Lmao. I saw the video a long time ago and I came across it again now. I forgot about the question about the impossibility of the existence of such function, but then I don't know why, I had a lapsus and tried to prove that "there is no function f:R->R satisfying f(t)-f(t-1)=t". I just spent some hours on it, even arrived at some nice formulas for f such as f(n)=f(-n-1) and f(n)-f(-n)=n for all natural n, but I couldn't find any contradiction. It is now 4:30 AM and I decided to look at the solution, but I was really reluctant to do so. Lately I am struggling with mathematics so being unable to prove such a simple-looking fact was really frustrating. Now I am glad that I wasted my time instead of being utterly bad at it xD
@Panure5 жыл бұрын
This deserves more views 👀👀
@DreamzAnimation5 жыл бұрын
This seems more like proof by counterexample - let t = 1/2.
@XanderGouws5 жыл бұрын
That's an interesting way of going about it. I hadn't thought of that :P
@SoppingWetDog5 жыл бұрын
cant wqit fore the next vid!!! i know ill love it :) .
@lordlix64834 жыл бұрын
Well, what if it was a function with {1/2} as domain?
@XanderGouws4 жыл бұрын
Then we could substitute in t=1/2 f(1/2) - f(1 - 1/2) = 1/2 f(1/2) - f(1/2) = 0 = 1/2 So we still have a contradiction.
@cubicardi80115 жыл бұрын
This is so cool!!!
@PackSciences5 жыл бұрын
Set theory, yes, please.
@lambda26932 жыл бұрын
thats is not a good enough proof. doing functional analysis you have to take into account for the domains and range, you have not put up condition. if you said f is continuous over all R then your statement would be considered true, but for example if the function blows up to infinite at the specified point and is obviously not continuous and differentiable then your proof would be wrong. All in all nice video but you took a pretty general case and did not specify the generality