due to discontinuities outside the domain of the video in the neighborhood of his head. (That is, hanging microphones.)

[@speedsterh]

@@iabervon Excellent !

[@calebbirnbaum4605]

The work is pleasingly clear and labeled

[@markfischer5044]

I tried to solve this from the thumbnail, so I missed that we were solving over the natural numbers. I got n = - W(-1, -ln(z) / z^7 ) / ln(z) where z is an integer and W is the Lambert W function aka the ProductLog. Apparently, we need the -1 branch. n. n^(1/n-7) 10.375 2 9 3 8.54777 4 8.31611 5 8.17246 6 8.07331 7 8 8 7.94315 9 7.89749 10 7.85982 11 7.82809 12 7.80089 13 7.77725 14 7.75646 15 7.73799 16...

[@riskassure]

From e < 4, we get 1 < 2 ln 2 by taking the natural log on both sides, and then dividing both sides by 2 yields the inequality.

[@kinshuksingh90]

An arguably quicker way of proving 2^n>128n could have been by plugging n=11 and checking that 2^n is indeed larger and then seeing that the slope of 2^n at n=11 is 1024*11 (chain rule) and that the slope of 128n is 128. Plus the slope of 2^n only keeps on growing whereas 128n has a constant (and lower) slope. Combining f(11)>g(11) and f'(x)>g'(x) for x in [11,∞), you know that f(x)>g(x) ,if f(x)=2^x and g(x)=128x

[@mathcanbeeasy]

I think it's easier how I solved it. So if n ^ (n-7) is natural, it means that n = k ^ (n-7), with k a natural number. It is immediately verified that n = 8 and n = 9 are solutions and k = 8 and k = 9. For n = 10, the only close k could be 2 becouse 2 ^ 3 = 8 and 3 ^ 3 is 27, much greater than 10. Therefore we observe that for n> = 11, n = 11, n = 2. By induction, P (m) => P (m + 1) So we have m

[@SQRTime]

Hi Adrian. If you are interested in math competitions, please consider our channel kzbin.info. Hope to see you 😊

[@RyanLucroy]

I did another induction: n < 2^(n-7) is true for n=11, then we can say: 2^(n+1-7) = 2*2^(n-7) > 2n = n+n > n+1 (for n>=11) So therefore: n+1

[@thewhitefalcon8539]

Observe that for n^(1÷m) to be a natural number all of n's prime factors must have multiplicity that's divisible by m. The smallest such number (other than 1) has prime factorization 2^m (because 2 is the smallest prime) which obviously grows faster than m+7 once you get past about m=4

[@thewhitefalcon8539]

... and obviously 1 isn't the mth root of any number m+7 because it's not the mth root of any number other than 1. Proving this is a waste of time. Unless they score your answer based on how fundamental your proofs go?

[@estolee5485]

This was my approach as well, and I'm surprised I haven't seen more comments about it. It seems way easier than any other solution I have seen

[@iooooooo1]

8:20 Another nice way to see 2^11 - 128(11) > 0 is to factor a 2^7 out of each term and observe it's true iff (2^4 - 11) > 0. (Even if it's not memorized, 128=2^7 was used earlier in the problem.)

[@lucas0m0james]

I just directly differentiated the original function extended to the reals, showed that was negative for n>=8 and then by exploration found that 8&9 were the only solutions.

[@rezhaadriantanuharja3389]

My solution substitute x = n - 7 and n = a^x for some integer a to obtain a^x = x + 7. Then obviously for x >= 4 we need a < 2 and we only need to evaluate x = 1,2,3. The only satisfactory pairs (a,x) are (8,1) and (3,2)

[@hogehoge1030]

If n^(1/(n-7)) is an integer, then n = m^(n-7) for some integer m greater than or equal to 2. Take a prime divisor p of n and let e be its exponent in the factorization of n. Since n = m^(n-7), the exponent of p in the factorization of m^(n-7) is greater than or equal to n-7 and hence so is a. It thus follows that 2^a - 7 ≦ n - 7 ≦ a. Therefore 2^a - a ≦ 7．Because 2^x - x is increasing on x ≧ 1 and 2^4 - 4 > 7, we have the inequality a ≦ 3; Hence n ≦ 10.

