Results 1 to 4 of 4

Thread: Intermediate Value Theorem

  1. #1
    Newbie
    Joined
    Mar 2009
    Posts
    24

    Intermediate Value Theorem

    Hi,

    I would like to know how to proof the following:
    Let f : (a,b) --> Z be continuous on (a,b) where Z is the integers.
    Show that f must be a constant function.

    I know we have to suppose f is nonconstant but next, how do we use the Intermediate Value Theorem to show the complete proof?
    May someone please help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    76
    Awards
    1

    Smile

    Quote Originally Posted by zxcv View Post
    I would like to know how to proof the following:
    Let f : (a,b) --> Z be continuous on (a,b) where Z is the integers.
    Show that f must be a constant function.
    We use proof by contradiction. Suppose that $\displaystyle f\colon(a,\,b)\to\mathbb{Z}$ is a nonconstant function that is continuous on its domain. Then there must exist $\displaystyle c$ and $\displaystyle d$ in $\displaystyle (a,\,b)$ such that $\displaystyle f(c)\neq f(d)$ (for, otherwise, $\displaystyle f$ would be constant). Choose any noninteger $\displaystyle k$ between $\displaystyle f(c)$ and $\displaystyle f(d).$ Then, since $\displaystyle f$ is a function of a real variable that is continuous on the interval $\displaystyle [c,\,d],$ we may apply the intermediate value theorem to conclude that there is some $\displaystyle e\in[c,\,d]$ such that $\displaystyle f(e)=k.$ But this would mean that the noninteger $\displaystyle k$ is in the codomain of $\displaystyle f,$ which is a contradiction. $\displaystyle \square$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2009
    Posts
    24
    Hi,

    I was just wondering if we can choose a non integer k between f(c) and f(d) since f is (a,b)-->Z and Z is the integers.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    76
    Awards
    1

    Smile

    Quote Originally Posted by zxcv View Post
    Hi,

    I was just wondering if we can choose a non integer k between f(c) and f(d) since f is (a,b)-->Z and Z is the integers.
    I am not sure I understand what you are asking. $\displaystyle f(c)$ and $\displaystyle f(d)$ must be distinct integers, and there are infinitely many real numbers between any two given integers. Pick one of them, and the intermediate value theorem says that $\displaystyle f$ must take on that value somewhere (which is impossible, since $\displaystyle f$ only gives integral values).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Intermediate Value Theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jan 26th 2011, 01:43 PM
  2. Intermediate Value Theorem
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Apr 18th 2009, 11:55 PM
  3. Intermediate Value Theorem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Apr 6th 2009, 08:49 AM
  4. Intermediate Value Theorem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Sep 22nd 2008, 12:33 PM
  5. intermediate value theorem/rolle's theorem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Dec 8th 2007, 01:55 PM

Search Tags


/mathhelpforum @mathhelpforum