1. Jan 21, 2006

### C0nfused

Hi everybody,
This may have been discussed before ( and by me) but I would like to see a proof of the following statement:
" (ab)+(ac)=ab+ac"

Ok let me explain a bit. When we want to set some axioms about the real numbers we say that we define two functions from RxR->R, addition(+) and multiplication, with the following properties:

...etc..... a(b+c)=ab+ac....etc

Does the above imply that multiplication precedes addition? I mean, we define how parentheses are used, but only by looking in the above property we can assume that multiplication precedes addition and that it's the same writing ab+ac and
(ab)+(ac)? Some have told me that the order of operations is simply a convention, that there's not much relationship with the axioms of algebra and that it could have been done in another way. But i think that almost anything,including the order of operations, can be derived from the axioms. A proof could be to interprete the right-hand side expression in all possible ways and see with examples that it's not equal to the left-hand side expression, but i think there must be another one,more formal.

Sorry for saying so much about this. Hope u won't find it as stupid as it sounds at first...

Thanks

Last edited: Jan 21, 2006
2. Jan 21, 2006

Pretend for the moment that instead of * (or nothing) for multiplication, we use $, and instead of +, we use &. Then your asking how to prove that: a$b&a$c=(a$b)&(a$c) But as you see now, it is ambiguous how to interpret the left-hand side of the equal sign, since these are all binary operations. It is simply a convention that expresions written as on the left are to be interpreted as on the right. 3. Jan 21, 2006 ### HallsofIvy No, it's not a stupid question- it's a very fundamental one. However, such fundamental properties tend to have very "deep" proofs- and depend on exactly how you define the number systems. If you are interested, here is a link to an essay I wrote on fundamental properties of our number systems. The proof of the (right)distributive law for the natural numbers is theorem 6 on page 6. Distributive laws for the integers, rational numbers, and real numbers are proved later on. http://academic.gallaudet.edu/courses/MAT/MAT000Ivew.nsf/ID/918f9bc4dda7eb1c8525688700561c74/$file/NUMBERS.pdf [Broken]

Last edited by a moderator: May 2, 2017
4. Jan 21, 2006

### Hurkyl

Staff Emeritus
Order of operations is a purely linguistic thing -- it has to do with how we write mathematics, as opposed to any particular mathematical theory.*

The easiest way to handle, and to internalize, is to say that fully parenthesized arithmetic expressions (FPAE's) are the only "right" way to write down an arithmetic expression -- omitting parentheses is simply shorthand, and we use the order of operations convention to figure out how the parentheses are supposed to go.

If you're really not happy with that, then things get considerably more abstract. "ab + ac" is merely a meaningless string of symbols from a particular grammar. One specification of this grammar is (I hope I have the syntax for this right: it's been a while):

Code (Text):

Expression :: Term | Expression + Term
Term       :: Factor | Term * Factor
Factor     :: (Expression) | Variable | Constant
Variable   :: a | b | c | ...
Constant   :: 0 | 1 | -1 | 2 | -2 | ...

This means that an expression is something that is either:
(1) A term
(2) An expression, followed by the symbols " + ", followed by a term.

A term is something that is either:
(1) A factor
(2) A term, followed by the symbols " * ", followed by a factor.

et cetera.

Given values for the variables, we can then use these grammar rules to recursively define an interpretation of an expression in, say, the theory of rings.

If we write v(a), v(b), ... for the values of the variables, then we define things like:
If T is a term, and F is a factor, then v(T * F) = v(T) * v(F).

And then for any expression E, we interpret it as the value v(E)

And then you can prove a theorem that this recursive definition is well-defined. (I think the above grammar is nice -- you can only parse an expression in one way. It's also context-free, which is nice)

You could define different grammars. For example, you could have one that allows omitting the * sign for multiplication.

But in any case, all of this has absolutely nothing to do with any sort of algebraic axioms.

*: Well, I suppose the study of (formal) languages is an exception.

