- #1
ArcanaNoir
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I'd like someone to check this proof out for violations of math law. It seems like hackery to me, but then so does a lot of what my professor says, so maybe it's not. If it's flawed, just tell me why. Please don't suggest other ways to prove this, that's my job this weekend.
Thanks :)
p.s. This is my first post using [itex] \LaTeX [/itex], since my discovery that it was there all along (used to see the raw text).
[itex] f: [0,1] \to [0,1] [/itex] and [itex]f[/itex] is continuous [itex] \Rightarrow \exists x \in [0,1] : f(x)=x [/itex]
Anything elementary calculus. Trying to keep it as simple as possible.
let [itex] g(x)=x [/itex]
Then [itex] g(0) \le f(x) \le g(1) [/itex]
or, [itex] g(a) \le f(x) \le g(b) [/itex] where [itex] [a,b]=[0,1] [/itex]
[tex] \lim_{a\to x^-}{g(a)} = \lim_{b\to x^+}{g(b)} = x [/tex]
therefore
[tex] \lim_{x\to x}{f(x)} = x [/tex]
Since [itex] f(x) [/itex] takes on the same value as [itex] g(x) [/itex] at some point [itex] x [/itex], [itex] f(x)=g(x)=x [/itex] at this point.
So, for some [itex] x \in [0,1], f(x)=x [/itex]
Thanks :)
p.s. This is my first post using [itex] \LaTeX [/itex], since my discovery that it was there all along (used to see the raw text).
Homework Statement
[itex] f: [0,1] \to [0,1] [/itex] and [itex]f[/itex] is continuous [itex] \Rightarrow \exists x \in [0,1] : f(x)=x [/itex]
Homework Equations
Anything elementary calculus. Trying to keep it as simple as possible.
The Attempt at a Solution
let [itex] g(x)=x [/itex]
Then [itex] g(0) \le f(x) \le g(1) [/itex]
or, [itex] g(a) \le f(x) \le g(b) [/itex] where [itex] [a,b]=[0,1] [/itex]
[tex] \lim_{a\to x^-}{g(a)} = \lim_{b\to x^+}{g(b)} = x [/tex]
therefore
[tex] \lim_{x\to x}{f(x)} = x [/tex]
Since [itex] f(x) [/itex] takes on the same value as [itex] g(x) [/itex] at some point [itex] x [/itex], [itex] f(x)=g(x)=x [/itex] at this point.
So, for some [itex] x \in [0,1], f(x)=x [/itex]
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