Question on continuous function.

[MathematicalPhysicist]

let f(x) be continuous for 0<=f(x)<=1. suppose that f(x) assumes rational vlaues only and that f(x)=1/2 when x=1/2. prove that f(x)=1/2 everywhere.

im having trouble with this simple question, here's what i did:
|x-1/2|<d implies |f(x)-1/2|<e. f(x) assumes only rational values so
( I am guessing that by 'assumes' it's input is rational values, so
|x-x0|<d |f(x)-f(x0)|<e f(x0)=p/q
if p/q>1/2 then |f(x)-p/q|<|f(x)-1/2|<=e
and if p/q<1/2 |f(x)-1/2|<|f(x)-p/q|<e, and thus the limit of f(x) at x0 always converges to 1/2.
is this right?

 

[Triss]

If you know a little topology, consider the following:
If $f:X\to Y$ is continuous and X is a connected set, then f(X) is also connected.

You have $f:[0,1]\to\mathbb{Q}$. What are the connected sets in $\mathbb{Q}$?

 

[Hurkyl]

What about the intermediate value theorem?

 

[VietDao29]

( I am guessing that by 'assumes' it's input is rational values...

No, it's output is rational, not input.
Say if there exists some $$x_0 \neq \frac{1}{2}$$, such that: $$f(x_0) \neq f \left( \frac{1}{2} \right) = \frac{1}{2}$$, then is there any value x₁ in the interval $$\left] \min \left(x_0, \frac{1}{2} \right) , \ \max \left( x_0 , \frac{1}{2} \right) \right[$$, such that f(x₁) is irrational?

 

[MathematicalPhysicist]

What about the intermediate value theorem?

i also thout about it.
but it states:"If f is continuous on a closed interval [a,b] , and c is any number between f(a) and f(b) inclusive, then there is at least one number x in the closed interval such that f(x)=c." (mathworld)
it doesn't say that for every x f(x)=c (in this case c=1/2).
what it does states (from the problem) is that for every x f(x)'s output ( as viet dao corrected me, i didnt understand what assumes means here) is rational values, but how do i use the intermediate theorem here to prove that for every x f(x)=1/2.
i know that for every x f(x)=p/q, and by the theorem there's at least one x (x=1/2) such that f(x)=1/2 f(0)<=f(1/2)<=f(1), then how do i infer that for every x in [0,1] f(x)=1/2?

 

[HallsofIvy]

Oh, come on! Use proof by contradiction. Suppose f(x) is not 1/2 for all x. Then there exist at least two distinct values of f: y₁ and y₂. Now what does the intermediate value theorem tell you?

 

[MathematicalPhysicist]

it tells us that there's at least one t such that f(t)=y1 and at least one t' such that f(t')=y2 where they are between f(a) and f(b), but i don't see how does this contradict the assumption that they are distinct?

 

[MathematicalPhysicist]

wait a minute i just reread the theorem, y1 is any number between f(a) and f(b) and so is y2, so they could also be equal.

thanks halls.

 

[VietDao29]

wait a minute i just reread the theorem, y1 is any number between f(a) and f(b) and so is y2, so they could also be equal.

thanks halls.

No, what's a, and b? You haven't even defined them!
You still seem not to understand the theorem correctly. Roughly speaking, the theorem is that if the function f is continuous, and returns 2 different values y₁, and y₂, then it can return any value between y₁ and y₂.
Say if g(x) = sin(x), g(x) is continuous, right? g(0) = 0, and $$g \left( \frac{\pi}{2} \right) = 1$$, then there exists some x value between 0 and $$\frac{\pi}{2}$$, such that $$g(x) = \frac{\sqrt{3}}{2}$$, ($$\frac{\sqrt{3}}{2}$$ is between g(0), and $$g \left( \frac{\pi}{2} \right)$$). That value is $$x = \frac{\pi}{3}$$.
Can you get this?
Now read the theorem, and all the posts above to see if you can finish the problem. :)

 

[MathematicalPhysicist]

let me know if this is any good.

suppose f(x) is not 1/2 everywhere (so only once f(x) gets the value 1/2), then there are y1,y2 which are values of f(x) which arent equal to each other.
because f is conitnuous on an interval [0,1] there's a value 1/2 between f(x1) and f(x2) (f(x1) and f(x2) are rationals), when x1<x2<1/2 and thus there's at least one x such that f(x)=1/2 where x!=1/2.

i don't think this proof is correct cause i haven't shown why 1/2 must be between f(x1) and f(x2), i know that they are rationals but it doesn't say that those are rationals between 0 and 1, the values of the domain are between 0 and 1.

