I suppose I'm looking for more of an intuitive answer here.
Recall if we raise a number $x$ to an $a$ power where $a$ is an irrational, then $x^a$ can defined as the limit of $x^{q_n}$ with $a$ as the limit of the sequence {$q_n$}. Since we know intuitively what it means to raise a number to a rational power.
Can multiplication by an irrational number be defined similarly? (Since we know what it means to multiply a number by a rational)
Yup. We can define $\alpha\cdot\beta$ - for $\alpha,\beta$ possibly irrational real numbers - as the "double limit" $$\lim_{p\rightarrow\alpha,q\rightarrow\beta, p,q\in\mathbb{Q}}p\cdot q.$$ This of course raises a couple questions:
We need to prove that this double limit always exists.
We need to prove that in case $\alpha,\beta$ are rational, this agrees with multiplication as usually defined.
But these aren't hard to prove.
And of course we can do the same with addition, etc.
In fact, there's a general abstract principle at work here: in general, since $\mathbb{R}$ is the completion of $\mathbb{Q}$, any "very continuous" function $f:\mathbb{Q}^n\rightarrow\mathbb{Q}$ extends uniquely to a continuous function $\hat{f}:\mathbb{R}^n\rightarrow\mathbb{R}$. Defining "very continuous" here isn't trivial - note that mere continuity isn't enough (the function sending rationals $<\sqrt{2}$ to $0$ and rationals $>\sqrt{2}$ to $1$ is continuous with respect to the topology on $\mathbb{Q}$). The point is that addition, multiplication, etc. are "very continuous" in the right sense, so definitions a la the above actually work.
Incidentally, all of this presupposes a rigorous definition of real numbers - if you're not familiar with this already, the two usual approaches are Cauchy sequences and (less commonly and less broadly useful, but cooler in my opinion) Dedekind cuts.
In essence that is how we define multiplication on the reals.
The rationals is an order field with ordered field axioms and definitions. It is the most basic of ordered fields as it is precisely the field one generates with only the axioms and definition. (I.E. with the elements $1$ and $0$ and the axioms)
But the rationals do not have the least upper bound property. The definition of the reals is the "smallest" ordered field with the least upper bound property with $\mathbb Q$ as a subfield.
But before we define them as such we must prove such a field exist and define what it's elements are and what the operations are.
I say, in essence, the real numbers are the limits of cauchy sequence or rationals and addition and multiplication the limits of sums and products of the terms (sort of) but that's getting ahead of ourselves as the definition of limits comes after the definition of the reals.
Instead we must define the reals and multiplication and addition in terms of the rationals and the least upper bound property. Traditionally this is done by defining the reals as sets of rational numbers that are bounded above but have no maximum elements (Dedekind cuts). In other words the real are equivalent to these sets and when we think of the reals as numbers (rather than as sets) these are what the supremema of such sets would be. And multiplication is defined, surprisingly awkwardly, as the sets of products of elements of two such sets.
Essentially, and with violation of circularity of concepts, $x$ is defined as $\{q\in \mathbb Q| q < x\}\cong \sup \{q\in \mathbb Q| q < x\}$.And $xy = \sup \{q\in \mathbb Q| q < x\}*\sup \{p\in \mathbb Q| p < y\}=\sup \{qp; q,p\in\mathbb Q| q< x; p < y\}$
Which essentially is definition by limits.... well, least upper bounds.
