Proving Equivalence Relation of Complex Numbers Using Roots and Factoring

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You have to be careful about the cases when some of the numbers are zero, but you can handle that with a bit of care.
  • #1
fishturtle1
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Homework Statement


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.

Homework Equations

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
 
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  • #2
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? .
 
  • #3
fishturtle1 said:

Homework Statement


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.

Homework Equations



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
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.
 
  • #4
SammyS said:
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.
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.
 
  • #5
andrewkirk said:
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? .
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...
 
  • #6
Yes that's it.
 

Related to Proving Equivalence Relation of Complex Numbers Using Roots and Factoring

1. What is a transitivity question?

A transitivity question is a type of question used in scientific research to explore the relationship between three variables or concepts. It typically asks whether there is a direct or indirect relationship between two variables, mediated by a third variable.

2. How is a transitivity question different from a mediation question?

A transitivity question specifically focuses on the relationship between three variables, while a mediation question may also involve exploring the indirect effects of other variables on the relationship between two variables.

3. What is the purpose of asking a transitivity question in scientific research?

Transitivity questions are used to gain a deeper understanding of the underlying mechanisms and relationships between variables. By examining the transitive relationship between three variables, researchers can identify potential causal pathways and make more accurate predictions about the outcomes of their study.

4. How do you formulate a transitivity question?

A transitivity question should start with the phrase "Does X have a direct or indirect effect on Y, mediated by Z?" where X, Y, and Z represent the three variables of interest. The question should be specific and testable, with all variables clearly defined.

5. What are some limitations of using transitivity questions in scientific research?

One limitation is that transitivity questions can only address the relationship between three variables, and may not capture the complexity of real-world situations. Additionally, the use of transitivity questions relies on the assumption of linearity and may overlook nonlinear relationships between variables.

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