1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Finding unions of sets

  1. Mar 5, 2009 #1
    1. The problem statement, all variables and given/known data
    X = {x: x=4n+1, n[tex]\in[/tex][tex]N[/tex]}
    Y = {y: y=m[tex]^{2}[/tex]+m+1, m[tex]\in[/tex][tex]N[/tex]}

    Find: Y[tex]\cap[/tex]X

    3. The attempt at a solution

    First I tried to set x=y
    [tex]\Rightarrow[/tex] n[tex]^{2}[/tex]+n+1=4n+1
    [tex]\Rightarrow[/tex] n[tex]^{2}[/tex]-3n=0
    [tex]\Rightarrow[/tex] n(n-3)=0
    [tex]\Rightarrow[/tex] n=0,3

    Subbing this in to either equation yields {1,13}.
    I thought I was doing okay.
    However I was comparing assignments with a classmate and she had this massive set that just keeps on going: {1, 13, 21, 57, 73, 133, 157, 241, 273, ...}
    After I thought about it, this made sense, as my solution only found the elements that are in both sets when n is the same for both equations, but set equality relies only on its elements, not at what stage they were reached.
    She'd done this the long way round, scripting a little program to calculate all elements for both sets then compare the elements and find the common ones.
    What I'd like to know is, how can I go about solving this algebraically?

    Thanks in advance,
  2. jcsd
  3. Mar 6, 2009 #2
    write the equation as it is.

    m^2 + m + 1 = 4n + 1
    --> m^2 + m = 4n
    --> m(m+1) = 4n

    now rhs is a multiple of 4... and therefore, so must the lhs. But if m is even, m + 1 is odd...
    do you see where i am going with this? can you proceed?
  4. Mar 9, 2009 #3
    Yes, Thankyou!

    So basically the solution is something like {x: x=m^2 + m + 1 ; m/4[tex]\in[/tex][tex]N[/tex] or (m+1)/4[tex]\in[/tex][tex]N[/tex]}
  5. Mar 10, 2009 #4
    yep... thats right!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook