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## How do we prove the distributive property of multiplication?

Guess I have a lot to learn.
After reading through, it seems like there's no actual proof of commutativity.

By the way, why are IMG codes disabled?

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 Quote by greswd After reading through, it seems like there's no actual proof of commutativity.
What would be an "actual proof"?

 Quote by greswd By the way, why are IMG codes disabled?
I don't know. Maybe something went wrong when the forum changed to a new style.

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 Quote by Fredrik What would be an "actual proof"?
Good question. I really don't know.

I've always assumed that mathematical proofs are final and irrefutable.

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 Quote by greswd How to prove that 3 x 2 = 2 x 3?
a 2 by 3 rectangle of six squares groups the squares in two ways - 3 groups of 2 and 2 groups of three.

The distributive law can be one with cubes.

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 Quote by lavinia a 2 by 3 rectangle of six squares groups the squares in two ways - 3 groups of 2 and 2 groups of three. The distributive law can be one with cubes.
yeah, the intuitive geometric proof.

but it could be that we describe geometry in commutative terms, and not the other way round.

 Recognitions: Gold Member Science Advisor Staff Emeritus How you prove a basic property like "commutativity" depends on your basic definitions. As I said before, we can start with "Peano's axioms" and define the positive integers to be a set, N, together with a function, s, such that: There exist a unique member of S, "1", such that s is one-to-one and onto from S to S-{1}. If a subset, X, of S contains 1 and has the property that if $x\in X$ then $s(x)\in X$, then X= S. We define addition, x+ y, by x+ 1= s(x) if y is not 1, then there exist z such that y= s(z) and x+ y= s(x+ z). We define multiplication, x*y by x*1= x If y is not 1, then there exist z such that y= s(z) and x*y= x*z+ x. Let X= {x| x*1= 1*x} It is certainly true that 1*1= 1*1 so $1\in X$ If $x\in X$ then $s(x)*1= s(x)$ while $1*s(x)= s(1*x)= s(x)$ so that $s(x)\in X$. Therefore X= S. That is, x*1= 1*x for all positive integers. Given x, define X= {y| x*y= y*x} By the above, $1\in X$. If $y\in X$ then $x*s(y)= x*y+ x$ while $$s(y)*x= yx+ x$$. Since $y\in X$, those are equal. That is, if $y\in X$ then $s(y)\in X$ so that X= N. xy= yx for all positive integers x and y. Now, as I said before, we can define the integers as equivalence classes of pairs of positive integers using the equivalence relation (m, n)= (u, v) if and only if m+v= n+ u. We define addition and multiplication by choosing "representatives" for the classes. That is, if x contains the pair (m, n) and y contains the pair (u, v) then x+ y is the class containing (m+u, n+ v) and x*y is the class containing the pair (mu, nv). (e can think of the equivalence class containing the pair (m, n), if m> n, as "represented" by the positive integer, m- n, if n> m, by -(n- m). For example, the class containing (7, 4) is represented by 7- 4= 3, the class containing (4, 7) is -(7- 4)= -3. It is easy to see that the all pairs in which the two members are equal, (m, m), are "equivalent" and that class is represented by "0".) Then it is easy to see that multiplication of integers is commutative: y*x is the class containing the pair (um, vn) and we have already shown that multiplication of positive integers is commutative. Now we can define the rational numbers as equivalence classes of pairs of integers, the second member of the pair, not being 0, using the equivalence relation (m, n) is equivalent to (u, v) if and only if nu= mv. We define the sum of two rational numbers x: if x contains (m, n) and y contains (u, v) then x+ y contains (mv+ nu, nv). We define x*y as the equivalence class containing (mu, nv). Then, of course, y*x contains (um, vn) which is the same as x*y because multiplication of integers is commutative. (We can represent the class containing the pair (m,n) as the fraction $\frac{m}{n}$. For example, the class containing the pair (1, 2) can be represented by the fraction 1/2.) Finally, we define the real numbers as equivalence classes of sequences of rational numbers using the equivalence relation $\{a_n\}$ equivalent to $\{b_n\}$ if and only if $\lim_{n\to \infty} a_n- b_n= 0$. If x is the equivalence class containing $\{a_n\}$ and y is the equivalence class containing $\{b_n\}$ then we define x+ y as the equivalence class containing $\{a_n+ b_n\}$ and x*y as the equivalence class containing $\{a_n*b_n\}$. Again, that is the same class as $\{b_n*a_n\}$. (Since every rational number is a terminating decimal, we can choose a rational number in one of the sequences in the equivalence class, with whatever number of decimal places we like, to approximate the real number but typically have to use an arbitrary symbol to represent the number itself. For example, the class containing the sequence 3, 3.1, 3.14, 3.141, 3.14, .... would be represented by "$\pi$".)

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 Quote by greswd yeah, the intuitive geometric proof. but it could be that we describe geometry in commutative terms, and not the other way round.
I think it only needs the axiom that the measure of two disjoint sets is the sum of their measures.

You need an idea of product measure I guess.

You can define the integers as the free abelian group on 1 generator. Then you get distribution and cumutativity by definition.

Then define the rationals as the field of quotients of the rationals.

Then define a completion of the rationals under a metric. The resulting fields are easily seen to preserve the two laws.

Now take extension fields of the reals as quotients of the ring of polynomials in one variable by an irreducible polynomial.

 Recognitions: Gold Member All that just sailed over my head.
 Recognitions: Gold Member Science Advisor Staff Emeritus Well, that's the problem, isn't it? You asked a question about the fundamentals of the number system without knowing much about the fundamentals.

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 Quote by HallsofIvy Well, that's the problem, isn't it? You asked a question about the fundamentals of the number system without knowing much about the fundamentals.
We've all got to start somewhere. Nobody starts off knowing everything.

I was hoping someone could simplify it. Or at least you could set me on the right track.

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 Quote by greswd I was hoping someone could simplify it. Or at least you could set me on the right track.
I think post #17 does that. To fully understand this stuff, you need to study the basics of set theory and the basics of abstract algebra. Then you can use a book on set theory (e.g. Goldrei or Hrbacek and Jech) to find out how to use set theory to define natural numbers, integers, rational numbers and finally real numbers (a Dedekind-complete ordered field).

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 Recognitions: Gold Member What should I read as prep before Goldrei?

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 Quote by greswd What should I read as prep before Goldrei?
Not sure you will need anything to prep for that, but you may find some of the stuff recommended in this thread useful, in particular the book linked to in post #2 and the 10-page pdf linked to in #5.

I've been discussing similar things with a guy in this thread, and he seems to find both of those useful.

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 Quote by Fredrik Not sure you will need anything to prep for that, but you may find some of the stuff recommended in this thread useful, in particular the book linked to in post #2 and the 10-page pdf linked to in #5. I've been discussing similar things with a guy in this thread, and he seems to find both of those useful.
That's great, mate.

Damn, that thread stretched 8 pages.

 Recognitions: Gold Member I'm struggling with No.5 in Book of Proof.

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 Quote by greswd I'm struggling with No.5 in Book of Proof.
Chapter 5? Problem 5? You may need to be more specific.

If it's an exercise that you're stuck on, you can start a thread about it in the homework forum. If it's a concept, you can start a thread in the forum that seems the most appropriate, probably "general math" or "set theory, logic, probability, statistics". Make sure to include the link to the online version of the book and a statement about what specifically you're having difficulties with.