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Say you have 3 apples and 2 oranges, and you want to multiply these two groups of fruits together to obtain a desired result, for instance:

A. You want 3 apples for each orange, so you have 6 apples and 2 oranges.

B. You want 2 oranges for each apple, so you have 6 oranges and 3 apples.

Now, lets say you want both A and B, 3 apples per 2 oranges, and 2 oranges per 3 apples, so you have 6 of each which = 12 'fruits'/objects/items etc.

This seems to be the only three multiplicative operations that can be applied when dealing with 2 groups or varying objects, and as a result, 2*3 seems to produce a different outcome than 3*2. So when working in the real world, with actual, physical objects, is multiplication ever really commutative?

Edit: So by what Sloan has shown, (hey, that rhymes!), it seems that the first scenario can be modeled by (3a*2o)/o = (6ao)/o = 6a. Where the second the variables are just reversed, 2o*3a/a = 6o.

I was confused with the third scenario however, where you just multiply 3 apples and 2 oranges together to produce 6 of each fruit, and this is why multiplication seems to be commutative, since 3*2 = 2*3.

However, it seems to be that 3*2/1 can produce different outcomes, where the distinction only makes sense when the numbers are associated with variables, like a and o.

So, it just depends on what 1 represents when used as the denominator, which to me is a very interesting property of mathematics.

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    $\begingroup$ This question is unclear. How are $2*3$ and $3*2$ producing different outcomes? $\endgroup$ – Mike Pierce Jun 23 '15 at 19:43
  • $\begingroup$ That's explained via the situations presented. You can have 2 oranges for each apple (of which there are 3), so you have 2 oranges for apple 1, 2 for apple 2, 2 for apple 3, so you have 2+2+2 oranges. If you have 3 apples for each orange (of which there are 2), you have three apples for each orange, 3 for orange 1, 3 for orange 2, so there's 6 apples. 6 oranges is not the same as 6 apples. $\endgroup$ – Jim Jam Jun 23 '15 at 19:47
  • $\begingroup$ In your example you are also not just simply multiplying by two objects but you are seemingly including some other unit $\endgroup$ – Quality Jun 23 '15 at 19:51
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To be clear, $3 \times 2$ and $2 \times 3$ are yielding the same result of $6$. However, if we tamper with the units (as you have), we see that $3 (\frac{apple}{orange}) \times 2(orange)$ is a different multiplication than $2 (\frac{orange}{apple}) \times 3 (apple)$, so we should expect the results to be different.

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    $\begingroup$ Thanks Sloan, I think I was confused in regards to the varying ways an expression could be well, expressed. This is primarily because I have been doing maths for a while, but only conceptually, so now I need to learn how to apply maths in the real world, thanks again. $\endgroup$ – Jim Jam Jun 23 '15 at 19:57

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