Elimination of cubic roots can be done instantly by using Euler's identity: a + b + c = 0 => a^3 + b^3 + c^3 = 3abc. Then the cubic equation is monic so if x is rational it is also an integer, which is also divisible by p.
The end of the first problem becomes much slicker with some basic abstract algebra facts: x^3 - 3p^2 x - p = p^5 has a rational solution if and only if x^3 - 3p^2 x - p - p^5 has a rational root, if and only if x^3 - 3p^2 x - p - p^5 has a degree one polynomial as a factor. But p^2 does not divide -p - p^5, so this polynomial is irreducible by Eisenstein's criterion.
For the second problem, he doesnt actually exhaustively conclusively prove there are no other functions that could fit the criteria, he only finds two that do..
@@leif1075 I disagree. He first proved that either f(1) = 1 or f(1) = -1. Then, he proved that f(1) = 1 implies f(x) = x and f(1) = -1 implies f(x) = -x^2. There are no other possibilities.
For the 2nd problem: You can set x such that xy-1 = y + h. The x that satisfies this is 1+(h+1)/y. This lets you calculate f(y+h). You can then subtract f(y) , divide by h and take the limit as h->0. Then assuming that f is smooth enough, you can express the remaining f(1+(h+1)/y) as a MacLaurin expansion around y=1, plus a couple of terms depending on the 1st and 2nd powers of h/y. The limit can then be rearranged as lim (h . A(h,y) + B(y))/h. Since the derivative "should" converge, it follows that B(y) = 0, and A(y) is f'(y). Now you have a new functional equation, this time it's a differential equation :) Yes, this approach may be more complicated but I thought I'd share :)
There's a slick way to solve #1. Suppose x=p^(1/3)+p^(5/3) is rational. Then 1+px>0 and [p^3+x]/[1+px]=[p^3+p^(5/3)+p^(1/3)]/[p^(8/3)+p^(4/3)+1]=p^(1/3) is also rational which is not possible.
Actually, we can generalize the statement as follow: Suppose n is not a perfect cube and a b are nonzero integer then an^(1/3)+bn^(2/3) is not rational. Otherwise we would have (b^2n+ax)/(bx+a^2)=[b^2n+abn^(2/3)+a^2n^(1/3)]/[bn^(2/3)+abn^(1/3)+a^2] =n^(1/3) is rational which is a contradiction.
Nice problem!! I solved the first question in this way: if you note that (p^(5/3) * p^(1/3)) = p^2, you can prove the thesis by contradiction. If assume that p^(5/3) + p^(1/3) is rational. the same will hold for the square. but if the square is rational, and the double product in the square is rational too for what we observed before, then p^(10/3) + p^(2/3) = p^3 * p^(1/3) + p^(2/3) must be rational. Each combination of the two with rational coefficients must be rational too. for example: p^3 * p^(1/3) + p^(2/3) - ( p^(5/3) + p^(1/3)) / p= (p^3 - 1/p) p^1/3 is rational, and now all you need to do is show that p^1/3 is not rational to get the contradiction
Michael, have you ever noticed how many similarities there are between your math videos and rock climbing? 1. Both activities involve mostly the hands and brain 2. You get chalk all over your fingers 3. When you are done, it's considered good form to remove your tick marks 4. Both activities are fundamentally goal-oriented, even if you can sometimes also just explore stuff 5. You practice for it mostly indoors 6. The difficulty ranges from trivial to only a few humans being able to achieve the hardest goals 7. Slate is a material you are likely to encounter if you do one of these long enough 8. A key skill to develop is identifying a good place to stop.
I also noticed that the 2nd problem's functional equation can be solved with f(x) = -x, BUT if and only if y = 1/x and with f(x) = x^2, BUT if and only if y = 1/x.
I kinda get it, but it still amazes me at 20:00 when you've shown a relation ( f(x) = x) that's true under these conditions you set up (x=x; y=x), and so it applies to all conditions under that case f(1) = 1.
We can reduce the first problem to " *c := p ^ {1/3}* irrational" : *x := c + p c^2 in Q 1 + pc + p^2c^2 in Q* Notice the new expression looks a lot like " *(pc)^3 - 1* ". If we expand by " *pc - 1 > 0* " we get *( (pc)^3 - 1 ) / (pc - 1) = (p^4 - 1) / (pc - 1) in Q* That is equivalent to " *c in Q* ", but the cube-root of *p* is clearly irrational - contradiction!
