Results 1 to 5 of 5
Like Tree3Thanks
  • 1 Post By SlipEternal
  • 1 Post By romsek
  • 1 Post By SlipEternal

Math Help - How to prove that a function is one-to-one?

  1. #1
    Member
    Joined
    Nov 2011
    Posts
    86

    How to prove that a function is one-to-one?

    Hi everyone,

    I was wondering, how can I show mathematically that f(x) = x2, defined for x <= 0 is one-to-one?

    Can I do it like this:

    If f(x) is one to one, then f(x1) = f(x2) if x1 = x2

    so

    (x1)2 = (x2)2
    +-(x1) = +-(x2) but the domain is only defined for negative numbers so
    -(x1) = -(x2)
    therefore x1 = x2

    or is this completely off?

    (Sorry for the sloppy notation, I just thought it would be ok in this case, since the calculations are fairly short and simple)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,353
    Thanks
    498

    Re: How to prove that a function is one-to-one?

    The solution is correct, but I agree that the notation could be improved. I would write:

    Suppose f(x_1) = f(x_2). Then x_1^2 = x_2^2. Taking the square root of both sides, we find x_1 = \pm x_2. If x_2<0, then -x_2 is not in the domain, so x_1 = x_2. If x_2 = 0, then x_2 = -x_2.

    The difference is \pm x_1 = \pm x_2 implies either x_1 = x_2 or -x_1 = -x_2, both of which mean x_1 = x_2. But, that is not actually what is meant. The notation I used is x_1 = \pm x_2 which is more general, as now it means x_1 = x_2 or x_1 = -x_2.
    Last edited by SlipEternal; November 29th 2013 at 02:03 AM.
    Thanks from Nora314
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    1,507
    Thanks
    545

    Re: How to prove that a function is one-to-one?

    Quote Originally Posted by Nora314 View Post
    Hi everyone,

    I was wondering, how can I show mathematically that f(x) = x2, defined for x <= 0 is one-to-one?

    Can I do it like this:

    If f(x) is one to one, then f(x1) = f(x2) if x1 = x2

    so

    (x1)2 = (x2)2
    +-(x1) = +-(x2) but the domain is only defined for negative numbers so
    -(x1) = -(x2)
    therefore x1 = x2

    or is this completely off?

    (Sorry for the sloppy notation, I just thought it would be ok in this case, since the calculations are fairly short and simple)
    to show f is injective what you need to show is that if {f(x1) == f(x2)} then {x1 == x2}

    if {x1 == x2} the fact that {f(x1) == f(x2)} is trivial since the function is single valued.
    Thanks from Nora314
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Nov 2010
    Posts
    1,353
    Thanks
    498

    Re: How to prove that a function is one-to-one?

    Quote Originally Posted by romsek View Post
    to show f is injective what you need to show is that if {f(x1) == f(x2)} then {x1 == x2}

    if {x1 == x2} the fact that {f(x1) == f(x2)} is trivial since the function is single valued.
    The OP meant f(x_1) = f(x_2) only if x_1 = x_2. That was obvious from the method of proof used.
    Thanks from Nora314
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Nov 2013
    From
    California
    Posts
    1,507
    Thanks
    545

    Re: How to prove that a function is one-to-one?

    yep, I didn't get that far once I saw the basic error I commented on.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. How to prove x^2 is a function?
    Posted in the Pre-Calculus Forum
    Replies: 12
    Last Post: October 8th 2011, 03:18 AM
  2. Prove a function is onto
    Posted in the Discrete Math Forum
    Replies: 29
    Last Post: April 30th 2010, 11:46 AM
  3. Prove that f(x) is an odd function?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 21st 2009, 02:09 PM
  4. Prove a function is onto
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: April 12th 2009, 02:44 PM
  5. odd function prove
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 21st 2009, 04:24 PM

Search Tags


/mathhelpforum @mathhelpforum