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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 $a - b = b -a$ ?

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  • $\begingroup$ @marshalcraft: There is no reasonable sense in which multiplication and division are associative "together" -- as Mark Bennet's answer points out, $a/(b\cdot c) \ne (a/b)\cdot c$ in general. $\endgroup$ – Henning Makholm Feb 18 at 13:29
  • $\begingroup$ Indeed its an error on my part. $\endgroup$ – marshal craft Feb 18 at 13:54
  • $\begingroup$ So I need to say is multiplication is commutative and associative, division is not associative and not commutative. But when you compute the result of division you remove the division and thus can then proceed to use multiplicative property of commutivity and associativity. K. $\endgroup$ – marshal craft Feb 18 at 14:35
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    $\begingroup$ To would-be editors: please, please do not convert the $\div$ operators in this question into the fractional form $\frac ab.$ It changes the question completely. $\endgroup$ – David K Feb 18 at 14:46
  • $\begingroup$ Also due to the non commutivity, and non associativity of division, the notation used above is decidably ambiguous. That means there are a number of equally valid ways to interpret expressions ( this is why different calculators can give different results). You can explore the possibilities. In higer level math fractions, parents, etc. are used and there is no ambiguity. $\endgroup$ – marshal craft Feb 18 at 15:11
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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: $$a\times b\div c=a\times b\times d=a\times d\times b=a\div c\times b$$

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Hope this makes sense.

$$ a\times b ÷ c $$

$$=a\times\dfrac{b}{c}$$

$$=a\times b\times\dfrac{1}{c}$$

$$=(a\times\dfrac{1}{c})\times b$$

$$=\dfrac{a}{c}\times b$$

$$=a÷c\times b$$

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  • $\begingroup$ The first step is not elementary. The literal interpretation of the order of operations in $a\times b \div c$ is $(a\times b) \div c$. what gives us the ability to take the $b$ out of the first operation and put it into $\frac bc$? I think this answer would be better if it went straight from $a\times b \div c$ to $a\times b \times \frac1c$. $\endgroup$ – David K Feb 18 at 14:53
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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.

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  • $\begingroup$ I don't understand what you are trying to say. How is this any different than with addition and subtraction? If you interpret $a-b+c$ as $a-(b+c)$ then you get a different answer. But you shouldn't do that, so what? See also my answer. $\endgroup$ – Marc van Leeuwen Mar 4 at 13:29
  • $\begingroup$ @MarcvanLeeuwen You are right that it is essentially the same and also that you shouldn't do it - but the fact is that people do get these things confused. If you use BIDMAS (brackets, indices, division, multiplication, addition, subtraction) as a rule, applied without proper procedural knowledge, as I have seen in English Primary Schools, then in your example you would get the second, wrong, answer by doing the addition first and the subtraction afterwards. $\endgroup$ – Mark Bennet Mar 4 at 17:06
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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.

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  • $\begingroup$ While it is true that "division is the inverse of multiplication" the emphasis should be on what follows from it: That division is equivalent to multiplication with the inverse and can be replaced by it. $\endgroup$ – Peter A. Schneider Feb 18 at 8:14
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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 $56\times100 \times8^{-1}=56\times 8^{-1}\times 100$

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    $\begingroup$ (-1) this answer is almost certainly nearly useless; the set of people that would ask this question will be nearly disjoint from the set of people that understand this answer $\endgroup$ – DreamConspiracy Feb 18 at 9:05
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    $\begingroup$ Additionally, it's not even clear he asking over reals, could be over natural numbers in which case there is no group with multiplication as group operation. $\endgroup$ – marshal craft Feb 18 at 11:18
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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).

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    $\begingroup$ Now explain why putting parentheses around the last two terms in $100\times56\div8$ is OK but putting parentheses around the last two terms in $56\div8\times100$ is not OK. For that matter, why is it OK to swap the first two terms in $100\times56\div8$ but not the last two terms of $56\div8\times100$? $\endgroup$ – David K Feb 18 at 14:43
  • $\begingroup$ The short answer to adding the parentheses is, "because it doesn't change the outcome". But you're right, a more rigorous explanation is needed, and I don't have one. To swap the last two terms of 56 ÷ 8 × 100 does require parentheses around them, and if they aren't already there, they can't be arbitrarily added without changing the whole thing. $\endgroup$ – Jennifer Feb 18 at 14:51
  • $\begingroup$ A formal reason is that $56\times100\div8$ means $(56\times100)\div8$ by convention, and $(56\times100)\div8=(100\times56)\div8=(100\times56)\times\frac18=100\times(56\times\frac18)=100\times(56\div8).$ But that's making it more complicated than necessary. I think it's simpler if we start with $(56\times100)\div8=(56\times100)\times\frac18$. $\endgroup$ – David K Feb 18 at 14:58
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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.

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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.

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  • $\begingroup$ No, as Mark Bennet explained. $\endgroup$ – Henning Makholm Feb 18 at 11:51
  • $\begingroup$ Okay if you vote so. $\endgroup$ – marshal craft Feb 18 at 11:54
  • $\begingroup$ Mark Bennets answer is incorrect. You CAN read from either right or left, provided you interpret the same operator symbol the same. So input on left of operator goes to numerator, etc. If you don't change convention you can read the combined expression from either left or right. It's indisputable that there are two levels of right and left. This is not addressed. Yet fundamental to the combined notion of associative and commutative. $\endgroup$ – marshal craft Feb 18 at 12:07
  • $\begingroup$ You clearly lack the prerequisites to answer this question. That is all you're displaying for all to see? $\endgroup$ – marshal craft Feb 18 at 12:08
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    $\begingroup$ Mark Bennet's answer shows that $a\div (b\times c)$ is not generally equal to $(a\div b)\times c$. His arithmetic is completely correct. You seem to be taking a view in which the operators $\times$ and $\div$ are not binary operators but are merely flags of some kind within an $n$-tuple of factors. I don't think that view is widely shared. $\endgroup$ – David K Feb 18 at 14:37

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