Why are there so many theorems that are completely obvious/common sense? IVT/EVT are the first ones that come to mind. Do there really need to be theorems for those things? They just state the obvious. The logic behind them just can't be refuted. So what purpose do they serve?
yes there does
>>8093289
yes
>muh rigor
That's the only reason, OP. """"Higher level"""""" math is just mental S&M
>>8093289
>So what purpose do they serve?
There have been conjectures that were true for millions of values until someone found a counterexample. Look at Eulers conjecture for the sum powers, it took until 1966 to find a counterexample. The point is you can't take for granted the idea that "The logic behind them just can't be refuted" you still need a formal proof.
>>8093289
I believe it is the theorem of 2=2 that is the key to all off the great questions of science. I'm currently writing my doctoral thesis on it.
>>8093422
go fuck yourself you fucking pathetic sack of shit.
if you turned your brain on for a fraction of a second
>>8093503
Fucking this. Also well put, anon.
>>8093289
You mean that they're theorems instead of axioms?
Well, they can be proven from definitions of continuous functions of reals.
And their generalizations lead to the definitions of compactness and connectedness iirc.
They just happen to be important properties.
>>8093289
>Why are there so many theorems that are completely obvious/common sense? IVT/EVT are the first ones that come to mind.
Sometimes the important thing about these theorems is to understand what assumptions need to be made. For example if you are working over the field of rationals instead of the field of reals then the IVT need not be true even for continuous functions. For example
[math]f(x) = x^2[/math] is continuous on [math][1, 2] \subset \mathbb{Q}[/math] but [math]f[/math] never takes on the value 2 between [math]f(1) = 1[/math] and [math]f(2) = 4[/math].
The assumption that is missing here of course is that [math]\mathbb{Q}[/math] is not complete in the Banach space sense.
>>8093824
Similarly for the EVT, when working in [math]\mathbb{Q}[/math] the function [math]f(x) = x - x^3[/math] is continuous on [math][0, 1] \subset \mathbb{Q}[/math] but has no maximum in [math][0,1] \subset \mathbb{Q}[/math].
>implying IVT is even true.
A lot of non-constructive babies here
>>8094939
came here to say this.
https://en.wikipedia.org/wiki/Constructive_analysis#The_intermediate_value_theorem
>>8093289
Jordan curve theorem.
