# Can we make the set of all non-negative integers a field? [duplicate]

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.

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

• 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). – Tobias Kildetoft Sep 3 '13 at 12:20
• Pick any bijection of $\mathbb{N}$ with $\mathbb{Q}$ and transport the field operations to $\mathbb{N}$ via that. – Daniel Fischer Sep 3 '13 at 12:20
• @Tobias: Any infinite cardinality. Finite sets can only become fields if their size is a prime power. – Henning Makholm Sep 3 '13 at 12:22
• ohh, this was trivial. – aaaaaa Sep 3 '13 at 12:22
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 $$3\cdot 4 = \phi^{-1}(a_3a_4) = \phi^{-1}(\frac{1}{2}\frac{-1}{2}) = \phi^{-1}(\frac{-1}{4}) = 17.$$