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Take the natural numbers $\mathbb{N}_0 = \{0, 1,2,3,\dots\}$ and the integers $\mathbb{Z} = \{\dots, -1, 0, 1, \dots\}$.

Can we come up with a bijective linear mapping $T: \mathbb{N} \rightarrow \mathbb{Z}$?

Consider the mapping $0 \mapsto 0$, $1 \mapsto 1$, $2\mapsto -1$, $3\mapsto 2$, $4 \mapsto -2$, etc. This can be summarized by the function $$T(x) = \begin{cases} -\frac{x}{2} \text{ if } x \text{ even}, \\ \frac{x+1}{2} \text{ if } x \text{ odd}. \end{cases} $$

$T$ is not linear: if $x$ is even and $y$ is odd, then $x+y$ is odd and so $T(x+y) = \frac{x+y+1}{2}$, but $T(x)+T(y) = \frac{-x+y+1}{2}$.

Is it possible to come up with a linear mapping? Intuitively, a linear mapping should stretch, rotate, or translate a set, rather than "zig-zagging". I don't know if there is a way to do this to turn $\mathbb{N}$ into $\mathbb{Z}$?

In general: is there some explanation of when a linear mapping exists between two countably infinite sets, and when it doesn't exist?

Edit: it seems from some comments that a linear mapping exists between two sets when those sets are both vector spaces. Is this the answer?

blue_egg
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  • What about the linear map $T(x)=x$? – Dietrich Burde Aug 06 '24 at 17:04
  • Oh sorry, I meant to specify that it should be a bijection – blue_egg Aug 06 '24 at 17:04
  • The bijections here are not linear, I think. – Dietrich Burde Aug 06 '24 at 17:06
  • @DietrichBurde thank you. I added a sentence about what I think a linear mapping should intuitively look like. Since every bijection between $\mathbb{N}$ and $\mathbb{Z}$ is "zigzagging", it seems like no linear map between them exists? – blue_egg Aug 06 '24 at 17:08
  • Linear mappings are defined on vector spaces, so a rigorous definition of what you're looking for would help. For example, is it enough that $f(x+y) = f(x) + f(y)$? Or do you also want that $f(xy) = xf(y)$? – MartianInvader Aug 06 '24 at 17:30
  • @MartianInvader I do want the scalar multiplication property. Is the answer to the general version of the question "linear mappings only exist from vector space to vector space"? And the example $\mathbb{N} \rightarrow \mathbb{Z}$ has no linear map because $\mathbb{N}$ is not a vector space? – blue_egg Aug 06 '24 at 22:23
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    Suppose f(x)>0 is a member of N then f(-x)=-f(x) for linearity but then f(-x)<0 which is a contradiction. f(x)=0 for all x in Z should probably satisfy linearity but I'm not sure. – SeetheMoar Aug 07 '24 at 05:04

2 Answers2

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Suppose $T(0)=a$ and $T(1)=b$. Then $T^{-1}(a)=0$ and $T^{-1}(b)=1$, so by linearity $T^{-1}(a-b)=-1$, which is not in $\Bbb N$.

TonyK
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  • thank you! are you using the fact that there can be an arithmetic progression in $\mathbb{Z}$ whose image is not an arithmetic progression in $\mathbb{N}$ ? How do you show this? – blue_egg Aug 06 '24 at 17:13
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    @900edges I’m not sure if my reasoning is correct here, but $\mathbb{N}$ is not a vector space since it doesn’t posses additive inverses to each element. And by definition, linear maps are defined on vector spaces. – Paul Ash Aug 06 '24 at 17:28
  • @PaulAsh Thanks for pointing that out. Is the answer to the general version of the question "linear mappings only exist from vector space to vector space"? – blue_egg Aug 06 '24 at 22:24
  • @900edges I believe so, yes. (Btw, $\mathbb{Z}$ is also not a vector space). – Paul Ash Aug 06 '24 at 23:01
  • @PaulAsh: Well, I suppose that's right: a linear mapping is a mapping between vector spaces. But the axioms are just (i) $f(x)+f(y)=f(x+y)$ and (ii) $f(ax)=af(x)$. And these axioms make sense in wider contexts. Here, however, that $a$ is problematic: is it $a\in\Bbb Z$, or $a\in\Bbb N$? So perhaps the question is simply badly posed. – TonyK Aug 06 '24 at 23:48
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The term "linear mapping" is generally defined on vector spaces, and neither $\mathbb{N}$ nor $\mathbb{Z}$ is a vector space, so I'm going to try to rephrase your question to avoid that term:

Rephrased Question: Does there exist a bijective function $T:\mathbb{N} \rightarrow \mathbb{Z}$ such that for all $x,y \in \mathbb{N}$, $T(x+y) = T(x) + T(y)$?

The answer to this question is No. To prove this, we'll prove an even stronger statement:

Claim: Any $T:\mathbb{N} \rightarrow \mathbb{Z}$ such that $T(x+y) = T(x) + T(y)$ is of the form $T(n) = nz$, for some fixed $z \in \mathbb{Z}$.

Proof of Claim: First we see that $T(0) = T(0+0) = T(0) + T(0)$, so $T(0) = 2T(0)$. $0$ is the only integer equal to its double, therefore $T(0) = 0$.

Next, for any $n > 0 \in \mathbb{N}$, we can express it as the sum of $n$ copies of $1$. Thus:

$T(n) = T(1 + \dots+1) = T(1) + \dots + T(1) = nT(1)$.

Setting $z = T(1)$ proves the claim.

The claim implies the image of $T$ is either completely nonnegative or completely nonpositive, meaning it cannot be bijective.

MartianInvader
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