Jtdylktuy

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?

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 $xy = lim p_nq_n = lim q_np_n = yx$.

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 $f_x(n)=lfloor nxrfloor$ then $limlimits_{ntoinfty}frac{f_x(n)}n=x$ 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 $xneq y$ 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 $x=yiff f_xsim f_y$ 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 : $begin{cases}f(n)le a|n|+k\|f(mn)-mf(n)|le(|m|+1)k\|nf(m)-mf(n)|le(|m|+|n|+2)kend{cases}$

for any $(n,m)inmathbb Z^2$.

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

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]

$le(g(n)+n)k_1+(f(n)+n)k_2quad$ [lemma 2]

$le nk_3+nk_4le nk_5quad $ [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.

The following proof is sketchy, but in the next section I test it out with a Python program.

Consider the set of positive real numbers $x gt 0$ where we apply the 'forgetful functor' and only know how to add; so we have to define multiplication in $left( ,(0,+infty),+ ,right)$ and then show that it commutative. But we certainly accept all the other axioms and laws of the real numbers.

Now even thought there is no multiplication, we have no problem 'multiplying' a real number by a positive integer, since that is just shorthand for 'repeated addition'.
Also, there is a real number, call it $2^{-1}$ with the property that

$tag 1 2^{-1} + 2^{-1} = 1$.

And so we can keep taking powers of our 'bisection operator',

$tag 2 2^{-(n + 1)} = 2^{-n} circ 2^{-1}$.

If $x gt 0$ and $n gt 0$, we can write

$tag 3 x = m_{x,n} 2^{-n} + r_{x,n} text{ with } r_{x,n} lt 2^{-n} text{ and } m_{x,n} in mathbb N$

If $y gt 0$ we also write

$tag 4 y = m_{y,n} 2^{-n} + r_{y,n} text{ with } r_{y,n} lt 2^{-n} text{ and } m_{y,n} in mathbb N$

We define the multiplication

$tag 5 x y = limlimits_{n to +∞} m_{x,n}, m_{y,n}, 2^{-2n} $

Since the multiplication of integers is commutative, so is the multiplication of real numbers.

Python Program:

a = 72.987654

b = 87.123456



print(a, b, a*b)



div = 1





for i in range(1,25):

    div = div / 2

    x = divmod(a, div)

    m, g = x

    y = divmod(b, div)

    n, g = y

    print(m*div, n*div, 'seq = ', i, m*n*div*div )

Output:

72.987654 87.123456 6358.936661812225

72.5 87.0 seq =  1 6307.5

72.75 87.0 seq =  2 6329.25

72.875 87.0 seq =  3 6340.125

72.9375 87.0625 seq =  4 6350.12109375

72.96875 87.09375 seq =  5 6355.1220703125

72.984375 87.109375 seq =  6 6357.623291015625

72.984375 87.1171875 seq =  7 6358.1934814453125

72.984375 87.12109375 seq =  8 6358.478576660156

72.986328125 87.123046875 seq =  9 6358.791286468506

72.9873046875 87.123046875 seq =  10 6358.87636756897

72.9873046875 87.123046875 seq =  11 6358.87636756897

72.987548828125 87.123291015625 seq =  12 6358.915457069874

72.987548828125 87.1234130859375 seq =  13 6358.924366682768

72.98760986328125 87.1234130859375 seq =  14 6358.929684273899

72.98764038085938 87.12344360351562 seq =  15 6358.934570475481

72.98764038085938 87.12344360351562 seq =  16 6358.934570475481

72.9876480102539 87.12345123291016 seq =  17 6358.935792026168

72.98765182495117 87.12345504760742 seq =  18 6358.936402801555

72.9876537322998 87.12345504760742 seq =  19 6358.936568976358

72.9876537322998 87.12345504760742 seq =  20 6358.936568976358

72.9876537322998 87.12345552444458 seq =  21 6358.9366037795835

72.98765397071838 87.12345576286316 seq =  22 6358.936641953047

72.98765397071838 87.12345588207245 seq =  23 6358.936650653853

72.98765397071838 87.1234559416771 seq =  24 6358.936655004256