First one is very basic: Set y=p^(1/3) which is irrational* and assume y+y^5=y+py^2 is rational. Squaring this expression and multiplying with p yields py^2+2p^2y^3+p^3y^4 is rational. With y^3=p this requires py^2+p^4y to be rational, hence (p^4-1)y and y are rational, a contradiction. * if y can be written as a reduced fraction y=m/n, then m^3=n^3p. So p divides m as a prime and substituting m=pt gives t^3p^2=n^3 yielding the contradiction p divides n
I think the slightly faster way to do TMO2017 is to cube both sides immediately and to notice that the LHS becomes p + 3p^2*x + p^5 to get the same polynomial (i think it's a bit easier to spot than the substitution in the video). Then using a rational root theorem and the fact that the coefficient in the x^3 is 1 we get that the rational root must be an integer. But rearranging we get x(x^2-3p^2)=p(p^4+1) which implies p|RHS so p|x or p|x^2-3p^2 from which still p|x. But then p^2|LHS which clearly cannot be said about the RHS.
Great video, thanks. But I am confused when in the second problem case 1 at [14:25] you work on the particular case when x=xy and y=1 to show that f(xy) = f(x)f(y). Hasn't this only be shown to be true when y=1 so that it simplifies to an identity? It's the same thing in case 2 but in neither case do you then use f(xy) = f(x)f(y) so maybe it is just irrelevant?
Here's how an algebraist would solve the first problem: Note in the field Q(cuberoot(p)), every element has a unique representation as a+b*cuberoot(p)+c*cuberoot(p^2). If r=cuberoot(p)+p*cuberoot(p^2)=cuberoot(p)+cuberoot(p^5) for some rational r, then we have r-cuberoot(p)-p*cuberoot(p^2)=0, which by uniqueness, implies r=-1=-p=0, which is a contradiction. Therefore, it is irrational.
Very nice proof! I want to note that this does rely on the fact that cuberoot(p) is irrational, so that we know x^3 - p is the minimal polynomial of cuberoot(p), which tells us that {1,cuberoot(p),cuberoot(p)^2} really is a ℚ-basis for ℚ(cuberoot(p)). Showing cuberoot(p) is irrational is easy, though.
@@riadsouissi Hello, I just saw your comment. I was really saying r and -1 and -p must all be 0. This is because 0 is rational which will have a unique representation where a=b=c=0.
Not far from being elementary but... if p^(1/3) + p^(5/3) = t + pt^2, where t = p^(1/3), is rational, then t would be a root of a polynomial with rational coefficients of degree 2; It follows that the degree d of the extension Q(t)/Q, where Q is the field of the rational numbers, is less than or equal to 2. On the other hand, x^3 - p is the minimal polynomial of t and hence d = 3 and we thus have a contradiction. Likewise f(p^(1/3)) is irrational for a polynomial f with positive rational coefficients of degree greater than or equal to 1.
The c+c^2 derivation at 10:34 and the f(x^2 - 1) => f(x+1)*f(x-1) at 17:10 could have been explained better (by pointing at the rule/equivalence). Other than that you're explaining the steps nicely and the rules used to produce them.
have you seen much about linear involutions? they're quite rare but hold valuable properties! -x is the most essential linear involution as it is the basis for even functions: f(x) + f(-x) = g(x) such that g(x) = g(-x) {involution on the input causes no change} weird to think -x^2 and x share this identity. i wonder how many other neat identities there are like this XD
Great video, thanks! A little comment though: I might be the only one thinking this, but the stereo audio bothers me a little. If you're standing to the left, only my left ear gets audio which gets tiring after a while. If you agree, maybe change to mono audio if that's possible? If not, so be it! I still enjoy the content. Have a nice day!
let a=3rd-root(p) You can notice that the minimal polinomial over Q of third root of p is X^3-p (it is irreducible by einsenstein) that means that {1,a,a^2} is linearly independant over Q, and because a^3=p then a+a^5=a+pa^2 it is not rational, because if it were, then r/s+a+pa^2=0 for som rational number r/s. but {1,a,a^2} is linearly independant... so the coeficiento a equals 0, absurd!
product of [\cuberoot of p] and [\cuberoot of p^5] is rational. if the sum is also rational, you get a quadratic equation with rational coefficient with [\cuberoot of p] as a solution. hence proven. this technique is also used in proving that e\pi or e+\pi, one of them is irrational. the thing in our question is that we know that the product term is rational, therefore the sum term must be irrational.
Interesting that you’re a climber, the pioneer of bouldering John gill was a professor of complex analysis,. There may be some correlation between solving problems on the wall and on paper!