5. Jan 21, 2006

### C0nfused

Thanks for all your answers. From what u have written I got that we can accept the statement in question as a convention we have made to simplify our mathematical notation. I have thought of it and actually I have started doubting about my initial question. So I have come to this question:

"if we say that a(b+c)=(ab)+(ac) then it's very clear what we mean
but if we write a(b+c)=ab+ac then does this imply that multiplication precedes addition or should we say that the right-hand side has absolutely no meaning as long as we haven't defined how series of binary operators are interpreted? "

StatusX answers this question and I think he is probably right. HallsofIvy (->I will check your paper but I don't have much time now...) seems to agree and so does Hurkyl. As for the grammar stuff that Hurkyl mentioned, I am not familiar with it so I am not sure I understand much, though it reminds me of the way we study programming theoritically,u know (recursive) syntax diagrams etc

Anyway thanks again for your answers and I hope u help me clear this up soon.

6. Jan 21, 2006

### Hurkyl

Staff Emeritus
It should! The relevant areas of Mathematics and Computer Science are inseparable.

7. Jan 22, 2006

### C0nfused

I checked wikipedia and found this: "These precedence rules (which are mere notational conventions, not mathematical facts)..." It also defines distributivity like this: x*(y+z)=(x*y)+(x*z). So I guess that notation depends mostly on conventions made and not on mathematical facts (if we don't want to include grammar theories), right?

Thanks

Last edited: Jan 22, 2006
8. Jan 22, 2006

### D H

Staff Emeritus
x*(y+z) is not defined to be (x*y)+(x*z). Multiplication of two natural numbers is defined in terms of repeated addition. This definition is then extended to the integers, the rationals, the reals, .... That x*(y+z) is equal to (x*y)+(x*z) is provably true based on the everyday definitions of multiplication and addition.
The Wikipedia article properly says that an operation "*" distributes over an operation "+" defined on some set S if x*(y+z)=(x*y)+(x*z) for any elements x, y, z of S. (To be most correct, that is only left-distributivity.) It could just as well have defined distributivity in terms of the operations $\square$ and $\circle$. Since this relationship is indeed true for the everyday definitions of addition and multiplication we can say that multiplication distributes over addition. The notation depends very much on mathematical facts.

Last edited: Jan 22, 2006
9. Jan 22, 2006

### Hurkyl

Staff Emeritus
But the (left) distributive law (of * over +) is defined to be the equation

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

(or something equivalent)

which is something that may or may not hold in a mathematical theory.

10. Jan 22, 2006

### D H

Staff Emeritus
Hurkyl, that is my point. a*(b+c) is not defined to be a*b + a*c. Instead, this relation is provably true for "everyday" multiplication and addition (i.e., the operations as used by non-mathematicians). Thus multiplication of the integers/reals/complex numbers distributes over addition of the integers/reals/complex numbers because these operations over these sets satisfies the definition of distributivity.

The one thing none of us has done is to answer the original post. I will proceed to outline a proof that a*(b+c)=ab + ac for the natural numbers (i.e, 0, 1, 2, ...).

Addition of the natural numbers is defined recursively via successive applications of successor operand S:
• n+0 = n for any natural number n
• n+1 = S(n)
• n+1+...+1= n + m = S(...S(n)...)

Multiplication of the natural numbers is defined recursively via
• n*0 = 0 for any natural number n
• (n+1)*a = na + a for any two natural numbers n and a

Given any two natural numbers a and b, a*(b+0) = a*b + a*0 since a*b = a*(b+0), a*0 = 0, and a*b + 0 = a*b. Repeatedly applying the successor operation to the "0" inside the parentheses on the left side of the equation and adding a on the right side of equation yields a*(b+c) = a*b + a*c for any natural numbers a, b, and c.

11. Jan 23, 2006

### HallsofIvy

Yes, that is what is implied by the way it is written, not any deep mathematical reason.

If we agreed to use the convention that "addition precedes multiplication" we would still have the distributive law but would write it as
ab+c= (ab)+ (ac). It's notation, not mathematics.