Ordering the Operations

Date: 11/10/98 at 22:51:00
From: Amy Greenburg
Subject: PEMDAS

In my math class we are studying PEMDAS. Why does the order of 
operations have to be in that order? Who made up PEMDAS? 

Thank you for your time.

Date: 11/11/98 at 12:38:53
From: Doctor Peterson
Subject: Re: PEMDAS

Hi, Amy -

People generally say that the order of operations is nothing more than 
an arbitrary convention - that is, there had to be some rule so 
everyone would read an expression the same way, so they just chose a 
rule. I don't think any one person made the decision, but it just 
gradually developed as the modern symbols for algebra and arithmetic 
developed. But I think there is a good reason that the traditional 
order was agreed upon without any arguments.

That reason is the distributive rule, which we write as:

   a * (b + c) = a * b + a * c

If we reversed the order of operations, doing addition before 
multiplication, we would write it this way:

   a * b + c = (a * b) + (a * c)

Do you see the difference? In our usual form, we can say that the 
multiplication distributes over the terms in parentheses. The 
parentheses are required because the addition has to be done first. 
But in the reversed form, the parentheses aren't needed there, so the 
distribution isn't nearly as obvious.

For the same reasons, polynomials would be more awkward to write, since 
each term would require parentheses.

To put it more simply, we do multiplication before addition because 
multiplication distributes over addition; multiplication is in some 
sense "more powerful" by nature.

Similarly, exponentiation distributes over multiplication, so we do 
that first:

   (a * b)^c = a^c * b^c

would be written as:

   a * b ^ c = (a^c) * (b^c)

if we did multiplication before exponents, and that isn't as clear.

Note, by the way, that exponentiation distributes only in one 
direction. Because it is not commutative, it is not true that:

   a^(b * c) = a^b * a^c

but rather:

   a^(b * c) = (a^b)^c

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/