## swissgirl 2 years ago @KingGeorge

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1. swissgirl

2. swissgirl

Its my weekly question :)

3. swissgirl

If you dont like this one I can give you another one

4. KingGeorge

It's been a little while since I've done actual analysis :P

5. swissgirl

Ya so let me get you smth easier

6. KingGeorge

Well, the first part is pretty easy. It's basically just the definition of "not uniformly continuous"

7. swissgirl

BTW Congratulations. Where is ur offer from?

8. swissgirl

9. KingGeorge

U of Rochester. I'm still waiting to hear back from most schools I applied to. Unfortunately, I also just recently got a rejection from Princeton :(

10. swissgirl

awwwwww :( That is sad They are stupiidddddd

11. KingGeorge

Ah. This new one I can do. It's just the IVP (intermediate value theorem). To be honest, I wasn't expecting them to accept me.

12. KingGeorge

Let's see how much I remember. Last time proved this, we started with the assumption that $$f(a)f(b)\le 0$$. In this particular case, if $$f(a)=0$$ or $$f(b)=0$$, we're done. If $$f(a)f(b)<0$$, then one of $$f(a),f(b)$$ is negative, and the other positive. At this point, we had another theorem that proved there was then some $$c\in[a,b]$$ s.t. $$f(c)=0$$. Then define $$g(x)=f(x)+y$$. Then, from what we just proved, it immediately follows that $$g(c)=y$$. If we started with $$g(b)\le y\le g(a)$$, take $$f(x)=g(x)-y$$, so $$f(b)\le 0 \le g(a)$$. From this we get that $$f(b)f(a)\le 0$$ and we're done.

13. KingGeorge

However, I cannot immediately remember how we proved that if $$f(a)f(b)\le0$$, then there was some $$c\in[a,b]$$ s.t. $$f(c)=0$$.

14. swissgirl

hmmm gonna check my notes

15. swissgirl

hmmm I found a proof in my book using the supremum

16. KingGeorge

Well, I'm sorry, but I've got to go now. I need to get up early tomorrow, so I need to head to bed. Have a good night, and good luck with this!

17. swissgirl

Thanks @KG :)

18. swissgirl

Good night