Two "different" definitions of $\sqrt{2}$

Question

In Walter Rudin's Principles of Mathematical Analysis (3rd edition) (page 10), it is proved that

for every $x>0$ and every integer $n>0$ there is one and only one positive real $y$ such that $y^n=x$. (This is number $y$ is then written $\sqrt[n]{x}$.)

In particular, this implies the existence of $\sqrt{2}$.

On the other hand, if one considers the polynomial $f(x)=x^2-2$ as an element in the ring $\mathbf{Q}[x]$, one can adjoin a root of $f$ to $\mathbf{Q}$. The procedure (see, for instance, Michael Artin's Algebra (2nd edition) page 456) is to form the quotient ring $K = \mathbf{Q}[x]/(f)$ of the polynomial ring $\mathbf{Q}[x]$. This construction yields a ring $K$ and a homomorphism $F\to K$, such that the residue $\overline{x}$ of $x$ satisfies the relation $f(\overline{x})=0$.

In the real analysis case, $\sqrt{2}$ can be approximated (or defined, depending on how one constructs the real numbers) by a Cauchy sequence of rational numbers: $$ 1, 1.4, 1.41, 1.414, 1.4142, \cdots $$

In the abstract algebra case, the set of real numbers is absent; one does not even need to define it. And there is no way to "approximate" $\overline{x}$.

These two ways to define the object $\sqrt{2}$ seems to be somewhat different in that the defined object has rather different properties.

How should one understand the "discrepancy" here? Are there other relations/connections between these two definitions besides being a root of $f(x)=x^2-2$?

Answer 1

It helps to view each construction in the context of ordered (or orderable) fields.

The algebraic definition describes a field, but $F=\mathbb{Q}[x]/(x^2-2)$ is a bit more than just a field: it's an orderable field. There are exactly two ways to make $F$ into an ordered field, determined by which square root of $2$ we pick to be positive.

On the "geometric" side, an ordering is exactly the additional data provided by Cauchy sequences (or Dedekind cuts, or etc.)! There is a certain set $X$ of equivalence classes of Cauchy sequences such that when we look at "$\mathbb{Q}+X$" and forget the ordering, we get an algebraic structure isomorphic to $F$. So basically, when we add $\sqrt{2}$ to $\mathbb{Q}$ in "the geometric setting" - and add a few more points to get good closure properties - we wind up with strictly more information than is provided by the purely algebraic construction of $F$. Moreover, the "extra points" we need (basically $X\setminus\{\sqrt{2}\}$) are determined in a straightforward way, so it's not that much extra information.

The relevant theorem here is: "For any set $Y$ of equivalence classes of Cauchy sequences, there is a smallest set $X_Y$ of equivalence classes of Cauchy sequences such that $Y\subseteq X_Y$ and "$\mathbb{Q}+X_Y$" is a field, and moreover $X_Y$ has a snappy description" (that last bit being a good exercise). Above, $X=X_{\{q\in\mathbb{Q}: q^2<2\}}$.

In this sense, the "geometric" approach provides strictly more information than the algebraic approach. On the other hand, it's not much more information: the two orderings on $F$ are isomorphic in the obvious way, so up to automorphism $F$ is a uniquely orderable field. So ultimately the two approaches aren't that far apart.

Incidentally, it's worth noting just for fun that $\mathbb{R}$ is in fact a truly-uniquely-orderable field since we can recover the ordering from the algebraic structure: $a\le b$ iff $\exists c(c^2+a=b)$. This is not in general true, to put it mildly, but it's cool.

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The fundamental difference between them is that they generalize differently: Cauchy sequences generalize to arbitrary metric spaces without any required algebraic structure, while ring extensions/quotients generalize to arbitrary rings without any required geometric structure.

Answer 2

One way to reconcile the two constructions is to note that $K$ is isomorphic to $\mathbb Q(\sqrt2) \subset \mathbb R$ under an isomorphism that takes $\bar x$ to $\sqrt2 \in \mathbb R$.