Results 1 to 10 of 10

Math Help - Showing a function is one-to-one or onto.

  1. #1
    Member
    Joined
    Mar 2010
    Posts
    75

    Showing a function is one-to-one or onto.

    Can someone help me understand how to do this problem?

    Let F:N -> {0,1} be defined by F(n)= 0 if 3|n or 1 if 3|n

    a)Show F is NOT one-to-one.
    b)Show F is onto.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by matthayzon89 View Post
    Can someone help me understand how to do this problem?

    Let F:N -> {0,1} be defined by F(n)= 0 if 3|n or 1 if 3|n

    a)Show F is NOT one-to-one.
    b)Show F is onto.

    Look carefully at what you wrote after "..be defined by..." : it makes no sense.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2010
    Posts
    75
    Quote Originally Posted by tonio View Post
    Look carefully at what you wrote after "..be defined by..." : it makes no sense.

    Tonio
    Why does it make no sense? it does make sense....

    read it like this:


    Let the funtion that MAPS Natural numbers to 0,1 be defined by F(n)= 0 if 3|n or 1 if 3|n
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,394
    Thanks
    1478
    Awards
    1
    Quote Originally Posted by matthayzon89 View Post
    Why does it make no sense? it does make sense....
    Let the funtion that MAPS Natural numbers to 0,1 be defined by F(n)= 0 if 3|n or 1 if 3|n
    It still makes no sense.
    0 if 3|n or 1 if 3|n.
    That reads 0 if three divides n or 1 if three divides n.
    That is nonsense.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Jan 2009
    Posts
    145
    Sorry, but what are you using | to mean?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Mar 2010
    Posts
    75
    Quote Originally Posted by Plato View Post
    It still makes no sense.
    0 if 3|n or 1 if 3|n.
    That reads 0 if three divides n or 1 if three divides n.
    That is nonsense.

    SORRY!!
    This is what it was suppose to say, I found out that there was a TYPO on the assignment itself...

    It is suppose to read 0 if 3(does NOT)|n or 1 if 3|n
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,394
    Thanks
    1478
    Awards
    1
    Quote Originally Posted by matthayzon89 View Post
    SORRY!!
    It is suppose to read 0 if 3(does NOT)|n or 1 if 3|n
    Well then:
    a) if F(1)=F(2) it is not one-to-one.

    b) F(1)=0~\&~F(3)=0. So...?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Mar 2010
    Posts
    75
    Quote Originally Posted by Plato View Post
    Well then:
    a) if F(1)=F(2) it is not one-to-one.

    b) F(1)=0~\&~F(3)=0. So...?
    So b) is ONTO.

    p.S
    I need help trying to understand the question and knowing what to look for.... I don't really need help solving it. Hints don't help much either.
    Can someone please explain how to approach such a problem and what I need to look for in order to prove something is onto or one-to-one?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,394
    Thanks
    1478
    Awards
    1
    Quote Originally Posted by matthayzon89 View Post
    I need help trying to understand the question and knowing what to look for.... I don't really need help solving it. Hints don't help much either.
    Can someone please explain how to approach such a problem and what I need to look for in order to prove something is onto or one-to-one?
    Any function is a set of ordered pairs.
    A function is one-to-one is no two pairs have the same second term.
    A function is onto if the range is the same as the co-domain (the final set).
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Mar 2010
    Posts
    75
    So I know the answer b/c I went over it with my prof. but I still dont really understand it, can someone please explain?

    Here it is:

    F:N -> {0,1}
    F(n)= {0 if 3 does not |n, 1 if 3|n)

    a) is NOT one-to-one <- I know this is due to having more then one value mapped to the same second value, but I dont understand how this conclusion is figured out given the information of the problem.

    Proof:
    Consider n(base1)= 1 and n(base2)= 2
    Then F(n(base1)=F(n(base2))=0 but n(base1)DOES NOT EQUAL 2.



    Also,
    Is F onto? YES

    Proof:
    Let y E {0,1}
    if y=0,choose n=1 then, n E N and F(n)=y since F(1)=0

    If y=1, choose n-3 then, n E N and F(n)=y since F(3)=1.

    Can someone please explain how this proof was reached and the logic behind whats going on in detail so i understand it once and for all?

    Thank You in advance,
    Matt H
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: October 19th 2011, 04:49 AM
  2. Showing that a function is odd
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: March 31st 2011, 11:19 PM
  3. Showing a function divereges at C
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 2nd 2008, 03:41 PM
  4. showing function differentiable
    Posted in the Calculus Forum
    Replies: 8
    Last Post: February 1st 2008, 07:49 AM
  5. Showing a function does not have an inverse
    Posted in the Pre-Calculus Forum
    Replies: 10
    Last Post: January 28th 2008, 05:06 PM

Search Tags


/mathhelpforum @mathhelpforum