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This question already has an answer here:

Can we define any kind of addition and multiplication on the set of all non-negative integers such that it becomes a field. I think not. Can we prove this ? If we have only a finite collection with prime cardinality then only it may become a field.

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marked as duplicate by Watson, Jack's wasted life, Willie Wong, suomynonA, Claude Leibovici Nov 19 '16 at 8:57

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  • $\begingroup$ We can turn a set of any cardinality into a field (and for at most countable sets, this does not even require choice or anything like that). $\endgroup$ – Tobias Kildetoft Sep 3 '13 at 12:20
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    $\begingroup$ Pick any bijection of $\mathbb{N}$ with $\mathbb{Q}$ and transport the field operations to $\mathbb{N}$ via that. $\endgroup$ – Daniel Fischer Sep 3 '13 at 12:20
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    $\begingroup$ @Tobias: Any infinite cardinality. Finite sets can only become fields if their size is a prime power. $\endgroup$ – Henning Makholm Sep 3 '13 at 12:22
  • $\begingroup$ ohh, this was trivial. $\endgroup$ – aaaaaa Sep 3 '13 at 12:22
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    $\begingroup$ There is an interesting field structure on natural numbers, due to John Conway, I believe; google "nimbers". I might add an answer eventually about this. $\endgroup$ – knsam Sep 3 '13 at 12:47
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Here is what you could do: the rational numbers are countable, so you can find a bijection $\phi: \mathbb{N}_0 \to \mathbb{Q}$ ($\mathbb{N}_0 = \{0,1,2,\dots\}$). Now $\mathbb{Q}$ is a field, so can define multiplication and addition making $\mathbb{N}_0$ into a field.

You can define, for example, addition $\mathbb{N}_0$ by: $$ a + b = \phi^{-1}(\phi(a) + \phi(b)) \\ $$

If you for example order $\mathbb{Q} = \{a_0, a_1, \dots \}$ like this site suggests: $$ a_0 =\frac 0 1, a_1 = \frac 1 1, a_2 = \frac {-1} 1, \frac 1 2, \frac {-1} 2, \frac 2 1, \frac {-2} 1, \frac 1 3, \frac 2 3, \frac {-1} 3,\\ \frac {-2} 3, \frac 3 1, \frac 3 2, \frac {-3} 1, \frac {-3} 2, \frac 1 4, \frac 3 4, \frac {-1} 4, \frac {-3} 4, \frac 4 1, \frac 4 3, \frac {-4} 1, \frac {-4} 3 \ldots $$ Then in $\mathbb{N}$ you would have $$ 3\cdot 4 = \phi^{-1}(a_3a_4) = \phi^{-1}(\frac{1}{2}\frac{-1}{2}) = \phi^{-1}(\frac{-1}{4}) = 17. $$

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    $\begingroup$ +1 for actually putting in a detailed calculation with an actual bijection. $\endgroup$ – Tobias Kildetoft Sep 3 '13 at 12:31

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