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I have an assignment that is asking to define a one-to-one correspondence between the sets $2\mathbb{Z}$ and $17\mathbb{Z}$... or in other words, define some bijective function on $$f:2\mathbb{Z}\to 17\mathbb{Z}$$

Note: I know that $\mathbb{Z}$ is the set of integers.. I'm just wondering what the number in front means.


Addendum:

Given that I now know what these sets represent... is this a satisfactory answer for the question?


A one-to-one correspondence between $2\mathbb{Z}$ and $17\mathbb{Z}$ could be as follows:

\begin{align*} 0&\mapsto 0\\ 2&\mapsto 17\\ -2&\mapsto -17\\ 4&\mapsto 34\\ -4&\mapsto -34\\ 6&\mapsto 51\\ -6&\mapsto -51\\ \end{align*} And so on...

In general, the function \begin{equation*} f:2\mathbb{Z}\to 17\mathbb{Z}:2x\mapsto 17x,\forall x\in\mathbb{Z} \end{equation*}

defines a one-to-one correspondence between the sets $2\mathbb{Z}$ and $17\mathbb{Z}$.

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11  
It means a short nap... –  copper.hat Mar 10 '13 at 2:51
2  
I suggest that instead of giving seven examples of your mapping and leaving it to the grader to infer the meaning of "and so on...", that you try to find some way to express the rule or process that gets you from the left side to the right, and that works for every possible example. –  MJD Mar 10 '13 at 3:03
    
@MJD How might I do that though? The best I can think of is some sort of recurrence relation, but I wouldn't know how to define one properly... Maybe $s_0=(0,0)$, $s_1=(2,17)$, $s_2=(-2,-17)$, $s_n=??$ How do I define the sequence when an ordered pair is involved? –  agent154 Mar 10 '13 at 3:13
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Can you think of any method for figuring out $f(20)$ without listing the items one by one? If not, then try listing the items one by one until you get to $f(20)$ and then see if anything suggests itself. –  MJD Mar 10 '13 at 3:15
    
See the definition I give below, agent154: for every integer k, (all integers), map $\;2k\mapsto 17k$. So given $k = 0,\;\; 0\mapsto 0,\;k=1,\;\; 2\cdot 1\mapsto 17\cdot 1,\;\;k=10,\;\; 20 \mapsto 170..., k= -2,\;\;2\cdot (-2) = -4 \mapsto 17\cdot(-2) = -34$ –  amWhy Mar 10 '13 at 3:15

5 Answers 5

up vote 7 down vote accepted

$\;2\mathbb Z\;$ denotes the set of all integer multiplies of $\,2$: $$2\mathbb Z = \{2k\mid k\in \mathbb Z\}$$

The set $\;17\,\mathbb Z\;$ denotes the set of all integer multiplies of $\,17$: $$17\,\mathbb Z = \{17k\mid k \in \mathbb z\}$$

You'll encounter the notation frequently: In general, $$\;n\mathbb Z = \{nk\mid k \in \mathbb Z\}$$


EDIT to answer added question:

For your bijection: Yes, you've got the idea: let your bijection $f: 2 \mathbb Z \to 17 \mathbb Z\,$ be defined by $\,2k\mapsto 17k\,$ for each $\,k \in \mathbb Z,\,$ and yes, that includes $0 \mapsto 0$.

Edit: you're map that you just added will work, sort of, but you'll need to be clear that $n$ is a regular old integer (add tag following definition of function): $\forall n \in \mathbb{Z}$, otherwise n will refer to an element of $2\mathbb{Z}$. But then you are really mapping from $\mathbb Z \to 17\mathbb Z$.

If you want $n \in 2\mathbb Z$ then use $$f: 2\mathbb{Z} \to 17\mathbb Z, \;\;f(n) = \dfrac 12 n \cdot 17, \;\forall n \in 2\mathbb{Z}.$$ That way you are mapping directly from an even number $n \in 2\mathbb Z \to f(n)\in 17\mathbb Z$

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Nice, you're map that you just added, but add (tag following definition of function: $\forall n \in \mathbb{Z}$, otherwise n will refer to an element of $2\mathbb{Z}$. If you want $n \in 2\mathbb Z$ then use $f: 2\mathbb{Z} \to 17\mathbb Z, \;\;f(n) = \dfrac 12 n \cdot 17$. That way you are mapping directly from an even number $n \in 2\mathbb Z$ –  amWhy Mar 10 '13 at 3:26
    
but if you use $n=3$, then you're not getting a proper element in $17\mathbb{Z}$.... –  agent154 Mar 10 '13 at 3:33
    
I think a slightly better notation would be $f:2\mathbb{Z}\to 17\mathbb{Z}:2x\mapsto 17x,\forall x\in\mathbb{Z}$ –  agent154 Mar 10 '13 at 3:35
    
Now that works fine, since you're specifying x refers to an integer $x \in \mathbb{Z}$ –  amWhy Mar 10 '13 at 3:38

$2\mathbb Z$ means the set $\{ 2\cdot n \mid n\in \mathbb Z\}$; that is, the set of even integers.

In general, $n\mathbb Z$ means the set of integer multiples of $n$.

Is your question asking for a bijection between $\{\ldots -6, -4, -2, 0, 2, 4, 6,\ldots\}$ and $\{\ldots -51, -34, -17, 0, 17, 34, 51,\ldots\}$?

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OK... I was going to ask if 0 is a member of these sets, but clearly it is by your definition. –  agent154 Mar 10 '13 at 2:54
    
It certainly is. –  MJD Mar 10 '13 at 3:01

The set of even integers. Generally: $$n\Bbb Z=\{nk\mid k\in\Bbb Z\}$$

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If $A$ is a subset of a vector space, the notation $\lambda A$ (where $\lambda$ is in the relevant field) generally means $\lambda A = \{ \lambda a \}_{a \in A}$.

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I believe you can think of this as simply a special case of the notation $f(A)$, where $A$ is a subset of the domain of $f$, meaning the set of all images of elements of $A$. The $2$ or $17$ are essentially acting as symbols for the functions $2x$ and $17x$.

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