# Extending the rationals using exponentiation

The set of integers can be constructed as an equivalence relation over the natural numbers using the the binary operation of addition, and a similar process yields the rationals from integers and multiplication. I'm curious what happens if I try to continue this sequence with next logical step: extending the rationals via an equivalence relation equating pairs of exponentials of rational numbers raised to to rational powers.

I know that after suitably defining the relation to exclude indeterminate forms like 0^0, the resulting set will be isomorphic to a proper superset of the rational numbers, since it includes the square root of 2, and a proper subset of the real numbers since numbers like pi can't be represented (it occurs to me I don't actually know how to prove this, though I do know how to prove pi is irrational). So my question is just what is this set of numbers? Does it have a name? And lastly, why isn't it mentioned in less advanced mathematics classes (presumably it lacks sufficiently useful algebraic properties, maybe?)?

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I don't understand your question. You may or may not be talking about a certain subring of the algebraic numbers (en.wikipedia.org/wiki/Algebraic_number). See also en.wikipedia.org/wiki/Transcendental_number . – Qiaochu Yuan Mar 15 '13 at 5:44
Can you explain the "suitable definition"? – Alex Becker Mar 15 '13 at 5:45
@AlexBecker Sure. In constructing the integers you first form the cartesian product of the naturals, and define the equivalence relation. But to construct the rationals, an equivalence relation is defined on Z-{0} instead of Z because there isn't a way to define zero to have a multiplicative inverse. All I meant by "suitable definition" was to exclude rational numbers which can't be used as bases of logarithms (at least I think that is sufficient for what I describe). – David H Mar 15 '13 at 6:02
@DavidHammett But what equivalence relation are you using? – Alex Becker Mar 15 '13 at 6:12

You'll have to be more careful about what you mean here.

The process by which we form the integers from the natural numbers really is the same in some sense as the process by which we form the rationals from the integers. In both cases, we have a commutative monoid ($\langle\mathbb{N},0,+\rangle$ and $\langle\mathbb{Z}^*,1,\times\rangle$), i.e. an associative, commutative operation with an identity element, and we form the Grothendieck group of the monoid, which is a fancy way of saying that we formally adjoin inverses to all elements.

But exponentiation on the rationals is very different from addition and multiplication. It isn't commutative or even associative. It has a right identity, 1 ($x^1 = x$ for all $x$), but no left identity (there is not $q$ such that $q^x = x$ for all $x$). And $\mathbb{Q}$ isn't even closed under exponentiation ($2^\frac{1}{2} \notin \mathbb{Q}$).

My point is just that continuing this sequence with $\mathbb{Q}$ and exponentiation might not be as logical a next step as you think.

Edit: Here are some specific questions to consider with respect to your proposed construction.

1. In the construction of $\mathbb{Z}$ from $\mathbb{N}$ and of $\mathbb{Q}$ from $\mathbb{Z}$, we consider pairs of numbers $(a,b)$. In the first case, $(a,b)$ is supposed to represent $a-b$, and in the second case, $(a,b)$ is supposed to represent $\frac{a}{b}$. In your case, what do you intend $(a,b)$ to represent?

2. One thing that makes these constructions useful is that we can extend the algebraic operations from the original structure to the new structure. In the construction of $\mathbb{Z}$, we can let $(a,b) + (a',b') = (a + a',b+b')$, since $(a - b) + (a' - b') = (a+a') - (b+b')$. In the construction of $\mathbb{Q}$ from $\mathbb{Z}$, we can let $(a,b) \times (a',b') = (aa',bb')$, since $\frac{a}{b}\times\frac{a'}{b'} = \frac{aa'}{bb'}$. We can also extend $+$ to $\mathbb{Q}$ using a more complicated "common denominators" rule. Which algebraic operations would you like to extend to your newly constructed set, and how would you extend them?

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