### Multiple Operators

As one who majored in MAE with a minor in math and computer programming, I did find it more intuitive to treat a/bc as a/(bc). I did not until recently swap camps when I realized that in several problems where multiple operators were present, the commutative and associative properties of multiplication would go away. Not only that, if we look at another formerly ambiguous expression (for those who forget that association and commuting do not work with subtraction):

a - b + c = (a - b) + c
a - b + c ≠ a - (b + c)

Likewise:
-x² = -(x²)
-x² ≠ (-x)²

So for if nothing more than the sake of consistency:

a/bc = (a/b)(c) [where distribution still works]
a/bc ≠ a/(bc)

We all agree that the two following statements are true:

1. a/b/c = a/(bc)
2. a/b × c = (ac)/b

If a/bc = a/(bc), then a/bc = a/b/c

Which seems nonsensical to me.

But if a/bc = (ac)/b, then a/bc = a/b × c

Which to me makes more logical sense since the two sides of the equation both deal with multiplication (one implied and one expressed)

Just my \$0.02

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