 

[VietDao29]

suppose f(x) is not 1/2 everywhere

This is fine. :)

(so only once f(x) gets the value 1/2)

It can get the value 1 / 2 twice, or three times, or blah blah blah, whatever. You just need to assume that there's one x₀, such that: $$f(x_0) \neq \frac{1}{2}$$. Can you get this? :)

, then there are y1,y2 which are values of f(x) which arent equal to each other.
because f is conitnuous on an interval [0,1] there's a value 1/2 between f(x1) and f(x2) (f(x1) and f(x2) are rationals), when x1<x2<1/2 and thus there's at least one x such that f(x)=1/2 where x!=1/2.

No, this part is not good. What are your x₁, and x₂? You haven't defined them.
Ok, so we have that:
$$f(x_0) \neq \frac{1}{2}$$, and
$$f \left( \frac{1}{2} \right) = \frac{1}{2}$$, right?
Now, is there any irrational number between the 2 distinct rational numbers f(x₀), and f(1/2)? There is, right? Do you remember this thread (I almost gave up on finding this thread, I thought it was in Calculus & Beyond! ). Let call that rational number a. Ok?
According to the Intermediate Value theorem, is there any x₁ between x₀, and 1 / 2, such that f(x₁) = a?
Now, can you go from here? Can you follow me? :)

 

[HallsofIvy]

let me know if this is any good.

suppose f(x) is not 1/2 everywhere (so only once f(x) gets the value 1/2), then there are y1,y2 which are values of f(x) which arent equal to each other.
because f is conitnuous on an interval [0,1] there's a value 1/2 between f(x1) and f(x2) (f(x1) and f(x2) are rationals), when x1<x2<1/2 and thus there's at least one x such that f(x)=1/2 where x!=1/2.

i don't think this proof is correct cause i haven't shown why 1/2 must be between f(x1) and f(x2), i know that they are rationals but it doesn't say that those are rationals between 0 and 1, the values of the domain are between 0 and 1.

You are completely misunderstanding the "intermediate value property"!

it tells us that there's at least one t such that f(t)=y1 and at least one t' such that f(t')=y2 where they are between f(a) and f(b), but i don't see how does this contradict the assumption that they are distinct?

No, it doesn't say anything like that! (especially since you haven't said what y1 and y2 are.) The intermediate value property says that if y is any value between f(a) and f(b), then there exist x between a and b such that f(x)= y. That is, f takes on all values between f(a) and f(b).

You are told that f(1/2)= 1/2. If f(x) is not identically 1/2, then there exist some x₁ such that y₁= f(x₁) is not 1/2. By the intermediate value property, if y is any value between 1/2 and y₁, there exist x such that f(x)= y. Now, can you think of a value between 1/2 and y₁ that will give a contradiction?

 

[MathematicalPhysicist]

i think i can follow, bacuase by the intermediate theorem, f assumes any value between 1/2 and f(x0) and thus also assumes an irrational number, but by the definition of the problem it only assumes rational numbers.

thanks for the help.

iv'e got another couple of questions:
1)prove that if f(x) is monotonic in [a,b] and satisfies the intermediate value property , then f(x) is continuous.
what i did is as follows:
assume f is monotonic increasing, x>y f(x)>f(y), and that between f(y) and f(x) the function receives the value e which correspond to the value x0 between y and x, i.e f(x0)=e, let's assume |x-y|<d, then f(x)>e>f(y)
e-f(x)>f(y)-f(x) i.e |f(x)-f(y)|<e-f(x)<e and thus the function is uniform continuous and thus also point conitinouos. is this right?

2)let f(x) be a polynomial f(x)=anx^n+...+a1x+a0
show that if n is odd, then f(x) has at least one real root, if an and a0 have opposite signs, then f(x) has at least one posirive root, and if n is even different than 0, then f(x) has a negative root as well.

for the first part here's what i did:
$$a_{2m-1}x^{2m-1}+...+a1x+a0=0$$
if we multiply by $$x=/=0$$ then $$a_{2m-1}x^{2m}+...+a_{1}x^2+a0x=0$$
if i substract both of the equations i get, $$a_{2m-1}x^{2m}+(a_{2m-2}-a_{2m-1})x^{2m-1}+...+(a_{1}-a_{2})x^2+(a_{0}-a_{1})x-a0=0$$ this equation is equivalent to the other equations (cause it's addition of one multiple of equations to another.
and this set of equations has a solution for x=1 and thus also the first equation has at least one real solution.

im not sure at all if this is correct.
i appreciate your help.

btw, I am quite sure that if in the first question f satisfies the intermediate property then it's already assumed to be conitnuous in the interval, so perhaps this part of the property isn't assumed here.

