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I'm reading the book Discrete Mathematics and its Applications 4th Edition by Susanne Epp. One of the questions is:

Indicate the elements in the following sets:

$ \{n \in \mathbb{Z} \mid n = (-1)^k, \text{ for some integer } k \} $

the answer given is the set $\{-1, 1\}$

Could someone explain to me how this works? I don't understand how you can find this set without knowing what k is. If it's even then n will have to equal 1, if it's uneven it will have to equal -1.

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    @TheChaz, thanks for that clarification.2012-04-22

1 Answers 1

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It is important to remember some fundamental properties of sets:

  • A set is considered equal to another set if it has the same distinct elements as the other set. As a consequence the repetition of distinct elements within a set doesn't matter, nor does the order (e.g $\{1,2,3,4\} = \{1,1,2,3,4\}$).
  • The elements of the set are defined by a "defining property", which all elements must meet

So, now let us consider your set:

$\left\{n \in \mathbb{Z} \mid n = (-1)^k, \text{ for some integer } k \right\}$

The defining property of your set is $n = (-1)^k, \text{ for some integer } k$. Like said in the comments, there is some ambiguity in the language here. This could be taken to mean "there exist a $K \in \mathbb Z$ such that $n = (-1)^k$" (or in logic notation: $ \exists k \left( (k \in \mathbb Z ) \wedge (n = (-1)^k)\right)$ ), or it could be taken to mean "for every $k \in \mathbb Z$: $n$, is defined as $n = (-1)^k$" (in logic notation: $\forall k \in \mathbb Z \left (n = (-1)^k \right )$). Given your answer of $\{1,-1\}$, it is reasonable the proper interpretation is "for every $k \in \mathbb Z$, $n$ is defined as $n = (-1)^k$" (note that most natural language proofs I have seen use "for some $k$" as "there exist a $k$ such that").

Now that we understand what our defining property is let us try to write the set using roster notation. There are three cases to consider: $(k \in \mathbb Z) < 0$, $(k \in \mathbb Z) > 0$, and $(k \in \mathbb Z) = 0$.

  • When $k$ is < 0 then we get: $ n_{-1} = (-1)^{-1} = \frac {1} {-1} = -1 $, $n_{-2} = (-1)^{-2} = \frac {1} {{-1}^2} = 1$, et cetera. The value of $n$ when $k$ is less than $0$ oscillates between $1$ and $-1$
  • When $k$ is > 0 then we get: $ n_{1} = (-1)^1 = -1$, $n_{2} = (-1)^2 = 1$, et cetera. This also oscillates between $1$ and $-1$
  • There is also the trivial case $(-1)^0 = 1$

So now that we have considered all the possible values of $k$ we can write our set using roster notation:

  • For $k$ is < 0 then we have $\{-1, 1, -1, ...\}$
  • Concatenating the values when $k$ is > 0 we have $\{-1, 1, -1, ..., 1, -1, 1...\}$
  • and lastly adding the trivial case: $\{-1, 1, -1, ..., 1, -1, 1,..., 1\}$

Lastly, because of the first property of sets that we recalled above, we must simplify this set to contain only distinct elements, and since the set $\{-1, 1, -1, ..., 1, -1, 1,..., 1\}$ is made of nothing but $-1$ and $1$ it can be written $\{-1,1\}$.