[@canaDavid1]

Set k = n-7 (k+7)^(1/k) in N, k > 0 (k+7) = m^k, m,k in N k = 1 and 2 are trivial solutions, with m = 8 and 3. Notice that (k+7)^(1/k) = m is strictly decreasing with k, so any other solution would have m = 2. 2^k = k+7 has a solution between 3 and 4, so not in integers. Hence, n in {8,9} are the only solutions.

[@chilling00000]

IMO is happening now, it would be fun to feature some of the questions from this year

[@ronbannon]

The sequence decreases on its domain and is asymptotic to 1. We know a_8 = 8, a_9 = 3, so you need to consider whether there's an n such that n^(1/(n-7)) =2; it is relatively easy to show that there are no natural number solutions to this equation.

[@Dusk-MTG]

First impression: Holy moly, that's gonna be hard Second impression: But wait, that's gonna be impossible very soon Solution: It's actually just the first 2 allowed numbers I'd say the perceived difficulty of this exercise is a strictly decreasing function of time.

[@KingstonCzajkowski]

It has a spike like x^-2, a discontinuity that occurs when he mentions that we're not using induction

[@RZMATHS]

Well if feel like doing a challenging prob You can try this kzbin.info/www/bejne/n5jHqmNrgb-db5I

[@Jivvi]

I had the opposite experience. My immediate impression (even before I noticed the ≥ 8 condition) was 8 works, 9 works, and then that's it. The more he explained why higher numbers don't work, the more I started to think that there's going to be some other number, probably a multiple of 7, where it equals 1, but that it would be really hard to prove.

[@hans-juergenbrasch3683]

The function f(x)=x^{1/(x-7)}=e^{ln x/(x-7)} is greater than 1 for x>7 and monotone decreasing, since f'(x)=f(x)/[x(x-7)^2]*[-7-x(ln x -1)]10 we have 1

[@utsav8981]

Me: Sees 1/n-7 *Ezee* Me: Sees n just below it *crap*

[@0xDEAD_Inside]

Same

[@84y87]

1/n-7 is integer only for n=6,8

[@alex-cm9fd]

Dorulo ili kvo

[@vedhase2370]

May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.

[@ethanyap8680]

@@vedhase2370 very late,but since 1/(n-7) is positive, n^(1/(n-7)) > n^(0) = 1

[@handanyldzhan9232]

Let n=b^x. Then we need to observe when b^[x/(b^x-7)] is an integer. That means b^x-7 divides x. For that to be possible, b^x-7 =< x. If x=1, b^x=8. If x>1, then 2^x-7 =< x so x

[@islandman9748]

Excellent! You did your number eight like me until I were 8 years old. I remember my teacher scolded me for it. So far, I do my eights "normally". It was just to annoy you but you're doing a great job!

[@islamgaziev1717]

Finally, something I managed to solve before watching the video. Beautiful solution.

[@徐瑞斌-i8o]

Please correctly state the question by adding: n ∈ N (natural number). Otherwise, there are more real number solutions for n, since the real function f(x) = x ^ (1/x-7) is a continuous function, and we have f(8)=8 and f(11)

[@Bodyknock]

0:50 "Similar things happen with n=5, 4, 3, 2, and 1..." A very minor correction, but actually 1⁻¹/⁶ = 1 so n=1 works. 🙂

[@gaufqwi]

The problem with underlining an inequality to emphasize its strictness is then you make it nonstrict 🤣

[@vedhase2370]

May I know why it isn't equal to 1?? I am not really good at math. Someone, please help me out.

[@paulchapman8023]

@@vedhase2370 Maybe he considered the fact that 1^n = 1 for all real values of n not worth commenting on.

[@kadenxu5142]

@@vedhase2370 well, any number to the zero ith power is equal to one, ex 3^0 = 1 and 7^0 = 1. The problem states that n is greater than or equal to eight. 1/(n-7) is always going to be greater than 0 if n is greater than eight. That means that the equation basically boils down to n^(a number greater than 0). In order for the whole equation to be less then or equal to one, n must be raised to the power of 0, or less, but since 1/(n-7) is always greater than zero due to the parameters of the question, we know that n^(1/(n-7)) is greater than one. Hope this clears things up

[@jongu71]

I thought of that too, he almost managed to confuse me!

