Brazil National Mathematical Olympiad 2010 Problem 1

#MathOlympiad #Irrational #FunctionalEquations
Here is the solution to Problem 1 in the Brazil National Mathematical Olympiad 2010!
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I share Maths problems and Maths topics from well-known contests, exams and also from viewers around the world. Apart from sharing solutions to these problems, I also share my intuitions and first thoughts when I tried to solve these problems.
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[@oskarjung6738]

Even though the question is so trivial, the proof is so beautiful. Its questions like these that gives you the 'aha' moments, and make you love maths.

[@JoseFernandes-js7ep]

I think the condition for irrational numbers has the porpoise of avoiding nice numbers like 0 or 1.

[@HagenvonEitzen]

If s² > 4p, then there are reals a,b with a b = p and a + b = s (namely, the roots of x² - s x + p).
Let p= -1 and s arbitrary irrational. Then a and b must both be irrational, hence the functional equation gives us f(s)=f(-1).
Now let p = -sqrt(2) and s any rational. Again, a and v must both be irrational, and the functional equation gives us f(s) = f(-sqrt(2)), and by the previous, we know already that f(-sqrt(2)) = f(-1).

[@letsthinkcritically]

Yes!!

[@danilonascimentorj]

Hi, could you please walk me through how did you conclude that f(x) is constant using this argument?

[@iurialmeida8979]

maaaan, i cant belive i was able to do it, thank you!

[@yaseengharehmohammadloo9955]

thanks for solving theses beautiful problems.

[@letsthinkcritically]

You’re welcome :)

[@matthismercier281]

How and with which support do you think I should train to solve this kind of problems ?
Love your channel btw, wondeful work !

[@matheuspedro3745]

Thanks for the solution, I'm Brazilian and I'm studying for this contest.

[@letsthinkcritically]

My pleasure!

[@NPC-pl5cy]

Orgem de Progresso

[@RZMATHS]

You can try this too if you are preparing for Olympiads
ua-cam.com/video/igdy05LZj90/v-deo.html

[@matheuspedro3745]

@མགུལ་སྙིང Thank you!

[@bernardinojosa3777]

Boa sorte

[@DaveyJonesLocka]

Beautiful work

[@user-yw6kl8tp2c]

Great video!
From you accent I guess you are from Hong Kong or similar places. Did I guess it right? Cuz I’m from Hong Kong and I absolutely love the channel

[@zakzaki9542]

Would it not be enough to say that f(a+b+0)=f(a+b)=f((a+b)0)=f(0), and that a+b can add to make any real number, therefore f(X)=k for all real numbers???

[@giovanicampos4120]

since 0 is not irrational, then we can't do f((a+b)+0) = f((a+b)0), since f(a + b) = f(ab) holds for all a and b that are irrationals.

[@zakzaki9542]

@@giovanicampos4120 yes you are correct. Thanks a lot

[@user-qn4fs2og7k]

thank you sir ~

[@aashsyed1277]

So awesome

[@quirtt]

Cool problem

[@djvalentedochp]

nice video I'm from brazil

[@pedrofarinazzo5669]

I don't get it. Why are you assuming all those functions are equal to f(0)?

[@Kyojuro-bf3bf]

Isnt it possible to first ignore the irrationality condition, then find all functions for all reals such that f(ab) = f(a+b). The only such function satisfying this is f(x) = c, a constant. So it also satisfies for any irrational a and b since the irrationals are a subset of the reals. ?

[@anasuit1111]

irrational :無理数

[@aviralsood8141]

Milk pudding

[@elobez]

{rational} + {irrational} = {irrational}.
Can we not let a = 1 and b be any irrational. Then:
f(b) = f(b+1), both sides mapping irrational to irrational and true for all b. Does this prove that f maps to a constant?

[@scion.]

The statement only says that f(a+b) = f(ab) when a and b are irrational, so you can't choose a=1.
On top of that, even if you could, f(b) = f(b+1) is not a sufficient condition to conclude that f is constant, f could be any function with period 1. For example, cos(2*pi*x) or frac(x) would be non-constant functions that satisfy your condition.

[@tsunningwah3471]

香港口音好親切

[@joelgerlach9406]

But if it holds for all real numbers, then it holds for all irrational numbers. So this condition is useless.

[@TechToppers]

But you don't know before you solve the problem.

[@angelmendez-rivera351]

It doesn't say it holds for all real numbers.

[@moonlightcocktail]

There could be a function that works only for irrational numbers but not for rationale, for example the piecewise f(x) = a solution proposed here if x irrational and 1 otherwise

[@TtTt-ur5hd]

You stupid?

[@krishnasimha8097]

e^a e^b=e^a+b

[@landsgevaer]

Assuming you mean e^a e^b = e^(a+b) , this is an example of f(a)f(b) = f(a+b), but not f(ab) = f(a+b) as asked. Moreover, it wouldn't show these were *all* such functions.

[@OOO-ef1ps]

A familiar accent🤔

[@taopaille-paille4992]

Chinese + British maybe

[@OOO-ef1ps]

@@taopaille-paille4992 To me it's classical Hong Konger speaking English

[@taopaille-paille4992]

@@OOO-ef1ps Nope that's not hk accent

[@taopaille-paille4992]

He has some British accent that for me means he has spent multiple years in UK

[@finmat95]

What a terrible calligraphy

[@omenuawo]

Cool problem