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The set of squares of rational numbers is inductive

  1. Apr 13, 2012 #1
    1. The problem statement, all variables and given/known data
    The set of squares of rational numbers is inductive

    2. Relevant equations
    definition of an inductive set

    3. The attempt at a solution
    sorry i know this is probably very easy to most but I am just learning analysis. Okay, so we can see that 1 is in the set because it is rational and 12 exists. Am I correct in thinking that it doesn't matter whether or not 12 is rational? I mean obviously it is, but aren't we really just checking to make sure that S(1) is defined? Then for the second part, we can take S(x) to be x2 whenever x=m/n, where m,n are integers and either m or n is odd. Again, aren't we basically proving that if S(x) exists, or (m2/n2) exists, then S(x+1), or [(m+n)2/n2] also exists. But then don't I also need to show that if x is rational, then so is x+1? Maybe i am reading way too much into this, but again I am new to these and I'm trying to understand exactly what I need to show.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 13, 2012 #2
    No, you're pretty much correct. Let S be the set of the squares of rationals. To show it's inductive you need to show that 0 is in S (it is because 0 is [itex]0^{2}[/itex] and 0 is rational). Then you need to show that if x is in S, then x+1 is in S.

    Then for two integers m,n [itex]x=\frac{m^{2}}{n^{2}}[/itex] so [itex]x+1=\frac{m^{2} + n^{2}}{n^{2}}[/itex]. The question then boils down to showing that this can be written in the form [itex]\frac{p^{2}}{q^{2}}[/itex], which isn't the same as [itex]\frac{(m+n)^{2}}{n^2}[/itex]. They're rational by your definition of x, but the latter can't be assumed to be rational squared, unless you show it.
     
  4. Apr 13, 2012 #3
    I'm fairly sure it's not inductive by the way.

    Take [itex]x=\frac{1}{4}[/itex]. Then [itex]x+1=\frac{5}{2^{2}}[/itex]. This cannot be written in the form [itex]\frac{p^2}{q^2}[/itex].

    Every time you multiply the fraction by some multiple [itex]\frac{5}{5}[/itex] you get that the index of 5 increases to an even number on the top but then an odd number on the bottom. So you can never have an even index of 5 on the top and bottom. So the fraction is not part of the rationals squared.
     
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