[@knvcsg1839]

That thinking of searching for n>11 is really brilliant.

[@nasrullahhusnan2289]

I think it is easier to grasp by using Taylor series to show that ln(2)

[@nasrullahhusnan2289]

Sorry the last sentence must be ln(2) < ½ instead.

[@judecarter6095]

By thinking of it as an n-7th root you can also show that n>=2^(n-7) as n!=1 and hence 8

[@jakubdutkiewicz5816]

Unless n is in natural, there has to be a solution for f(n) = n^(1/(n-7)) = 2. Note that f is monotonic, continous in (7,+inf> and has an asymptote at f = 1.

[@AntonBourbon]

11:19 : After reading Richard Feynman's brilliant "Surely You're Joking, Mr. Feynman!" I can't forget that ln 2 = 0.69... which is clearly bigger than 0.5

[@scottrichmond3548]

Me after seeing thumbnail: "Oh that's easy, it's 8" Me after clicking on video: "Ah, the old bait and switch"

[@GhostyOcean]

I believe you could also use binomial expansion for 2^m to show the inequality as well. Use 2^m = SUM( m choose k) from k=0 to m and use however many terms you want to show that 2^m > 128m

[@yangli3144]

Love the channel, interesting problems and crystal clear explanations! for proving ln 2 > 1/2, maybe it is a bit more straightforward to do: ln 2 = 1/2 * 2 * ln 2 = 1/2 * ln (2^2) = 1/2 * (lg 4 / lg e) > 1/2 ?

[@ddognine]

But, now you have to prove that ln 4 is greater than 1.

[@yangli3144]

@@ddognine I think that's quite straightforward as e=2.7*** < 4. of course, for formal proof one should probably claim on monotonicity of lg function etc.

[@RobertMOdell]

n = 1 is a soln

[@dingo_dude]

trivial

[@roverdover4449]

good point, missed that.

[@AnkhArcRod]

The problem states that n>=8.

[@RJSRdg]

@@AnkhArcRod But he says the reason the problem says n>=8 is that it doesn't work for n

[@ireallyhatemakingupnamesfo1758]

AnkhArcRod yea, but he said that n=1 wasn't a solution when he was talking about why the bounds of the problem were n>=8

[@JamesLee-pk8oj]

you need not use derivative of f(x). it's enough that just prove f(n+1) > f(n) for all n > 7，which is much easier: 2^(n+1)-128(n+1)-(2^n-128n)=2^n-128 > 0

[@lunascapes]

Yes, that’s the proof by induction.

[@IoT_]

Actually he mentioned the method in the video (proof by induction)

[@84y87]

He mentioned PMI and said he would prove it using a "different" method.

[@patrickgambill9326]

I used a slightly different approach. I took the derivative of x^(1/(x-7)), and showed it was decreasing for all x greater than or equal to 8. Since we know 1 is our lower bound, and the n=11 case is less than 2, then we know we only need to check 8, 9, and 10

[@avin9106]

Thanks for the great videos! Simpler final proof (edit: oops, it's the same): ln(2)>1/2 2>e^(1/2) 4>e

[@MarcoMate87]

Lol, it's the same proof he used, nothing different.

[@avin9106]

@@MarcoMate87 Oh, you're right. I just saw that he started differently and wrote this down quickly.

[@dominikstepien2000]

6:55 I don't think it is equivalent because you could take such a function that it would be positive on integers but not always positive for real numbers but anyway that doesn't change the solution

[@ChristianRosenhagen]

One of the best: creative and not too hard!

[@joaosousapinto3614]

Here's how I solved it, which is probably slightly simpler: if n is a natural number for which n^(1/(n - 7)) is an integer, then there exists a positive integer k greater than 1 for which k divides n^(1/(n - 7)) [as clearly the latter is never 1]. In particular, this implies that k^(n - 7) divides n (by taking powers on both sides of the divisibility condition), which in turn implies that k^(n - 7) must not exceed n. Now, k >= 2, so 2^(n - 7) 11, the left-hand side doubles every step, while the right-hand side only increases by 1). So one need only analyze the cases where n is 8, 9, 10; clearly, of these, only 8 and 9 satisfy the given equation. In the end it's not too different from your solution, but at least I found it easier and perhaps more intuitive.

