One of the responses to the rather large thread on PEMDAS got me thinking about this. I teach middle school math, and we always teach order of operations to 6th graders. I have always explained that PEMDAS is just the set of rules for evaluating mathematical expressions; no student has ever asked me "why are these rules the way they are?"

Is it just arbitrary that multiplication and division come before addition and subtraction in PEMDAS, or is there a reason? In algebra, when we get to 2-step equations (3X + 12 = 24), I have always suggested undoing addition and subtraction first, then undoing multiplication and division. But PEMDAS teaches "doing" multiplication and division first. No wonder many kids get confused.

When you **evaluate** an expression, you "do" PEMDAS (mult and div first), but when you **solve** an equation, you "undo" (add and sub first). Is there a logical way to explain this to students?

Evaluating an expression means you have to reduce it to lowest terms. You do this by taking apart the most complicated, highest, parts first (driving into the parenthesis and other groupings and evaluating them first, bringing back a number to combine with other numbers); then the middle parts (exponents being repeated multiplication, then multiplication being repeated addition and division repeated subtraction) are dismantled, and finally the parts you've found are summed.

It's like that tower of disks on the post puzzle. Add and subtract are on the bottom, multiply and divide on that, exponentiation on that, and parenthesis on the top. You can't take off the bottom disk until you have taken off the others.

We use polynomials a lot. Without *EMDAS* we'd have to write: *p(x) = (a(x*^{3}))+(b(x^{2}))+(cx)+d. With *PASMDE* we'd still need to write the same expression. Only *PEMDAS* allows us to get rid of the parentheses and write: *p(x) = ax*^{3}+bx^{2}+cx+d.

In addition to that the distributive law of multiplication would look weird:

*a*b+c = (a*b)+(a*c)*. Even more so with implied multiplication:

*ab+c = (ab)+(ac)*.

Maybe I didn't understand you correctly but IMHO *PEMDAS* has nothing to do with the order of solving this equation.

Division first, then subtraction:

3X + 12 = 24

X + 4 = 8

X = 4

Subtraction first, then division:

3X + 12 = 24

3X = 12

X = 4

Kind regards

Thomas

[quote]

Even more so with implied multiplication:

*ab+c = (ab)+(ac)*.

[/pre]

You confused me there. I think you would rather have implied addition if addition had precedence over multiplication:

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

I just came across this on youtube. Interesting.

www.youtube.com/watch?v=y9h1oqv21Vs&feature=youtube_gdata

Don

Interesting read on the **Math Forum** at this url...

http://mathforum.org/dr.math/faq/faq.order.operations.html

*The order of operations in which one is to interpret a mathematical expression such as "2+3 * 5" is a convention. This means that a long time ago, people just decided on an order in which operations should be performed. It has nothing to do with magic or logic. Some people decided to adopt a way, and it has stuck ever since. It just makes communication a lot easier.*

Another way of saying this is that rather than being inherent in the structure of mathematics, the concept of "order of operations" is a matter of mathematical notation. Order of operations refers to which operations should be performed in what order, but it's just convention. The notation tells you which operations to do first, not something about the underlying mathematics.

..continued at this url...

http://mathforum.org/library/drmath/view/57289.html

*The PEMDAS convention currently taught in schools, ... tells us ... as long as we're working with people who also follow that convention, we can leave the grouping symbols out. That doesn't seem like such a big benefit, but if you're writing hundreds or thousands of expressions each day, then those little savings can add up.*

*
This particular convention is optimized to make it possible to write polynomials without grouping symbols.*

SlideRule

Wouldn't that rather be *a*bc = (a*b)(a*c)*?

Guys! This forum is serious, wake up pls!

a*bc = ab*c

while (a*b)(a*c)=a^2*bc

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this is the clearest of the explanations, and i'd hope would be the end of the matter. but what of the millions (or billions?) of pocket calculators made over the years, telling all who will listen that 1+2*3=9?

yes, this is a loaded question; 1+2*3=9 because in the early years of the calculator, it was (from an engineering perspective) much easier to leave out operator precedence. yet long after the engineering issues were banished, manufacturers continued to build calculators that get it wrong. certainly for at least 30 years, there has been no excuse for 1+2*3 to equal anything other than 7.

so why do we still manufacturer and distribute to the mass population calculators that don't follow what we so desperately attempt to teach in schools?

xD

No, obviously the formatting options here are not user friendly.

I meant: a^{2}bc = a^{2}*b*c

Distributive law of multiplication:

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

With *PEMDAS*:

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

With *PEMDAS* and *implied multiplication*:

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

With *PASMDE*:

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

With *PASMDE* and *implied multiplication*:

*ab + c = (ab) + (ac)*

With *PASMDE* and *implied addition*:

*a*bc = (a*b)(a*c)*

HTH

Thomas