Hello Michael! I have a severe problem with the second question. My result was, there is no solution at all. I came to the point, where I found f(y)=+/-1. Then I argued in the following way: In the functional equation the goal function f occurs three times with different variables, namely (xy-1), x and y. But every time there is only one arguement for the function f. Therefore the three occurences of f(...) refer to the same function. If I obtain any result for f(y) beeing some expression comprising y, the function f(x) must be the same expression but y being substituted by x in every occurence. For example: I obtain f(y)=y*sin(y) then f(x)=x*sin(x) and f(xy-1)=(xy-1)*sin(xy-1). This is my understanding of the functional equation, as the name of the variable is completely irrelevant. I can rename the variable of the function as I want. Consequently for f(y)=+/-1 I got f(x)=+/-1 and f(xy-1)=+/-1 Substituting this for the three occurences of f(...) leads to an equation with the constant +/-3 at the left hand side and 2xy-1 on the right hand side, which must be true for every x and every y. Of course this is impossible and there is no solution. I have a big problem in testing your solutions. Usually this is done by substituing the solution for the f(...) in the given functional euqation. You can do this for f(x) and f(y), but what is f(xy-1) if f(x)=x and f(y)=1?
Here is an alternative solution to the first problem: suppose that a := x + x^5, where x := p^(1/3), is rational; then pa^2 - a = p(x^2 + 2x^6 + x^10) - (x + x^5) is also rational. Using the fact that x^3 = p, one can simplify that expression to pa^2 - a = 2p^3 + (p^4 - 1)x. Solving for x, one concludes that x = (pa^2 - a - 2p^3)/(p^4 - 1) is also rational, which contradicts the fact that p^(1/3) is irrational if p is prime (or, more generally, if p is not a perfect cube). Therefore, a = p^(1/3) + p^(5/3) is irrational.
Is it enough to show that cbrt(p) + cbrt(p^5) = (p + p^5)/( cbrt(p^2) - p^2 + cbrt(p^10) ) and then say that the numerator is an integer and the denominator is not an integer, and hence the expression is irrational?
Always remember solving like this always adds extra soln but never removes any. So, you have to check against the given question to verify whether something is the solution or not. In this question since you got only two possible soln that is x, and -x^2. You can be sure that the given question has at most these both as soln.
A few people have pointed out that noting that x^3-3p^2 x-(p+p^5) is Eisenstein shortens the solution to #1. Here's an alternative way of doing #2. The change of variables z=xy-1 turns the given equation into f(z)+f(x)f((z+1)/x)=2z+1. The right-side is independent of x, so f(z)+f(x)f((z+1)/x)=f(z)+f(1)f(z+1) hence f(x)f(Y/x)=f(1)f(Y) for all x and Y, or more simply f(1)f(xy)=f(x)f(y) for all x and y. Now f(1)=+/-1 as in the video, so this is one of Cauchy's functional equations, and its solution is f(x)=+/-x^c for some real c (as can be easily shown). Plugging this back into the original equation and setting y=1, we can show (e.g. by comparing coefficients or repeated differentiation) that the only two solutions are f(x)=+x and f(x)=-x^2.
Can't help but feel that there's another approach to the first problem. If you have a cube root with a Rational answer you must have a Rational input. With a prime number that means it must be an integer. But any prime number that has a Rational cube root is not prime, as it has the factor of cbrt(p). So it must be irrational by the fact that it's prime, what am I missing?
@@dalex641 In fact, in this case it is true, but the reason is that cuberoot(p^5)=cuberoot(p^2)*p, so the (unique) vector space representation of x=cuberoot(p)+cuberoot(p^5) is actually cuberoot(p)+p*cuberoot(p^2) and since cuberoot(p) and cuberoot(p^2) are in fact Q-linearly independent (because Q(cuberoot(p)) is a degree-3-extension of Q, if p is not a cube) this indeed proves that x is not contained in the Q-vector-subspace of Q(cuberoot(p)) generated by 1, i.e. Q.
I know this is a nuke but you can use the field trace en.wikipedia.org/wiki/Field_trace with the decomposition field of the third root of p. The trace of the sum is zero so it cant be the degree of the extension times the that number that you want to see if it is irrational or not so, because if it was a rational number then it would be the same number times the degree of the extension.
One was very ez! Divide by p. Let x = p^(2/3) + 1/p^(2/3). x^3 = p^2 + 1/p^2 + 3x, so if x = a/b, a(a^2 - 3b^2)/b^3 = (p^4 + 1)/p^2 Because both fractions are in simplest form (as gcd(a,b) = 1), denominators are equal. Thus b^3 = p^2 implying p is a cube, CONTRADICTION!