[@NWRIBronco6]

Agreed, although I think he basically spent the whole video proving slightly more rigorously this statement in your proof: "Now, k >= 2, so 2^(n - 7) 11, the left-hand side doubles every step, while the right-hand side only increases by 1)."

[@mohamednejighnimi4860]

it's viable also for n=1 as 1^any number=1 which is an integer.

[@stefanocorno4649]

You can also take the derivative of the function itself. It takes a bit of implicit differentiation but it's not too complex and you get y((x-7)/x-ln(x))/(x-7)^2=y'. Then since you know y is always positive and so is (x-7)^2 (at least whenever x>8, which is the relevant range anyways) you can show (x-7)/x-ln(x) is always negative since (x-7)/x approaches 1 and ln(x) approaches infinity (a bit trickier than just that but that's the idea). Positive * positive * negative = negative. Since the derivative is always negative, and you know for x=10, y8 or whatever the function is always greater than 1, you know the value will always be between 1 and 2 and therefore can't be an integer.

[@abhishekkumar-os5zk]

take x= n*power(1/n-7) take log both side logx= 1/n-7 * log n , t= logx n-7* t = logn

[@sujitsivadanam]

0:47 Actually, n = 1 will work (if not for the restriction on n)

[@dneary]

I proved the inequality by swapping n-7=k and proving that 2^k>k+7 for k>=4 - which is easy by expanding (1+1)^k using the binomial theorem and keeping only the kC0, kC1, kC2, kC(n-1)and kCn terms , giving the inequality 2^k>=2+2k+(k)(k-1)/2 > 8+2k for k>=4 >k+7

[@asa4766]

Wonderful video, always a pleasure to watch. I was wondering if you Could make a video on the “points in the space” type of problem (I believe it’s called misc but I’m not sure). An example to this type of problem is “There are a 1000 points in the space, prove you could always pass a line such that half the points would be on one side and the other half on the other side”.

[@bcwbcw3741]

isn't it easier to take both sides to the power n-7 so we're looking for x^(n-7)=n . x positive integer. since x=1^(n-7) is always 1 and the smallest x is then 2 and 2^(n-7)>n for all n>11 is pretty obvious with no heavy duty math. The induction is just that for n+1, 2*(previous>n)>n+1 .

[@danieldomert1022]

A suggestion for a quick and easy proof (it feels almost too easy - are there any flaws in my reasoning?) : f(n)=n^(1/(n-7) >= 8^(1/(n-7)) = 2^(3/(n-7) > 1. When does f(n) get >=2? 2^(3/(n-7))>=2 3/(n-7) >=1 n

[@lord_hamza3550]

I think we can also say that n is between 8 and 10 because the first integer to accept a 4th root is 16 and n=11 gives us 4th root of 11 which is impossible.

[@septagram9491]

I really like the ending you make in each video. That's a good place to stop. It feels like maths has an infinite domain one can explore. It's like the opposite of claustrophobia.

[@SQRTime]

Hi Septagram. If you are interested in math competitions, please consider our channel kzbin.info. Hope to see you 😊

[@shreyathebest100]

One of the few number theory problems I silved on my own

[@ivoh2384]

Proof that ln 2 > 1/2: 4 > e. [known] ln 4 > ln e. [ln is increasing] 2 ln 2 > 1. [log property] ln 2 > 1/2. [divide by 2 > 0] Done.

[@cognomen9142]

Why is 4 > e "known"? I mean, I know ln 2 is about 0.693 just as certain as I know e is about 2.718. Why is one of them "known" and not the other? This is the flaw of his proof. He needs to prove 4 > e as well. This can be done from f.ex. the definition of e as the limit of f(x) = (1 + x)^(1/x) as x tends to 0 noting that f'(x) < 0 (it's an ugly but straightforward derivative and it's easy to see it's negative) and that f(-1/2) = 4 implying f(x) < 4 for any x > -1/2 and hence also at the limit.

