How is rearranging $56\times 100\div 8$ into $56\div 8\times 100$ allowed by the commutative property? So according to the commutative property for multiplication:
$a \times b = b \times a$ 
However this does not hold for division
$a \div b \neq b \div a$ 
Why is it that in the following case:
$56 \times 100 \div 8 = 56 \div 8\times 100$
It seems like division is breaking the rule. There is something I am misunderstanding here. 
Is it because $a\times b\div c=a\div c\times b$ is allowed since $b\div c$ are not being rearranged so that $c\div b$? 
If this is the case are you allowed to rearrange values in equations so long as no values have the form $a \div b = b \div a$ and $a - b = b -a$ ?
 A: Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.
So $a\div b \times c$ is being interpreted from left to right as $(a\div b)\times c=\cfrac {ac}b$, but done from right to left $a\div (b\times c)=\cfrac a{bc}$ and the two results are not the same.
Likewise with $a\div b \div c$ we have $(a\div b)\div c=\cfrac a{bc} \neq a\div (b\div c)=\cfrac {ac}b$.
So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.
A: Notice that you have always $\div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $\div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$a\times b\div c=a\times b\times d=a\times d\times b=a\div c\times b$$
A: Because division is the inverse of multiplication, that is: $$X \div Y =X\cdot \frac 1Y$$
So you have: $$56\cdot 100 \cdot \frac18 =56\cdot\frac18\cdot100$$
Which is obvious.
A: Hope this makes sense.
$$ a\times b ÷ c $$
$$=a\times b\times\dfrac{1}{c}$$
$$=(a\times\dfrac{1}{c})\times b$$
$$=\dfrac{a}{c}\times b$$
$$=a÷c\times b$$
A: Start with 56 × 100 ÷ 8.
Swap the first two factors which is definitely allowed: 100 × 56 ÷ 8.
Now put parentheses around part of it: 100 × (56 ÷ 8).
Again, swap: (56 ÷ 8) × 100.
Take off the parentheses: 56 ÷ 8 × 100.
And there you go!  transforming one into the other using only multiplicative commutation (plus the fact that one of the factors commutated can be a product, not just a simple value).
A: The non-zero real numbers form an abelian (commutative) group under multiplication. The notation $a\div b$ is shorthand for $ab^{-1}$. 
So $ab^{-1}\ne ba^{-1}$ but $56\times100 \times8^{-1}=56\times 8^{-1}\times 100$
A: The exact same situation occurs with addition and subtraction instead of multiplication and division. Do you find it hard to justify rewriting say $a+b-c$ to $a-c+b$? And if not, which laws are you using to justify this? I would say associativity, commutativity, and the special equivalence $x-y=x+(-y)$ (which can be construed to be the definition of subtraction, although historically and in teaching subtraction is introduced before introducing negative numbers). In detail one has inserting parentheses to make explicit the one implied by convention)
$$\eqalign{
  a+b-c &= (a+b)-c \\&= (a+b)+(-c) \\&= a+(b+(-c)) \\&
= a+((-c)+b) \\&= (a+(-c))+b \\&= (a-c)+b &= a-c+b}
$$
By perfect analogy, let us do that with '$\times$' replacing '$+$', and '$\div$' replacing '$-$':
$$\eqalign{
  a\times b\div c &= (a\times b)\div c \\&= (a\times b)\times (\div c) \\&= a\times (b\times (\div c)) \\&
= a\times ((\div c)\times b) \\&= (a\times (\div c))\times b \\&= (a\div c)\times b &= a\div c\times b}
$$
Note how I just invented the unary use of '$\div$', where $\div x$ of course means the multiplicative inverse of$~x$, just as $-x$ means its additive inverse. Now why did nobody think of that before? It could have avoided the ugliness of$~x^{-1}$, which has to borrow exponentiation and additive inverse to get multiplicative inverse.
A: With multiplication and division alone (no addition or subtraction) they are associative with respect to one another, BUT division itself is NOT commutative.
So basically you can do the multiplication or division in either order, BUT you must respect which way you interpret the inputs to the left or right of the operator's sign. If you flip this it has the effect of flipping the inputs which is equal precisely when the operator is commutative. As you noted, with division, it is not commutative.