Sir... I think you should take action about the some comments w/ ¶0Rπ content quickly, otherwise it'll grow and be annoying like in some other youtubers. Thank you ❤️
2020 TMO is a joke... My friend did all 10 problems in 20mins (It took him 2seconds to do question 1) Edit: the one who did is gold medal TMO2019 but it's still too easy if they can do all of the question in less than an hour.
You escaped a point in the problem 1. You should check x=0 before writing it as a/b. Although it is easy in competitions these mistakes can result in breaking points 😉
@@nicolasguereca3559 Why not? You may say I am lying or I am just showing my "talent" off, but I must say that I am actually 13 years old this year exactly. And I already know this, and even more I already know 90% of calculus. Even more surprising, I watched almost every recent videos in this channel and I understood every step in every problem.
@@ritam8767 Lmao, you are just not believing in me. In my knowledge, the four main topics of calculus are limits, derivatives, integrals and differential equations. I can differentiate almost every function easily(the only thing that may be difficult is the chain rule part). Unlike derivatives, integrals require higher thinking. Sometimes I am confused by not knowing which integration technique to be used. The rest of the 10% are the second order differential equations, and the applied calculus, that is in order words, word problems.
Elimination of cubic roots can be done instantly by using Euler's identity: a + b + c = 0 => a^3 + b^3 + c^3 = 3abc. Then the cubic equation is monic so if x is rational it is also an integer, which is also divisible by p.
The end of the first problem becomes much slicker with some basic abstract algebra facts: x^3 - 3p^2 x - p = p^5 has a rational solution if and only if x^3 - 3p^2 x - p - p^5 has a rational root, if and only if x^3 - 3p^2 x - p - p^5 has a degree one polynomial as a factor. But p^2 does not divide -p - p^5, so this polynomial is irreducible by Eisenstein's criterion.
Indeed that is correct
For the second problem, he doesnt actually exhaustively conclusively prove there are no other functions that could fit the criteria, he only finds two that do..
@@leif1075 I disagree. He first proved that either f(1) = 1 or f(1) = -1. Then, he proved that f(1) = 1 implies f(x) = x and f(1) = -1 implies f(x) = -x^2. There are no other possibilities.
@@ricardocavalcanti3343 But what if you plug in other values into the function besides 1?
@@ricardocavalcanti3343 yea you literally forgot about tbe infinte other values if x is not 1..what about 2 or pi or any other number besides 1?
For the 2nd problem: You can set x such that xy-1 = y + h. The x that satisfies this is 1+(h+1)/y.
This lets you calculate f(y+h). You can then subtract f(y) , divide by h and take the limit as h->0. Then assuming that f is smooth enough, you can express the remaining f(1+(h+1)/y) as a MacLaurin expansion around y=1, plus a couple of terms depending on the 1st and 2nd powers of h/y. The limit can then be rearranged as lim (h . A(h,y) + B(y))/h. Since the derivative "should" converge, it follows that B(y) = 0, and A(y) is f'(y). Now you have a new functional equation, this time it's a differential equation :)
Yes, this approach may be more complicated but I thought I'd share :)
Daily dose of Michael Penn
I'm from Thailand and like watching your channel so much. After Covid-19 crisis, hope you visit Krabi again.
There's a slick way to solve #1. Suppose x=p^(1/3)+p^(5/3) is rational. Then 1+px>0 and
[p^3+x]/[1+px]=[p^3+p^(5/3)+p^(1/3)]/[p^(8/3)+p^(4/3)+1]=p^(1/3) is also rational which is not possible.
Nice
So p is not necessarily prime. It is enough that it's not a perfect cube
Actually, we can generalize the statement as follow:
Suppose n is not a perfect cube and a b are nonzero integer then an^(1/3)+bn^(2/3) is not rational. Otherwise we would have
(b^2n+ax)/(bx+a^2)=[b^2n+abn^(2/3)+a^2n^(1/3)]/[bn^(2/3)+abn^(1/3)+a^2] =n^(1/3) is rational which is a contradiction.
26:21 นี่เป็นจุดที่ควรหยุดแล้ว
Breaking the character a bit... Anyone else with a computer science background?
นี่เป็นจุดที่ควรหยุดแล้ว is more appropriate. I am Thai so yeah.
@@paerrin Edited, thanks!
Good Place To Stop np!
Mathematics, physics and computer science.
โอเค
At 4:06, the polynomial x^3-3p^2x-p-p^5 is p-Eisenstein. That proves it right there.
a good thing to notice
Nice problem!! I solved the first question in this way: if you note that (p^(5/3) * p^(1/3)) = p^2, you can prove the thesis by contradiction. If assume that p^(5/3) + p^(1/3) is rational. the same will hold for the square.
but if the square is rational, and the double product in the square is rational too for what we observed before, then p^(10/3) + p^(2/3) = p^3 * p^(1/3) + p^(2/3) must be rational.
