Proof of multiplicative commutativity for all real numbers I have seen proofs for commutativity for all integers, and these can be extended to rationals easily because a rational number is just the ratio of two integers. However, I have yet to see a proof that multiplication of real numbers is commutative. How would you prove this one?
 A: Let $x, y \in \mathbb R$.  Then there exist two sequences of rational numbers $\{q_n\} \rightarrow x$ and $\{p_n\}\rightarrow y$.
It's a standard exercise to prove that if $\lim p_n = x$ and $\lim q_n = y$ then then $\lim p_n*q_n = \lim q_n*p_n = x*y$
....
$p_nq_n - xy = (p_n -y)(q_n - x) + y(q_n - x) + x(p_n - y)$
$(p_n -y)(q_n - x) = p_nq_n - xy - y(q_n-x) + x(p_n-y)$
For $\epsilon > 0$ let $n > N$ imply $|p_n - y| < \sqrt{\epsilon}$ 
and $n > M$ imply $|q_n - x| < \sqrt{\epsilon}$.  
So for $n > \max(N,M) = K$ we have $|(x - q_n)(y-p_n)| < \epsilon$
So $\lim (q_n -x)(p_n -y) = 0$.
$\lim p_nq_n - xy - \lim y(q_n-x) + \lim x(p_n-y) = 0$
So $xy = \lim p_nq_n = \lim q_np_n = yx$.
A: 
This is done very detailed in this paper for the construction of $\mathbb R$ from the ring of integers via quasi-homomorphisms : Street : The efficient real numbers

So if you are ok with commutativity for integers, there is not even the need to go through rationals. All others properties of $+,\times$ are also prooved in theorem 10.


Here is a quick summary if you don't want to read everything :
The motivation of this construction of $\mathbb R$, is that if we define $f_x(n)=\lfloor nx\rfloor$ then $\lim\limits_{n\to\infty}\frac{f_x(n)}n=x$ so we would like to map $f_x$ to the corresponding real $x$, but of course this would be biting our tail to do it via the floor function.
If $x\neq y$ you can notice that $|f_x(n)-f_y(n)|\to+\infty$ this is the property of $\mathbb R$ to be archimedian. So we will say that $x=y\iff f_x\sim f_y$ if their difference is bounded.

Also, notice that if $i$ is an integer then $f_i(n)+f_i(m)=f_i(m+n)$, so $f_i$ is a homomorphism.
So, instead we will say that if $|f(m+n)-f(m)-f(n)|\le k$ is bounded then this quasi-homomorphism $f$ can be assimilated to a real.
The next step is to show three lemmas : $\begin{cases}f(n)\le a|n|+k\\|f(mn)-mf(n)|\le(|m|+1)k\\|nf(m)-mf(n)|\le(|m|+|n|+2)k\end{cases}$ 
for any $(n,m)\in\mathbb Z^2$.

Now addition is defined ($f+g)(n)=f(n)+g(n)$ and multiplication is defined as $(fg)(n)=f\circ g(n)=f(g(n))$.

The sketch for commutativity of multiplication, then goes as the following :
$|nf(g(n))-ng(f(n))|=|nf(g(n))\underbrace{-g(n)f(n)+f(n)g(n)}_{\text{commutativity in } \mathbb Z=0}-ng(f(n))|$
$\le|nf(g(n))-g(n)f(n)|+|ng(f(n))-f(n)g(n)|\quad$ [triangular inequality]
$\le(g(n)+n)k_1+(f(n)+n)k_2\quad$ [lemma 2]
$\le nk_3+nk_4\le nk_5\quad $ [lemma 1]
So in the end we have $|f(g(n))-g(f(n))|\le k_5$ which means that $fg=gf$ $(*)$ 
$(*)$ remember that quasi-homomorphisms are equal when they are in the same equivalence class, which is that their difference is bounded in this context.
