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We find a bijection $\phi$ between $\mathbb N$ and $\mathbb Q$. Hence, defining addition and multiplication correspondingly on $\mathbb N$ (such that $\phi$ becomes a homomorphism) transforms $\mathbb N$ into a field.

Is there a mistake in my reasoning?

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    $\begingroup$ Sure, it’s just that the operations won’t be the usual ones. $\endgroup$ – user208649 Aug 18 '20 at 5:39
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    $\begingroup$ Yeah. So if $\phi(7) = 0; \phi (23)=1; \phi(18)=2; \phi(9)=\frac 12$ then we'd have $7$ is additive identity, $23$ is the multiplicative identity, and $18$ is the multiplicative inverse of $9$. And $23+23=18\times 23 = 18$. Looks weird but there is nothing wrong with it. $\endgroup$ – fleablood Aug 18 '20 at 5:46
  • $\begingroup$ FYI this question is very close to a duplicate of this one; in my opinion it's a bit different, though, in that this question starts with the idea of transporting structure along bijections already and is about making that concept precise. That said, you might find the discussion at that other question worthwhile. $\endgroup$ – Noah Schweber Aug 18 '20 at 5:47
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A field isn't just a set, it's a set together with some additional structure (the two field operations). So it's not quite true that $\mathbb{Q}$ is a field - rather, $(\mathbb{Q};+,\times)$ is a field.

Bijections let us "transport structure:" if $\oplus,\otimes$ are binary operations on some set $A$ such that $(A;\oplus,\otimes)$ is a field and $f:A\rightarrow B$ is a bijection, we can give $B$ the structure of a field in a natural way: consider the operations $\hat{\oplus}$ and $\hat{\otimes}$ given by $$x\hat{\oplus} y=f(f^{-1}(x)\oplus f^{-1}(y))\quad\mbox{and}\quad x\hat{\otimes}y=f(f^{-1}(x)\otimes f^{-1}(y))$$ for $x,y\in B$. But the set $B$ itself is not a field; rather, the structure $(B; \hat{\oplus},\hat{\otimes})$ is a field.

In particular, when we lift the usual $+$ and $\times$ along your favorite bijection $h:\mathbb{Q}\rightarrow\mathbb{N}$, we get operations $\hat{+}$ and $\hat{\times}$ such that $(\mathbb{N};\hat{+},\hat{\times})$ is a field, but these operations will be very odd-looking - in particular, they'll be totally different from the usual addition and multiplication of natural numbers we're used to. So there's no tension between this result and the fact that $(\mathbb{N};+,\times)$ is clearly not a field.

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    $\begingroup$ OK, admittedly people do write things like "$\mathbb{Q}$ is a field" as opposed to "$(\mathbb{Q};+,\times)$ is a field." This abuse of notation is justified on the grounds that the extra structure is clear from context, but it is technically incorrect. $\endgroup$ – Noah Schweber Aug 18 '20 at 5:53
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There is no mistake. In fact, any infinite set can be turned into a field. Note that the operations that you define on $\mathbb N$ this way will necessarily be different from the usual addition & multiplication of natural numbers (because with the usual operations the natural numbers are not a field).

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Using the usual diagonal mapping but alternating between positive and negative values and skipping the duplicate representations of "fractions not in lowest terms" we can have the bijection of which the first several terms are:

$$1\mapsto 0; 2\mapsto 1;3\mapsto -1; 4\mapsto 2;5\mapsto -2; 6\mapsto \frac 12; 7\mapsto -\frac 12; 8\mapsto 3;9\mapsto -3;10\mapsto \frac 13;11\mapsto -\frac 13; 12\mapsto 4;13\mapsto -4; 14\mapsto \frac 32; 15\mapsto -\frac 32; 16\mapsto \frac 23; 17\mapsto -\frac 23; 18\mapsto \frac 14;19\mapsto -\frac 14... etc...$$

Now this is a field. The additive identity is $1$ and $1 + k = k+1 = k$ for all $k \in \mathbb N$.

Every value, $k$ has an additive inverse, $-k$ so that $k+(-k)= 1$. Forexample the additive inverse of $4$ is $-4 =5$ and $4+5 = 1$. Likewise $-11 = 10$ and $11 + 10 = 1$.

The multiplicative identity is $2$ and $2\cdot k = k\cdot 2 = k$ for all $k \in \mathbb N$.

And for every value $k$ except $1$, will have a multiplicative inverse $\frac 1k$ where $k\cdot \frac 1k = 2$. For example $\frac 14 = 6$ and $4\cdot 6 = 2$.

And so on.

This all makes sense because all I did was replace the "usual" rational numbers with what mapps into them. If I make notes of $k \color{blue}{\mapsto m}$ to represent what I "really" mean and cut and paste what I wrote above it would be:

...........

Now this is a field. The additive identity is $1\color{blue}{\mapsto 0}$ and $1\color{blue}{\mapsto 0} + k = k+1\color{blue}{\mapsto 0} = k$ for all $k \in \mathbb N$.

Every value, $k$ has an additive inverse, $-k$ so that $k+(-k)= 1\color{blue}{\mapsto 0}$. Forexample the additive inverse of $4\color{blue}{\mapsto 2}$ is $-4\color{blue}{\mapsto 2} =5\color{blue}{\mapsto -2}$ and $4\color{blue}{\mapsto 2}+5\color{blue}{\mapsto -2} = 1\color{blue}{\mapsto 0}$. Likewise $-11\color{blue}{\mapsto -\frac 13} = 10{\mapsto \frac 13}$ and $11\color{blue}{\mapsto -\frac 13} + 10\color{blue}{\mapsto \frac 13} = 1\color{blue}{\mapsto 0}$.

The multiplicative identity is $2\color{blue}{\mapsto 1}$ and $2\color{blue}{\mapsto 1}\cdot k = k\cdot 2\color{blue}{\mapsto 1} = k$ for all $k \in \mathbb N$.

And for every value $k$ except $1\color{blue}{\mapsto 0}$, will have a multiplicative inverse $\frac 1k$ where $k\cdot \frac 1k = 2\color{blue}{\mapsto 1}$. For example $\frac 1{4\color{blue}{\mapsto 2}} = 6\color{blue}{\mapsto \frac 12}$ and $4\color{blue}{\mapsto 2}\cdot 6\color{blue}{\mapsto \frac 12} = 2\color{blue}{\mapsto 1}$.

And so on.

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