Each combination of the two with rational coefficients must be rational too. for example: p^3 * p^(1/3) + p^(2/3) - ( p^(5/3) + p^(1/3)) / p= (p^3 - 1/p) p^1/3 is rational, and now all you need to do is show that p^1/3 is not rational to get the contradiction
Michael, have you ever noticed how many similarities there are between your math videos and rock climbing?
1. Both activities involve mostly the hands and brain
2. You get chalk all over your fingers
3. When you are done, it's considered good form to remove your tick marks
4. Both activities are fundamentally goal-oriented, even if you can sometimes also just explore stuff
5. You practice for it mostly indoors
6. The difficulty ranges from trivial to only a few humans being able to achieve the hardest goals
7. Slate is a material you are likely to encounter if you do one of these long enough
8. A key skill to develop is identifying a good place to stop.
😂😂
This post is underrated.
I also noticed that the 2nd problem's functional equation can be solved with f(x) = -x, BUT if and only if y = 1/x and with f(x) = x^2, BUT if and only if y = 1/x.
I kinda get it, but it still amazes me at 20:00 when you've shown a relation ( f(x) = x) that's true under these conditions you set up (x=x; y=x), and so it applies to all conditions under that case f(1) = 1.
We can reduce the first problem to " *c := p ^ {1/3}* irrational" :
*x := c + p c^2 in Q 1 + pc + p^2c^2 in Q*
Notice the new expression looks a lot like " *(pc)^3 - 1* ". If we expand by " *pc - 1 > 0* " we get
*( (pc)^3 - 1 ) / (pc - 1) = (p^4 - 1) / (pc - 1) in Q*
That is equivalent to " *c in Q* ", but the cube-root of *p* is clearly irrational - contradiction!
First one is very basic: Set y=p^(1/3) which is irrational* and assume y+y^5=y+py^2 is rational. Squaring this expression and multiplying with p yields py^2+2p^2y^3+p^3y^4 is rational. With y^3=p this requires py^2+p^4y to be rational, hence (p^4-1)y and y are rational, a contradiction.
* if y can be written as a reduced fraction y=m/n, then m^3=n^3p. So p divides m as a prime and substituting m=pt gives t^3p^2=n^3 yielding the contradiction p divides n
I think the slightly faster way to do TMO2017 is to cube both sides immediately and to notice that the LHS becomes p + 3p^2*x + p^5 to get the same polynomial (i think it's a bit easier to spot than the substitution in the video). Then using a rational root theorem and the fact that the coefficient in the x^3 is 1 we get that the rational root must be an integer. But rearranging we get x(x^2-3p^2)=p(p^4+1) which implies p|RHS so p|x or p|x^2-3p^2 from which still p|x. But then p^2|LHS which clearly cannot be said about the RHS.
Yep, that’s exactly how I did it. But in the end, it remains quite equivalent to Michael approach.
Great video, thanks. But I am confused when in the second problem case 1 at [14:25] you work on the particular case when x=xy and y=1 to show that f(xy) = f(x)f(y). Hasn't this only be shown to be true when y=1 so that it simplifies to an identity? It's the same thing in case 2 but in neither case do you then use f(xy) = f(x)f(y) so maybe it is just irrelevant?
Thanks Michael :)
Here's how an algebraist would solve the first problem:
Note in the field Q(cuberoot(p)), every element has a unique representation as a+b*cuberoot(p)+c*cuberoot(p^2). If r=cuberoot(p)+p*cuberoot(p^2)=cuberoot(p)+cuberoot(p^5) for some rational r, then we have r-cuberoot(p)-p*cuberoot(p^2)=0, which by uniqueness, implies r=-1=-p=0, which is a contradiction. Therefore, it is irrational.
Very nice proof! I want to note that this does rely on the fact that cuberoot(p) is irrational, so that we know x^3 - p is the minimal polynomial of cuberoot(p), which tells us that {1,cuberoot(p),cuberoot(p)^2} really is a ℚ-basis for ℚ(cuberoot(p)). Showing cuberoot(p) is irrational is easy, though.
Can you explain how you deduced r=-1 step
@@riadsouissi Hello, I just saw your comment. I was really saying r and -1 and -p must all be 0. This is because 0 is rational which will have a unique representation where a=b=c=0.
Incredible presentation, michael!
