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.

  • 1
    $\begingroup$ 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

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.

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.