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Transitivity question

  1. Oct 4, 2017 #1
    1. The problem statement, all variables and given/known data
    If a, b ##\epsilon \mathbb{C}##, we say that ##a## ~ ##b## if and only if ##a^k = b^k## for some positive integer ##k##. Prove that this is an equivalence relation.

    2. Relevant equations


    3. The attempt at a solution

    Proof:

    (Reflexivity): Suppose ##a \epsilon \mathbb{C}##. Then ##a = x + iy## for some ##x,y \epsilon \mathbb{R}##. We know ##(x+iy)^k = (x+iy)^k##. So a ~ a. So ~ is reflexive.

    (Symmetric): Suppose a ~ b. Let ##a = x + iy## and ##b = s + it## for some ##x,y,s,t \epsilon \mathbb{R}##. Then ##(x+iy)^k = (s+it)^k, k \epsilon \mathbb{N}##. So
    ##(s+it)^k = (x+iy)^k## is true. So b ~ a. So ~ is symmetric.

    (Transitivity): Suppose a ~ b and b ~ c. Let ##a = x + iy, b + s + it, ## and ##c = m + ni##. Then ##(x+iy)^k = (s+it)^k## for some ##k \epsilon \mathbb{N}## and
    ##(s + it)^g = (m + ni)^g## for some ##g \epsilon \mathbb{N}##.

    So we have to show there exists ##h \epsilon \mathbb{N}## such that ##a^h = c^h##.

    Consider 2 cases:

    Case1: k > g.

    We know ##a^k = b^k##
    ##a^k b^g = b^k c^g##
    ##a^k b^g b^{-k} = b^k b^{-k} c^g##
    ##a^k b^{g-k} = (1) c^g##
    ....

    I think the transitivity will have to do with the roots and factoring out b's somehow, can someone point me in a direction please
     
  2. jcsd
  3. Oct 4, 2017 #2

    andrewkirk

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    Think in terms of ratios. The condition ##a^k=b^k## is the same as ##\frac{a^k}{b^k}=1##, provided ##b\neq 0##. Can you think of a way of chaining ratios together to get the result you want for transitivity? .
     
  4. Oct 4, 2017 #3

    SammyS

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    It seems to me that you have made this much more complicated than necessary.

    No need to even mention the express complex numbers in the form, ##\ x+yi\ ##.

    The reflexive and symmetric behavior of the ~ relation follow directly from the reflexive and symmetric behavior of equality .

    Transitivity is not terribly difficult either.

    To start, consider that ##\ a\text{~}b \ ## implies that there exists some ##\ k\,##∈ℤ+ such that ##\ a^k=b^k\ ##.

    It then follows that ##\ \left(a^k\right)^p=\left(b^k\right)^p\ ## for every ##\ p\,##∈ℤ+

    etc.
     
  5. Oct 4, 2017 #4
    Yea I see your point that the expansion into a + bi is unnecessary.. Here is the updated proof,

    Proof:
    (Reflexivity) Suppose ##a \epsilon \mathbb{C}##. Then ##a^k = a^k##, so a ~ a. So ~ is reflexive.

    (Symmetry) Suppose a ~ b. Then ##a^k = b^k## so ##b^k = a^k##. So b ~ a. So ~ is symmetric.

    (Transitivity) Suppose a ~ b and b ~ c. Then ##a^k = b^k## and ##b^n = c^n## for some ##k, n \epsilon \mathbb{Z^{+}}##
    By raising the first equation by n, we have ##a^{nk} = b^{nk}## and by raising the second equation by k, we have ##b^{nk} = c^{nk}##. So ##a^{nk} = b^{nk} = c^{nk}## where ##nk \epsilon \mathbb{Z^{+}}##. Thus a ~ c, so ~ is transitive.

    We conclude ~ is an equivalence relation.
     
  6. Oct 4, 2017 #5
    Thank you for the suggestion.
    So using this and the next post, my transitivity step would look like..

    (Transitivity) Suppose a ~ b and b ~ c. Then ##\frac {a^k} {b^k} = 1## and ## \frac {b^n} {c^n} = 1## for some ##k, n \epsilon \mathbb{Z^{+}}##. Raising the first equation by n and the second equation by k, we have ##\frac {a^{kn}} {b^{kn}} = 1## and ## \frac {b^{kn}} {c^{kn}} = 1##. We know ## \frac {b^{kn}} {c^{kn}} = 1 = \frac {c^{kn}} {b^{kn}}##. Then ##\frac {a^{kn}} {b^{kn}} = \frac {c^{kn}} {b^{kn}}## which simplifies to ##a^{kn} = c^{kn}## where ##kn \epsilon \mathbb{Z^{+}}##. Thus, a ~ c. So ~ is transitive.

    Not sure if this is what you meant though...
     
  7. Oct 4, 2017 #6

    andrewkirk

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    Yes that's it.
     
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