Michael Penn: goes to Thailand to do rock climbing
Thailand:
Not far from being elementary but... if p^(1/3) + p^(5/3) = t + pt^2, where t = p^(1/3), is rational, then t would be a root of a polynomial with rational coefficients of degree 2; It follows that the degree d of the extension Q(t)/Q, where Q is the field of the rational numbers, is less than or equal to 2. On the other hand, x^3 - p is the minimal polynomial of t and hence d = 3 and we thus have a contradiction.
Likewise f(p^(1/3)) is irrational for a polynomial f with positive rational coefficients of degree greater than or equal to 1.
The c+c^2 derivation at 10:34 and the f(x^2 - 1) => f(x+1)*f(x-1) at 17:10 could have been explained better (by pointing at the rule/equivalence). Other than that you're explaining the steps nicely and the rules used to produce them.
have you seen much about linear involutions? they're quite rare but hold valuable properties! -x is the most essential linear involution as it is the basis for even functions: f(x) + f(-x) = g(x) such that g(x) = g(-x) {involution on the input causes no change}
weird to think -x^2 and x share this identity. i wonder how many other neat identities there are like this XD
Great video, thanks! A little comment though: I might be the only one thinking this, but the stereo audio bothers me a little. If you're standing to the left, only my left ear gets audio which gets tiring after a while. If you agree, maybe change to mono audio if that's possible? If not, so be it! I still enjoy the content. Have a nice day!
That is my bad. I must have forgotten to convert to mono. Oops!
@@MichaelPennMath Ah no problem! Thanks for replying
Thanks michael
let a=3rd-root(p) You can notice that the minimal polinomial over Q of third root of p is X^3-p (it is irreducible by einsenstein) that means that {1,a,a^2} is linearly independant over Q, and because a^3=p then a+a^5=a+pa^2 it is not rational, because if it were, then r/s+a+pa^2=0 for som rational number r/s. but {1,a,a^2} is linearly independant... so the coeficiento a equals 0, absurd!
product of [\cuberoot of p] and [\cuberoot of p^5] is rational. if the sum is also rational, you get a quadratic equation with rational coefficient with [\cuberoot of p] as a solution. hence proven. this technique is also used in proving that e\pi or e+\pi, one of them is irrational. the thing in our question is that we know that the product term is rational, therefore the sum term must be irrational.
Thanks a lot
Ty
Interesting that you’re a climber, the pioneer of bouldering John gill was a professor of complex analysis,. There may be some correlation between solving problems on the wall and on paper!
For the first problem can the rational root theorem be used to contradict the statement?
Great video, thanks. One question on the second question, how do you prove that you have found ALL functions?
Greeting from Thailand !
Hello Michael!
I have a severe problem with the second question. My result was, there is no solution at all. I came to the point, where I found f(y)=+/-1. Then I argued in the following way:
In the functional equation the goal function f occurs three times with different variables, namely (xy-1), x and y. But every time there is only one arguement for the function f. Therefore the three occurences of f(...) refer to the same function. If I obtain any result for f(y) beeing some expression comprising y, the function f(x) must be the same expression but y being substituted by x in every occurence. For example: I obtain f(y)=y*sin(y) then f(x)=x*sin(x) and f(xy-1)=(xy-1)*sin(xy-1). This is my understanding of the functional equation, as the name of the variable is completely irrelevant. I can rename the variable of the function as I want. Consequently for f(y)=+/-1 I got f(x)=+/-1 and f(xy-1)=+/-1 Substituting this for the three occurences of f(...) leads to an equation with the constant +/-3 at the left hand side and 2xy-1 on the right hand side, which must be true for every x and every y. Of course this is impossible and there is no solution.
I have a big problem in testing your solutions. Usually this is done by substituing the solution for the f(...) in the given functional euqation. You can do this for f(x) and f(y), but what is f(xy-1) if f(x)=x and f(y)=1?
Ok so that just begs the question: how did you get to f(y) = +/- 1 because MP didn't get that result.
@@Packerfan130 Sorry I don't know any more as I haven't saved the calculation.
I loved these problems :D
Ambos son problemas muy interesantes.
6:08 mp
Here is an alternative solution to the first problem: suppose that a := x + x^5, where x := p^(1/3), is rational; then pa^2 - a = p(x^2 + 2x^6 + x^10) - (x + x^5) is also rational. Using the fact that x^3 = p, one can simplify that expression to pa^2 - a = 2p^3 + (p^4 - 1)x. Solving for x, one concludes that x = (pa^2 - a - 2p^3)/(p^4 - 1) is also rational, which contradicts the fact that p^(1/3) is irrational if p is prime (or, more generally, if p is not a perfect cube). Therefore, a = p^(1/3) + p^(5/3) is irrational.
Shalom! when are the next lectures on Vertex Algebra
What are hard problems that kind of stunned you for awhile.
