1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Show that a multivariate function is bijective

  1. Dec 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that ##f: \mathbb{Z}^{+} \times \mathbb{Z}^{+} \longrightarrow \mathbb{Z}^{+} ##where ##\displaystyle f(m,n) = \frac{(m+n-2)(m+n-1)}{2}+m## is bijective

    2. Relevant equations


    3. The attempt at a solution
    First, we show that ##f(a,b) = f(c,d) \implies a=c \land b=d##.

    ##f(a,b) = f(c,d)##

    ##\displaystyle \frac{(a+b-2)(a+b-1)}{2}+a = \frac{(c+d-2)(c+d-1)}{2}+c##

    After simplification, we find that

    ##(a+b)^2 + a - b = (c+d)^2 + c - d##

    Comparing sides, then ##a + b = c+d, ~a-b = c-d##
    Then ##a=c,~b=d##

    Second, we show that for all positive integers a, there exist an m and an n such that f(m,n) = a.
    We see that ##f(m,2-m) = m##, so the function maps to every element in the codomain

    Does this show that the function is bijective?
     
    Last edited: Dec 3, 2016
  2. jcsd
  3. Dec 4, 2016 #2
    For the second part 2-m might not necessarily be a positive integer which seems to be a problem since the function is defined for pairs of positive integers.

    For first part this conclusion
    doesn't necessarily follow from the previous line. It might be correct but needs some additional justification me thinks.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Show that a multivariate function is bijective
  1. Bijective function (Replies: 2)

  2. Multivariable Function (Replies: 1)

Loading...