 

[Office_Shredder]

assume f is monotonic increasing, x>y f(x)>f(y), and that between f(y) and f(x) the function receives the value e which correspond to the value x0 between y and x, i.e f(x0)=e, let's assume |x-y|<d, then f(x)>e>f(y)
e-f(x)>f(y)-f(x) i.e |f(x)-f(y)|<e-f(x)<e and thus the function is uniform continuous and thus also point conitinouos. is this right?

e-f(x) < 0, so how can it be greater than the absolute value of something?

I don't really follow your logic here (ignoring the absolute value error), or see where you're trying to go... why can't the function be discontinuous at some point a > y, or < x? Or somewhere between x and e?

I would try arguing by contradiction... if the function WASN'T continuous at a point, what does that mean about the points around it?

 

[StatusX]

for the first part here's what i did:
$$a_{2m-1}x^{2m-1}+...+a1x+a0=0$$
if we multiply by $$x=/=0$$ then $$a_{2m-1}x^{2m}+...+a_{1}x^2+a0x=0$$
if i substract both of the equations i get, $$a_{2m-1}x^{2m}+(a_{2m-2}-a_{2m-1})x^{2m-1}+...+(a_{1}-a_{2})x^2+(a_{0}-a_{1})x-a0=0$$ this equation is equivalent to the other equations (cause it's addition of one multiple of equations to another.
and this set of equations has a solution for x=1 and thus also the first equation has at least one real solution.

In effect, you've just multiplied the polynomial by (x-1), and claimed it must have a real root because this new polynomial has the real root 1. Try to see exactly where your argument breaks down.


One way to do this question is, once again, using the intermediate value theorem. A polynomial is dominated at large x by the largest power of x, and in the case of an odd degree polynomial, this term will have different signs depending on the sign of x (for example -x^3+3x+4 is negative for large positive x and positive for large negative x). Can you figure out what to do from here?

Alternatively, you could use the property that all complex roots of a polynomial with real coefficients come in complex conjugate pairs, so when there are an odd number of roots, there is at least one root which has to be its own complex conjugate (although this argument needs to be made more formal)

 

[MathematicalPhysicist]

if the function wasnt conitnuous then there exists e>0 and d>0 such that |x-x0|<d |f(x)-f(x0)|>=e it is monotone, then suppose it's monotonic increasing then for x>y f(x)>f(y), by the intermediate property, any value value between f(x) and f(y), say c, such that x0 between x and y, and f(x0)=c f(x)>f(x0)>f(y) f(x)-f(x0)>=e, if we let e=f(x)-f(y) then f(x)-f(x0)>=f(x)-f(y) which is not possible.

i don't think this proof is correct cause in the first place i mentioned there exists an e>0, so it could be of any postive value not necessarily the value I've given.

 

[MathematicalPhysicist]

when i think about again, perhaps i do have something, as before we assume that f isn't continuous then |f(x)-f(x0)|>=e for |x-x0|<d with the intermediate value property we have that for every x'>x0 in the interval [a,b] and any number c between f(x')>f(x0) tere exists x1 which is between x0 and x' such that f(x1)=c.
if we employ this property over again with x1 and x0 and then getting x2 etc, we are getting closer and closer to x0, and thus the difference between the functions at x0 and xi i=1,...,n approaches zero, which is contradiction to f not being continuous.

is this a good idea?

 

[VietDao29]

You are making it a mess. I doubt that you can really understand what you wrote. Mathematics is not a collection of, may I say, meaningless symbols, and signs. Be sure to express your ideas correctly so that others can understand. If you cannot understand what you wrote, then you cannot suppose us to be able to interprete them.

if the function wasnt conitnuous then there exists e>0 and d>0 such that |x-x0|<d |f(x)-f(x0)|>=e

There's no such definition for discontinuity! What if f(x₀) is undefined?

it is monotone, then suppose it's monotonic increasing then for x>y f(x)>f(y), by the intermediate property, any value value between f(x) and f(y), say c, such that x0 between x and y, and f(x0)=c f(x)>f(x0)>f(y) f(x)-f(x0)>=e, if we let e=f(x)-f(y) then f(x)-f(x0)>=f(x)-f(y) which is not possible.