Casually mentions his Thailand visit. Then goes on to demolish the question
อ่าวต้นไทร จ.กระบี่ มีปีนเขาด้วยเหรอ
Is it enough to show that cbrt(p) + cbrt(p^5) = (p + p^5)/( cbrt(p^2) - p^2 + cbrt(p^10) ) and then say that the numerator is an integer and the denominator is not an integer, and hence the expression is irrational?
you need to prove that denominator is irrational to prove original problem
Can someone explain me why we can say, that f( (x+1)(x-1) ) = f(x+1)f(x-1) ???
Great video, but the midroll ads are waaaay too disruptive.
In the second problem how do we say that these are the only two solutions do we have to prove that or is it evident in the solution itself?
Always remember solving like this always adds extra soln but never removes any. So, you have to check against the given question to verify whether something is the solution or not.
In this question since you got only two possible soln that is x, and -x^2. You can be sure that the given question has at most these both as soln.
A few people have pointed out that noting that x^3-3p^2 x-(p+p^5) is Eisenstein shortens the solution to #1. Here's an alternative way of doing #2.
The change of variables z=xy-1 turns the given equation into f(z)+f(x)f((z+1)/x)=2z+1. The right-side is independent of x, so f(z)+f(x)f((z+1)/x)=f(z)+f(1)f(z+1) hence f(x)f(Y/x)=f(1)f(Y) for all x and Y, or more simply f(1)f(xy)=f(x)f(y) for all x and y. Now f(1)=+/-1 as in the video, so this is one of Cauchy's functional equations, and its solution is f(x)=+/-x^c for some real c (as can be easily shown). Plugging this back into the original equation and setting y=1, we can show (e.g. by comparing coefficients or repeated differentiation) that the only two solutions are f(x)=+x and f(x)=-x^2.
Can't help but feel that there's another approach to the first problem.
If you have a cube root with a Rational answer you must have a Rational input. With a prime number that means it must be an integer.
But any prime number that has a Rational cube root is not prime, as it has the factor of cbrt(p).
So it must be irrational by the fact that it's prime, what am I missing?
Yes, both of the terms are obviously irrational, but it doesn't mean that their sum is irrational also.
@@dalex641 In fact, in this case it is true, but the reason is that cuberoot(p^5)=cuberoot(p^2)*p, so the (unique) vector space representation of x=cuberoot(p)+cuberoot(p^5) is actually cuberoot(p)+p*cuberoot(p^2) and since cuberoot(p) and cuberoot(p^2) are in fact Q-linearly independent (because Q(cuberoot(p)) is a degree-3-extension of Q, if p is not a cube) this indeed proves that x is not contained in the Q-vector-subspace of Q(cuberoot(p)) generated by 1, i.e. Q.
I know this is a nuke but you can use the field trace en.wikipedia.org/wiki/Field_trace with the decomposition field of the third root of p. The trace of the sum is zero so it cant be the degree of the extension times the that number that you want to see if it is irrational or not so, because if it was a rational number then it would be the same number times the degree of the extension.
how much time do you take to solve one of these problems?
Noticed for 1st math problem he got a depressed cubic. Probably irrelevant but makes me think of Cardano's cubic solution that way.
Good
p^(1/3) + p^(5/3) is irrational. PROOF: I am incapable of reasoning about the expression. Q.E.D.
LOL!
One was very ez!
Divide by p. Let x = p^(2/3) + 1/p^(2/3).
x^3 = p^2 + 1/p^2 + 3x, so if x = a/b,
a(a^2 - 3b^2)/b^3 = (p^4 + 1)/p^2
Because both fractions are in simplest form (as gcd(a,b) = 1), denominators are equal.
Thus b^3 = p^2 implying p is a cube, CONTRADICTION!
ooh! that waw clean, but can you write a more rigorous proof on why a(a^2-3b^2)/b^3 is in simplest form?
So this irrationality proof is just an extension of the proof of the rational root theorem itself?
I have one nice problem
The first one is nice
Thailand land is briliant
nice problem as usual :)
Cool
Sir... I think you should take action about the some comments w/ ¶0Rπ content quickly, otherwise it'll grow and be annoying like in some other youtubers. Thank you ❤️
ไทย จ๋าาาา
Can you make timestamps like chapters in the video timeline? (For the different problems)
crabby town?
Not crappy town I hope
NICE PROBLEM :)
2020 TMO is a joke... My friend did all 10 problems in 20mins
(It took him 2seconds to do question 1)
Edit: the one who did is gold medal TMO2019 but it's still too easy if they can do all of the question in less than an hour.
โหดครับ
Yeah sure kid.
