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In terms of the basic algebraic operations -- addition, negation, multiplication, division, and exponentiation -- is there any gain moving from $\Bbb Q$ to $\Bbb R$?

Say we start with $\Bbb N$: $\Bbb N$ is closed under addition and multiplication. But then we decide we'd like a number system that's closed under negation as well, so we construct $\Bbb Z$. Great. But then we decide we'd like to extend this number system further to be closed under division and so we construct $\Bbb Q$. The next step is closure under exponentiation - but when we construct that number system, we don't get $\Bbb R$, we get a subset of $\Bbb C$ which I'll call $\Bbb Q_{\exp}$.

Now clearly when constructing $\Bbb R$ from $\Bbb Q$ we do gain completeness, but our gain is then analytic, not necessarily algebraic. Do we gain any algebraic advantage in constructing $\Bbb R$ from $\Bbb Q$ similar to what we get at each of the other constructions I mentioned?

4 Answers4

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For instance, we get to solve all equations of the form $ax^n=b$, for $a,b \in \mathbb N$.

We also get to solve all polynomial equations of odd degree, or at least find one solution.

More generally, we get to solve equations, even transcendental ones, using the bisection method. One just needs an interval where the function takes values of different signs at the extremes. This is all very constructive, as long as you can find the sign of the value of a function.

On the other hand, it can be argued that the bisection method also works in $\mathbb Q$, and delivers increasingly smaller intervals that contain a root.

lhf
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Algebraically speaking, whatever you gain can be gained on a countable field. Both $\Bbb R$ and $\Bbb C$ are uncountable. To see this, just pick any operation, like exponentiation, and just enlarge $\Bbb Q$ by adding all the complex numbers that you get from the last step. After finishing all the finitely indexed steps, you have a field closed under exponentiation.

So in some sense, they are not the closure of $\Bbb Q$ under any reasonable algebraic operation.

But $\Bbb R$ is one of the fundamental objects of modern mathematics, so we continue to investigate it, and build a lot of mathematics around it. In some parts of mathematics we care more about algebraically closed fields, so we move to $\Bbb C$ instead, which is the closure of $\Bbb R$.

Not everything is about algebraic operations. Analysis, and topology have a lot to do later in mathematics, especially when you try to tame very large objects.

Asaf Karagila
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  • Yes, I believe the motivation for $\mathbb R$ comes more from analysis & topology. Although, there are some theorems concerning complete linearly ordered fields and "real-closed" fields. I just can't understand them at the moment. – Forever Mozart Jun 27 '14 at 03:13
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I thinks the motivation for $\mathbb{R}$ is not algebraic, but rather it corresponds to our geometric intuition about which numbers are possible. Also it was founded at a time when we viewed the universe as continuous space.

Rene Schipperus
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  • I'll agree that we no longer consider the universe to be some $\Bbb{R}^n$ but not continuous? – guest196883 Jun 27 '14 at 03:12
  • I wasnt implying that was the case, but rather that ideas about the universe now are much more involved and based on much more sophisticated mathematical objects, for example, energy is not always continuous. – Rene Schipperus Jun 27 '14 at 03:18
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As Asaf mentioned, you cannot obtain $\mathbb{R}$ as a completion of $\mathbb{Q}$ under these operations. However, there are still algebraic things to be said about $\mathbb{R}$.

For example, every irreducible polynomial over $\mathbb{R}$ has degree less than or equal to two, which is trivial to check since $\mathbb{C}$ is algebraically closed, contains $\mathbb{R}$ and $[\mathbb{C}:\mathbb{R}]=2$.

This implies, for instance, that $x^4+1$ can be factored over $\mathbb{R}$. And the factorization is $(x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)$.

J. Loreaux
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