Watched Khan Academy, now I’m confused by division (noncommutative) being the inverse of multiplication (commutative) I was watching a khan academy video about division which described division in two different ways and I’m now confused by my interpretation of division as the inverse of multiplication.
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I understand why multiplication is commutative: 12 objects could be counted as 3 sets of 4 or counted as 4 sets of 3 with the same result:
My interpretation of how 3x4=12 AND 4x3=12
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However, when he introduced division he asked ‘what is 8÷2?’, and said it was the answer to the question ‘how many groups of two can be made out of eight?’, and he drew:
8 objects being grouped into groups of 2
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4 groups of 2 could be made, so the answer is 4.
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But, he then immediately said ‘Or it’s like the answer to the question - split 8 into 2 equal groups, how many are in each group?’, and he drew:
8 objects being grouped into 2 equal groups
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4 objects were in each group, so the answer is 4.
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I was confused to see two different ways to model division, since there’s only one way to write it.
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With multiplication there are two ways to write the expression, and they correspond to two ways to model the situation - and it all demonstrates the commutative nature of multiplication.
Thinking of division as the inverse of multiplication, it’s like the two models of division are recovering the two models of multiplication and it all fits together - the commutativity of multiplication is still ‘in’ division somewhere. However, whereas multiplication has notation for its two commutations, it seems like there’s a discrepancy with the division notation. It’s like there’s information lost somewhere.
It feels like division wants commutativity, it just doesn’t know how.
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Q1: What have I misunderstood, how can the same notation ‘a÷b’ describe two different processes?
Q2: Does mathematics talk about information being lost by having something which is noncommutative as the inverse of something which is commutative?
Q3: What area(s) of mathematics have I touched on/where can I go for more information?
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Thanks so much for your time!
 A: There is no "information being lost" here. The fact that division describes the result of two different processes actually corresponds directly to the first fact you mentioned: two different processes give the same result under multiplication. To put it another way, the two different processes describing division come from commutativity of multiplication, not some "missing" commutativity of division.
When you ask "what is $8$ divided by $2$?", you are asking "what number, when multiplied by $2$, gives $8$?" That is, $8 \div 2 = \mathord{?}$ means the same thing as $8 = \mathord{?} \times 2$, i.e., how many groups of size $2$ do you need to make $8$? The answer is $8\div 2 = 4$, because $4\times 2 = 8$, i.e., $4$ groups of $2$ make $8$. 
But we also know that $\mathord{?}\times 2 = 2\times \mathord{?}$. If we take some number $\mathord{?}$ of groups of size $2$, we'll have the same number of things as if we take $2$ groups of that same size $\mathord{?}$. 
So we could equally well say that $8 \div 2 = \mathord{?}$ means the same thing as $8 = 2\times \mathord{?}$, i.e., $2$ groups of what size make $8$? The answer is $8\div 2 = 4$, because $2\times 4 = 8$, i.e., $2$ groups of $4$ make $8$.
A: I will address $Q1$. I don't think I know the answer for the other two questions. The simbol $a \div b$ denotes a number and this same number is obtained in the two procedures you described. If you wish to be more simbolic we can denote the first division by $a \div b$ and the second one by $a \div' b$. Let's show that this two numbers are the same. Notice that by your definition of multiplication we have 
$$a=b(a \div b) = b(a \div' b)
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and by associativity we conclude that
$$b\big((a \div b)-(a \div' b)\big) = b(a \div' b) - b(a \div b) =0.
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Therefore, we have:
$$b\big((a \div b)-(a \div' b)\big)=0.
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Now, $b \neq 0$ and therefore the only way for us to obtain a $0$ in the above identity is if
$$(a \div b)-(a \div' b)=0.$$
From that we obtain
$$a \div b = a \div' b.
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So, the two procedures give you the same number.