I forgot to mention he's a gold medal 2019 TMO
OH and the audio is kinda trippy imo
You escaped a point in the problem 1. You should check x=0 before writing it as a/b. Although it is easy in competitions these mistakes can result in breaking points 😉
Can you give me the timestamp for this ?
4:17
@@electrovector7212 Thanks
@David Schmitz true.
But if p is a prime so is a positive intetger so x is always positive not equal to 0 . Im wrong ???
why am i here, im 13
If you are gifted, even 13 can understand this.
@@2070user no
@@nicolasguereca3559 Why not? You may say I am lying or I am just showing my "talent" off, but I must say that I am actually 13 years old this year exactly. And I already know this, and even more I already know 90% of calculus. Even more surprising, I watched almost every recent videos in this channel and I understood every step in every problem.
@@2070user 90% of calculus, as in? Name some topics.
@@ritam8767 Lmao, you are just not believing in me.
In my knowledge, the four main topics of calculus are limits, derivatives, integrals and differential equations.
I can differentiate almost every function easily(the only thing that may be difficult is the chain rule part). Unlike derivatives, integrals require higher thinking. Sometimes I am confused by not knowing which integration technique to be used.
The rest of the 10% are the second order differential equations, and the applied calculus, that is in order words, word problems.
2017: Let 3 root p = u and u + u^5 = r, with r being rational. Then u^4 + u^8 = pr, therefore also rational. Letting v = u^4, v^2 + v - pr = 0 makes it so that v can be written as a*sqrt(b)+c, with a, b and c rational. If we call such numbers "square rooted", then u^4/u^3 = u is also square rooted, but the cube root of a prime can't be such a number.
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2014: f(xy-1) + f(x)*f(y) = 2xy -1
" + f(kx)*f(y/k) = "
f(kx) / f(x) = f(y) / f(y/k)
f(-1) + f(0)*f(y) = -1
f(0)*f(y) = -1 - f(-1)
Since f(y) can't be a constant, f(0)=0.
f(y-1) + f(1)*f(y) = 2y-1
f(y) = [2y-1 - f(y-1)] / f(1)
f(1) = 1/f(1) = +-1
If f(1) = 1:
f(1) = 1 - f(0)
f(2) = 3 - f(1)
f(n) = 2n-1 - f(n-1)
f(n) = ...(5 - (3 - (1-0))...)
= 2n-1 + 2n-5 + ... - (2n-3 + ...)
n=2k+1 -> 2n-1+ ... +1 - (2n-3+ ... +3) = [(2n-2)/4]*2 + 1 = n
n=2k -> 2n-1 + ... + 3 - (2n-3+ ... + 1) = (2n/4)*2 = n
f(y+1) = 2y+1 -f(y)
f(y+2) = 2y+3 -f(y+1)
f(y+n) = 2y+2n-1 -f(y+n-1)
f(y+n) = (2y+2n-1-(...(2y+3-(2y+1- f(y)...) = [(2y+2n-1) + (2y+2n-5) + ...] - [(2y+2n-3) + (2y+2n-7) + ...]
n=2k+1 -> [(2y+2n-1) + (2y+2n-5) + ... + (2y+1)] - [(2y+2n-3) + (2y+2n-7) + ... + (2y+3) + f(y)] = (2y+2n-1) - f(y) - 2*(2n-2)/4 = (2y+2n-1) - f(y) - (n-1) = (2y+n) - f(y)
n=2k -> [(2y+2n-1) + (2y+2n-5) + ... + (2y+3) + f(y)] - [(2y+2n-3) + (2y+2n-7) + ... + (2y+1)] = f(y) + 2*2n/4 = f(y) + n
A rational number of x/y gives the result x/y, because f(x) / f(x/y) = f(y) / f(1) = 1.
Let x and y be two irrational numbers whose product is rational:
f(xy-1) + f(x)*f(y) = 2xy -1
f(x)*f(y) = 2xy -1 - (xy-1) = xy
xy-1 = x => f(x)*f((2x+1)/x) = 2(x+1) -1 = 2x+1
Let x be 1 and y irrational:
f(y-1) + f(y) = 2y -1 => f(x) = x.
If f(1) = -1:
f(y) = [2y-1 - f(y-1)] / f(1) = f(y-1) - 2y + 1
f(1) = f(0) - 2*1 + 1
f(2) = f(1) - 2*2 + 1
f(n) = f(n-1) - 2n + 1
f(n) = f(0) - 2*1 + 1 - 2*2 + 1 ... - 2n + 1 = 0 - (n+1)n + n = -n^2
With the same reasoning as if f(1)=1, f(n) = -n^2 can be generalized.