Proof by Contradiction: Why This Function is Impossible

In this video, we cover a very basic type of proof: proof by contradiction. I'm *very* excited about the next video, and I wanted to give myself as much time as possible to put it together!

It should be noted that for this problem 0 and 1 would be in the domain of f. Also, this problem can be solved more generally by setting t=x, and t=1-x, and adding the two equations to show that 0=1. However, for the purpose of illustration, I think using 0 and 1 works better.

Further 'reading':

[SoppingWetDog]

cant wqit fore the next vid!!! i know ill love it :)



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[DiegoMathemagician]

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

[DreamzAnimation]

This seems more like proof by counterexample - let t = 1/2.

[lordlix]

Well, what if it was a function with {1/2} as domain?

[lambda]

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