[@gmchess7367]

Amazing explanation

[@rous914]

A much easier way to prove it is the following: Notice that n^(1/(n-7)) = [n * 1*1*1*.......*1]^(1/(n-7)), where we have multiplied by (n-8) 1's. Then applying the AM-GM inequality we get: n^(1/(n-7)) 13, we are interested only in the values of n

[@pawelartymowicz1617]

You forgot about solution n=1 while discussing n

[@antman7673]

Also the fact that the minimum prime factorization can be log2(n) indicates that the forth root is only available at the integer 16 at minimum is sort of a proof.

[@jackboyce]

The original problem qualified that n is an integer. Otherwise you have 7 distinct real solutions that make n^(1/(n-7)) a natural number, corresponding to values from 8 down to 2, inclusive. For example, n = 8.316108164429... evaluates to 5.

[@dorijancirkveni]

0:24 0:48 For n=1, n^(1/(n-7))=1^(-1/6)=1

[@TIMS3O]

Define f(x)=x^(1/(x-7)). Another way to prove it is to notice that f(11)1 when x>7. Then we are done if we can show that f is strictly decreasing when x>7 which by differentation amounts to showing that x-7-ln(x)x7. But this is clearly true since ln(x)>ln(7)>1.

[@pcarlisi]

0:52 A wonderful example of why you have to be careful making generalizations in mathematics, as this is a false statement. n=1 could be a potential solution (if it were not excluded by the n greater than or equal to 7 condition), as 1^(1/(1-7))=1^(-1/6)=1/(1^6)=1 is in fact an positive integer.

[@AntonBourbon]

0:45 : I can't see why the obvious solution n=1 is ignored, as 1^(1-7) = 1^(-6) = 1 is definitely a natural number. I don't really appreciate fiddling with derivatives and natural logarithms instead of the induction proof (which is trivial even for me, and 90% of the things on this channel are way too complicated for me). Otherwise, a great video, thank you!

[@vinc17fr]

Just a thought… Proving the result by proving that f(11) > 0 and f'(x) > 0 could be regarded as some form of "continuous induction".

[@oximas]

ya kinda

[@Alan-zf2tt]

Well 14:14 may be a good place to stop. It is also an excellent place to start. Why? There are interesting ways forward that include asymptotic approximations concepts where for given epsilons and deltas there are Ns where n>N provide acceptable values. Admitted not pointwise answers other than N but mathematically acceptable (?) answers to tolerances appreciated by mathematicians, physicists and engineers alike. It seems good in the round to bring these things together and Michael's video is an excellent venue for that purpose alone. It also seem healthy introducing real analysts to ordinary world of numerical readings and associated tolerance values or error bounds? Debatable? I used Qualculate v4.9.0 a free software calculator with graphic facilities that make it easy to graph and change input range and so forth. So for answers accurate to 1 within error bound 10^(-4) at n=20000 and n=20001 yielded a interesting descending staircase graph. Obtained by plotting the graph where n is replaced by x and real x is approximated by setting sampling rate at 1001 with minimum value on x = 20000 and maximum value of x = 20001. Of course these equate to n=20000 and n=20001 respectively. An interesting change also happens between x=20000 and x=20002 I chose to present that range as it is the first time I noticed graph switch from smooth line to digital steps. In this case descending staircase. Important concepts? numerical methods, graphical methods, practical application of epsilons and deltas within acceptable error bounds. Thus! (I claim) fuller answer is 8, 9, infinity and N for infinitely many n>N where N defines finite range of n directly according to unacceptable error bound and infinitely many n>N within acceptable error bound.

[@alex95sang52]

Why doesn't 1 work? Edit : I know the question says "n more than 8", but he explains at 0:48 that all integers between 1 and 7 are not solutions anyway

[@jussari7960]

You are correct it would be a solution, but he probably didn't think it was necessary to mention it

[@RexxSchneider]

He was simply wrong to claim that 1 would not be a solution.