Err, what?
---------------------
Ok, let have a look back at Office_Shredder's hint:

I would try arguing by contradiction... if the function WASN'T continuous at a point, what does that mean about the points around it?

Monotone function means that it can either be an increasing, or decreasing function. So, say if there's a discontinuity at x₀. f(x₀) is either missing (or undefined), or there'll be a jump there. So what happens, then?

 

[MathematicalPhysicist]

i beg to differ i know that if a function is not continouous in a particular point then there exists e>0 and d>0 such that for |x-x0|<d |f(x)-f(x0)|>=e, this is the negation of continuity.

 

[MathematicalPhysicist]

does this negation only applies to uniform continuity?

btw, i understand that if at point x0 it is discontinuous then the intermediate value doesn't apply cause f(x0) is not defined or missing as you put it, but still I am quite sure about what i wrote that if f isn't continuous at x0 then |f(x)-f(x0)|>=e, isn't this the negation of continuity?

 

[nocturnal]

i beg to differ i know that if a function is not continouous in a particular point then there exists e>0 and d>0 such that for |x-x0|<d |f(x)-f(x0)|>=e, this is the negation of continuity.

This isn't quite the negation of continuity. A function f is discontinuous at a point x0 in dom(f) if there exists an e>0 s.t. for all d>0 there is an x s.t. |x-x0| < d and |f(x) - (fx0)| >= e.

Do you see the difference? Notice that when you negate a statement containing quantifiers the "for alls" become "there exists" and vice versa. You should understand why this works.

For this problem it is fine to assume WLOG that f is an increasing function, because if f is decreasing then -f is increasing. We can conclude that f is continuous if -f is continuous since f = (-1)(-f). Recall the theorem that says if a function g is continuous then so is cg where c is a constant.

The theorem can be proven directly (no contradiction necessary). Note that f([a,b]) is an interval, call it I. You can show this function is continuous at x0 for three cases.
1) x0 is in (a,b)
2) x0 = a
3) x0 = b (this proof is very similar to case 2)
hint for case 1: there exists an E>0 s.t. (f(x0)-E, f(x0)+E) is a subset of I. Let e>0 and assume e < E. Show there exists a d>0 s.t. |x-x0|<d implies
|f(x) - f(x0)|<e.

 

[MathematicalPhysicist]

i tried to appraoch this probelm directly and indirectly but so far instead of helping me you got me confused even more.

i know that AxFx=~Ex~Fx where E and A are the quantifiers and F is a predicate so i know that i needed to rephrase it as you stated.

anyway, one user tell me my definition is incorrect and another one just reinforces me.

anyway, for the first case x0 is in (x0-E,x0+E) such that f(x0) is in (f(x0-E),f(x0+E)) for any x in (x0-E,x0+E) x0-E<x<x0+E x-x0<E=d
but how do i deduce that |f(x)-f(x0)|<e?

 

[nocturnal]

anyway, for the first case x0 is in (x0-E,x0+E) such that f(x0) is in (f(x0-E),f(x0+E)) for any x in (x0-E,x0+E) x0-E<x<x0+E x-x0<E=d
but how do i deduce that |f(x)-f(x0)|<e?

if e<E then $$f(x_0) - e, f(x_0) + e \in (f(x_0) - E, f(x_0) +E) \subseteq I$$
Thus there exists $x_1, x_2$ s.t. $$f(x_1) = f(x_0) - e$$ and $f(x_2) = f(x_0) + e$. Note that $x_0 \in (x_1, x_2)$. The choice of d should be apparent.

 

[MathematicalPhysicist]

what happens is that f(x)<f(x0)+e.

but if this is such easy why no one suggested this way before you?

 

[nocturnal]

I never said this was easy. There were a couple of mistakes in my last two posts (major ones). I have fixed them now. I had (x0-E, x0+E) but it should have been (f(x0)-E, f(x0)+E). Sorry for the confusion.

 

[matt grime]

Looking at the polynomial question, the poly x-i is odd degree and has no real solutions. Moral of the story: write all of the conditions of the question out properly. Note, for instance, that as stated you original question omits some very important information about the domain of the function f. It is prefectly possible to have continuous functions that take only rational values and are not constant if you choose the domain properly.