Math Is Fun Forum

xsw001

Member

Registered: 2010-10-23

Posts: 7

prove that f(a) < f(c) < f(b)

Let f:[a,b]->R be continuous and one-to-one such that f(a)<f(b).
Let a<c<b.  Prove that f(a)<f(c)<f(b)

My first instinct is to apply intermediate value theorem. Let me know whether my proof makes sense or not.

Proof:
Since f: f:[a,b]->R is continuous and one-to-one.
Therefore f is strictly increasing function.
Suppose a<c<b
According to Intermediate Value Theorem
There exists f(c) such that f(a)<f(c)<f(b)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: prove that f(a) < f(c) < f(b)

Hi xsw001,

If you're happy that you've proved it's an increasing function then for all x in the range f(a) < f(x) and f(x) < f(b) because that's what increasing means  so put x = c and it's done.

Or assume, to the contary that a<c<b but f(b) < f(c)

Consider the interval [c,b].  f(c) < f(b) makes a contradiction.  Similarly, if f(c) < f(a).

Does that make sense?  I hate these 'proofs' where common sense tells you it's obvious.  I feel like the proof is just so many grains of sand that are slipping through my fingers.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

xsw001

Member

Registered: 2010-10-23

Posts: 7

Re: prove that f(a) < f(c) < f(b)

I know, I agree with you Bob.  It is so obvious for any given continuous injective function, it is strictly monotone, either strictly increasing or decreasing, with the given domain [a,b], then it has to be strictly increasing instead of strictly decreasing, or else the domain would have stated [b,a], isn't it?

Here is the sketch of the proof.
Assume by contradiction that a<c<b, and case 1) f(a)<f(b)<f(c) or case 2) f(c)<f(a)<f(b)
Since the function is one-to-one, therefore the graph of the continuous function can't oscillate, so it is strictly monotone, either strictly increasing or decresing.
case 1) if a<c<b but f(a)<f(b)<f(c)
then f(a)<f(b)<f(c) => a<b<c if it is strictly increasing which is a contradiction.
also f(a)<f(b)<f(c) => c<b<a if it is striclty decreasing which is also a contradiction.
case 2) if if a<c<b but f(c)<f(a)<f(b)
then f(c)<f(a)<f(b) => c<a<b if it is strictly increasing which is a contradiction.
also f(c)<f(a)<f(b) => b<a<c if it is striclty decreasing which is also a contradiction.
Hence if a<c<b, then f(a)<f(c)<f(b), and the function is strictly increasing.

Is it about right?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: prove that f(a) < f(c) < f(b)

hi xsw001

Looks good to me but , I confess, it's years since I did this sort of thing.  Was it you who posted a request to find functions with given domains and ranges?  I put one idea back but haven't thought about the other two since.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob