R is a field. By definition, the additive and multiplicative identities of a field are distinct. Somewhat less flippantly, the Peano axioms suffice, since 0 is not the successor of any natural number, while 1 is the successor of 0.
@@j9dz2sf "If 0=1 then Ø={Ø}" That's even less convincing than just "0 ≠ 1" Jokes aside, if you really wanna prove that 0≠1, then yes, that's probably the way to go, since I think one of the most fundamental definitions of numbers are in terms of elements in sets. Correct me if I'm wrong.
@@DeJay7 Yes, in Set Theory ZF, everything is a set: 0 is defined as Ø, 1 is defined as {Ø}, 2 is defined as {Ø,{Ø}} and so on en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers#Definition_as_von_Neumann_ordinals .
There is actually another way of proving continuity of x^2, i.e. that the limit of x^2 = a^2 as x goes to a. |x^2 - a^2| = |x - a| * |x + a| = |x - a| * |x - a + 2a|
My Calculus Professor (Tony Tromba, UC Santa Cruz 1981) dropped the Dirichlet Function on us at the end of a Friday lecture to give something to discuss at Happy Hour.
This was amazingly accessible, thank you. Could you do a video explaining constructive logic and how to prove there? How to rationalize sequential continuity without L.E.M
Here is a nice related problem: it can be shown that function f(x)=0, x is irrational, 1/n if x=m/n, gcd(m,n)=1 is continuous in all irrational points and not continuous in all rational. The question therefore is: is there a reverse function, not continuous in all irrational points and continuous in all rational?
I've found very interesting and brilliant equation that I can't solve. The command for the task: Solve cos(cos(cos(cos(x)))) = sin(sin(sin(sin(x)))). x is a real number. This problem comes from Russian Math Olympiad, 95. Michael, I believe you can overcome this task :)
I am working on this problem now. At x=0, using his proof with our definition of f(x)=x, our f(xo)=f(a) at zero so it is continuous at that point only. I believe this works for the definition of continuity which uses neighborhoods of epsilon around f(xo) as well.
*0:00** Good place to start*
12:38
12:00 “Because 0 is not equal to 1” Proof ? 😛
If 0=1 then Ø={Ø}, then Ø has an element (Ø), which is a contradiction :-)
R is a field. By definition, the additive and multiplicative identities of a field are distinct.
Somewhat less flippantly, the Peano axioms suffice, since 0 is not the successor of any natural number, while 1 is the successor of 0.
@@j9dz2sf "If 0=1 then Ø={Ø}"
That's even less convincing than just "0 ≠ 1"
Jokes aside, if you really wanna prove that 0≠1, then yes, that's probably the way to go, since I think one of the most fundamental definitions of numbers are in terms of elements in sets. Correct me if I'm wrong.
@@DeJay7 Yes, in Set Theory ZF, everything is a set: 0 is defined as Ø, 1 is defined as {Ø}, 2 is defined as {Ø,{Ø}} and so on en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers#Definition_as_von_Neumann_ordinals .
There is actually another way of proving continuity of x^2, i.e. that the limit of x^2 = a^2 as x goes to a.
|x^2 - a^2| = |x - a| * |x + a| = |x - a| * |x - a + 2a|
this is way better, thank you!
My Calculus Professor (Tony Tromba, UC Santa Cruz 1981) dropped the Dirichlet Function on us at the end of a Friday lecture to give something to discuss at Happy Hour.
This was amazingly accessible, thank you. Could you do a video explaining constructive logic and how to prove there? How to rationalize sequential continuity without L.E.M
in the first example you can take delta to be -a+sqrt(a^2+e) and than you have (x-a)(x+a) is less than a^2+e-a^2=e.
Here is a nice related problem: it can be shown that function f(x)=0, x is irrational, 1/n if x=m/n, gcd(m,n)=1 is continuous in all irrational points and not continuous in all rational. The question therefore is: is there a reverse function, not continuous in all irrational points and continuous in all rational?
I've found very interesting and brilliant equation that I can't solve. The command for the task: Solve cos(cos(cos(cos(x)))) = sin(sin(sin(sin(x)))). x is a real number. This problem comes from Russian Math Olympiad, 95. Michael, I believe you can overcome this task :)
here is an interesting fact cosx function iterated even times and sinx funtion iteterated for same number of times are not equal for any real x
@@angelmendez-rivera351 what can you say about general case?
Baby Penn be like : *High pitched sound*
Why you dont choose sqrt(-epsilon + a^2)
One of the questions we had to answer was: Show that the following function is continuous at 0.
f(x)=x for x in Q or f(x)=0 otherwise.
I am working on this problem now. At x=0, using his proof with our definition of f(x)=x, our f(xo)=f(a) at zero so it is continuous at that point only. I believe this works for the definition of continuity which uses neighborhoods of epsilon around f(xo) as well.
Can someone point me to the "last video" referred to in the intro? I'm on the real analysis play list and I can't find it anywhere.
i can't either :(
Thanks a lot for this marvelous video!!! The second example was the one what I was looking for
Very cool, thanks for the quality in all the aspects!!!
2:08 f(x)=x^2 is continous all all [sic!]
when i saw the function that u gonna present in the video i was genuinely amazed
I already knew the function, but yeah, it´s quite clever!
Excelente video , gracias por hacerlo.
Could you do a video about uniform continuity? Thanks :)
That is upcoming, maybe in another week I will be at that section.
very helpful.
very awesome can u make video on dirichlet and Thomas fun pls reply
If 'n'th is an odd possitive integer, prove that coefficients of the middle terms in the expansion of (x+y)^n are equal......
Should be easy using pascals triangle (and that it's symmetric)
Trivial using binomial coefficients and symmetry
I beg , please someone tell me , what is real analysis?
It's really just centered around mastering the art of proving things.
@VeryEvilPettingZoo some people still doesn't know about google ...
them bicesp
Yes, this is definitely not for me.