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Homework Help: 1:1 and onto proof of Z+ and Q+

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data
    After exploring this problem a bit out of curiosity I decided to attempt to write a full fledged proof of it for my final exam.
    I will try my best to explain what I am doing. i am pretty frustrated with it but I've put so much time into it I don't want to give up and do something else now.

    For this theorem Equality in Q+: [itex]\frac{a}{b}[/itex]=[itex]\frac{c}{d}[/itex] [itex]\Leftrightarrow[/itex] a=c, and b=d

    now I'm using the grid system, i.e.
    1/1 1/2 1/3 1/4 1/5 1/6..
    2/1 2/2 2/3 2/4 2/5 2/6..
    3/1 3/2 3/3 3/4 3/5 3/6..
    but adjusting it to represent the diagonals and defining it as follows
    1/1 [itex]\Rightarrow[/itex] p+q=2 (1 element)
    1/2 2/1 [itex]\Rightarrow[/itex] p+q=3 (2 elements)
    1/3 2/2 3/1 [itex]\Rightarrow[/itex] p+q=4 (3 elements)
    as you can see each list represent one diagonal of the grid

    next I went to the integers by observing the following number line and looking at the sums that represent the integers

    0 1 2 3 4 5 6 7 8 9 10 11 12

    Representing each section with a summation we get

    now I'm trying to show f:Z[itex]^{+}[/itex][itex]\rightarrow[/itex] Q[itex]^{+}[/itex]
    f(1) = 1/1

    let n[itex]\geq[/itex]2[itex]\in[/itex]Z[itex]^{+}[/itex]
    let S[itex]_{n}[/itex]={s[itex]\in[/itex]Z[itex]^{+}[/itex][itex]\leq[/itex][itex]\sum^{j}_{k=1}[/itex]k}

    then consider 2n [itex]\sum^{2n}_{k=1}[/itex]k= 1+2+...+n+n+1+...+2n
    thus n[itex]\leq[/itex][itex]\sum^{2n}_{k=1}[/itex]k
    therefore 2n[itex]\in[/itex]S[itex]_{n}[/itex] therefore S[itex]_{n}[/itex][itex]\neq[/itex][itex]\oslash[/itex]
    Then by the well ordering principle [itex]\exists[/itex] m[itex]\in[/itex]S[itex]_{n}[/itex] [itex]\exists[/itex] m[itex]\leq[/itex]j [itex]\forall[/itex]j[itex]\in[/itex]S[itex]_{n}[/itex]

    i.e exists a "least" element
    Call the least element M

    or in other words m[itex]\in[/itex] ([itex]\sum^{m-1}_{k=1}[/itex]k, [itex]\sum^{m}_{k=1}[/itex]k]

    so let p= n-[itex]\sum^{m-1}_{k=1}[/itex]k

    then choose a q such that p+q = m+1
    q= m+1-p

    now f(n) = p/q
    That ends the mapping

    Now show f:Q[itex]^{+}[/itex][itex]\rightarrow[/itex]Z[itex]^{+}[/itex] is 1:1 and onto.
    This is where I can't finish the proof but this is what I have of an attempt on both

    One to one:
    x[itex]_{1}[/itex][itex]\neq[/itex]x[itex]_{2}[/itex] [itex]\Rightarrow[/itex] f(x[itex]_{1}[/itex])[itex]\neq[/itex]f(x[itex]_{2}[/itex])
    f(x[itex]_{1}[/itex])=f(x[itex]_{2}[/itex])[itex]\Rightarrow[/itex] x[itex]_{1}[/itex]=x[itex]_{2}[/itex]

    suppose f(n[itex]_{1}[/itex])=f(n[itex]_{2}[/itex])

    then [itex]\frac{p_{1}}{q_{1}}[/itex]=[itex]\frac{p_{2}}{q_{2}}[/itex]
    I have no idea how to get from here to n[itex]_{1}[/itex]=n[itex]_{2}[/itex]

    consider p+q
    the diagonals the p/q is in has the property that the # of elements in that diagonal is p+q-1 (referring way back to the beginning)
    or in other words ([itex]\sum^{p+q-2}_{k=1}[/itex]k, [itex]\sum^{p+q-1}_{k=1}[/itex]k]
    then let n= p+[itex]\sum^{p+q-2}_{k=1}[/itex]k

    and then I get hung up again
  2. jcsd
  3. May 1, 2014 #2


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    2017 Award

    Staff: Mentor

    You seem to prove some unnecessary or trivial stuff while other parts are wrong.

    So ##\frac{1}{2}=\frac{2}{4} \Leftrightarrow## 1=2 and 2=4? That is wrong.

    What does that mean?

    I guess the sum is supposed to run up to n instead of j?
    Okay, so as an example:
    S3 = {1,2,3,4,5,6}
    It is obvious that Sn is not empty (as an example, it always contains 1) and that it has a smallest element (it always contains 1 and 1 is the smallest number in Z+).

    Why should ##n \leq \sum^{1}_{k=1} k = 1## be true?

    This plus the wrong way to start make it impossible to finish your mapping or show that it has the desired properties.
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