[@miketakeahike4741]

Yeah Sir please do IMO questions also like IMO 2003 P2 number theory and algebra questions

[@parmilakumari3146]

Fun problem

[@miketakeahike4741]

Yeah that would be great

[@cicik57]

check small values, you gen n = -+1,8,9. prove by induction, if for some n >9 n^(n-7) > n, look at n+1: (n+1)^(n-6)= (n+1)^(n-7) * (n+1) us n+1 what is compare (n+1)^(n-7) v 1, what is obviously bigger then one.

[@ussenterncc1701e]

For ln(2)>1/2 can't we just operate on both sides with e. Get 2>√e then square both for 4>e. That's true, so the starting inequality was true.

[@potatochipbirdkite659]

The title card asks when the expression is an INTEGER rather than a NATURAL, and doesn't bound it to n >= 8. Solving that, we can include both 1 and -1.

[@Marre2795]

We can include 1 (at n=1) and 0 (at n=0), but I don't think (-1)^(1/(-8)) is an integer. EDIT: n=0 is undefined because 0^(1/-7)=(1/0)^7, and that's dividing by 0. ALSO: The title card doesn't specify that n should be an integer, so if we assume that n can be irrational, we can express all integers greater than 1 as n^(1/(n-7)). f(n)=n^(1/(n-7)) is a continuous function for n>7. All integers greater than 8 will lie between 7

[@chrisrybak4961]

Ok, nice problem, but it is much quicker simply to consider (integer inputs to) the function y=f(x)=x^(1/(x-7)) . Quickly differentiating you can see that f’(x) is always negative for x>7, and using l’Hopital the limit as x->infinity is 1. We know that f(8)=8 and f(9)=3, and that the function crosses the value 2 at a non-integer x. Therefore (as it then asymptotically approaches f(x)=1) there are no other solutions. (Graphing the function on DESMOS makes this obvious) Anyway, it is the same basic proof, but the video could have been 2 mins long!

[@SQRTime]

Amazing. For math competitions, please consider kzbin.info/www/bejne/goTcoaduqriGjZY and other videos in the Olympiad playlist. Hope you enjoy 😊

[@sebastientraglia1351]

Wouldn't it be enough to see that the real function f=x^(1/(x-7)) is continuous for x>7 and approaches 1 from above as x goes to infinity? This proves that when x is large enough the inequality 1

[@jamirimaj6880]

after watching Numberphile: *I NOW KNOW AND UNDERSTAND MATH! I CAN SOLVE ANY MATH PROBLEMS NOW, I CAN DO IT!* after watching serious Math like this channel: 😭😭😭😭😭😭😭

[@mathadventuress]

Right? This is nuts

[@ruathak1106]

f(n) = (n)^(1/(n-7)) is a strictly decreasing function for n >= 8. f(8) = 8 f(9) = 3 Since f(n) is strictly decreasing for n >= 8, f(n) cannot equal 4, 5, 6, or 7 if n is an integer >= 8 Limit of f(n) = 1 as n --> infinity. Therefore f(n) cannot equal 1 if n is an integer >= 8. This leaves only one other possible case: f(n) = 2. f(10) > 2 f(11) < 2 So f(n) cannot equal 2 if n is an integer >= 8. And we're done.

[@ananyashukla2749]

Nicely explained and question was good

[@tolep]

On the left side: numbers with natural roots: 8 9 16 25 27 32 36 49 64 81 Now it's clear that 16 doesn't have a natural root of 9th degree, 25 doesn't have root of 18th degree and so on.

[@KiLLJoYYouTube]

I really don't know why he did the ln2 > 1/2 thing. "It's well known that e is 2.718..)" but wouldn't that also require a proof? I could just say "well my calculator says ln2 is 0.693..". Other than that, showing the gradient is always positive was good. edit: now I see it, he wanted to make it easy to understand

[@SinclairLocke]

Do those olympiads even allow calculators?

[@mohangohil9613]

ln2=0.3010

[@shqotequila]

Love number theory

[@TomFarrell-p9z]

Not clear whether the n >= 8 requirement was added or part of the original problem. It sounded like Michael said all n < 8 didn't work. Actually n = 1 also works.