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
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 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 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
Thanks a lot for this marvelous video!!! The second example was the one what I was looking for
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?
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.
Baby Penn be like : *High pitched sound*
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?
Very cool, thanks for the quality in all the aspects!!!
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.
2:08 f(x)=x^2 is continous all all [sic!]
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.
Why you dont choose sqrt(-epsilon + a^2)
very helpful.
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 :(
Could you do a video about uniform continuity? Thanks :)
That is upcoming, maybe in another week I will be at that section.
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 ...
Yes, this is definitely not for me.