How do we prove the distributive property of multiplication?

In summary, the conversation discusses the difference between the commutative and distributive laws for multiplication, and how the proofs for these laws vary depending on the number system being used. It is mentioned that Rudin's "Principles of Mathematical Analysis" has a proof for the commutative law for real numbers, but it is a complex process that involves constructing different number systems. The conversation also touches on different approaches to defining and proving the existence of real numbers.
  • #1
greswd
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How to prove that 3 x 2 = 2 x 3?
 
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  • #2
greswd said:
How to prove that 3 x 2 = 2 x 3?
This is not an instance of the distributive law but the commutative law for multiplication.
The distributive law says that a x (b + c) = a x b + a x c.

The proofs of these laws will be different depending upon which number system we talk about: natural numbers, integers, rational numbers, real numbers, complex numbers, and even others can occur, and also upon which axioms or constructions we use for these systems.
 
  • #3
Erland said:
This is not an instance of the distributive law but the commutative law for multiplication.
The distributive law says that a x (b + c) = a x b + a x c.

The proofs of these laws will be different depending upon which number system we talk about: natural numbers, integers, rational numbers, real numbers, complex numbers, and even others can occur, and also upon which axioms or constructions we use for these systems.

Let's say in the case of real numbers.
 
  • #4
You will have to be content with a reference then. Get Rudin's "Principles of Mathematical Analysis". The commutative law for real numbers is proven in the appendix of chapter 1.
 
  • #5
micromass said:
You will have to be content with a reference then. Get Rudin's "Principles of Mathematical Analysis". The commutative law for real numbers is proven in the appendix of chapter 1.

alright, I'll check it out. Is it complicated? :smile:
 
  • #6
greswd said:
alright, I'll check it out. Is it complicated? :smile:

Yes, it is quite complicated. There are actually various layers that you need to go through and Rudin only proves the final layer (which is the most complicated one). The first layer is the construction of the natural numbers and proving the distributive law for that. You can find this in many standard set theory books such as Hrbacek and Jech. The next layer is the construction of the integer and proving the distributive law there. Then you construct the rationals and prove the distributive law there. These two things are rather straightforward.
The most complicated layer is constructing the real numbers from the rational numbers and proving the distributive law for that. That is done in Rudin.

This is however just one approach to the real numbers. Many people also prefer to accept the real numbers axiomatically. The axioms that govern the real numbers are called the field axioms. One of the field axioms is the distributive law. In that case, the distributive law becomes an axiom and doesn't need a proof.
The problem with an axiomatic approach is that there doesn't need to be anything that actually satisfies the axiom. To prove that the real numbers actually exist, you really do need to construct them. But many books don't really bother with that.
 
  • #7
See, in mathematics distributing different numbers/factors, EVENLY, doesn't change anything. Eg.
1(100+10)=110 is the same exact as 2(100+10)=220. We distributed evenly throughout the equation.
In my own theory, we can do this for all mathematical equations, even e=mc^2. If we added a square to "e" it would be e^2=mc^2^2 since you have to add the square to both sides to balance the equation. If you were to square root it, you would return to nice ol' e=mc^2!

Hope this helped.
 
  • #8
42Physics said:
See, in mathematics distributing different numbers/factors, EVENLY, doesn't change anything. Eg.
1(100+10)=110 is the same exact as 2(100+10)=220. We distributed evenly throughout the equation.
In my own theory, we can do this for all mathematical equations, even e=mc^2. If we added a square to "e" it would be e^2=mc^2^2 since you have to add the square to both sides to balance the equation. If you were to square root it, you would return to nice ol' e=mc^2!

Hope this helped.

What?!?
 
  • #9
First, it makes no sense to talk about "adding" a square to both sides. What you mean is "square both sides". In that case, yes. If you have an equation- a= b where a and b can be any mathematical expression, then f(a)= f(b) where f is any function. I think that is what you were trying to say.

However, in your example you need parentheses: e^2= (mc^2)^2 which is the same as e^2= m^2c^4. What you wrote would be e^2= m(c^2)^2= mc^4.
 
  • #10
greswd said:
Let's say in the case of real numbers.
Then it depends on exactly how you define "real numbers"- and there are several different ways to do that.
 
  • #11
HallsofIvy said:
Then it depends on exactly how you define "real numbers"- and there are several different ways to do that.

Woah, there are? I've never thought about the definition of real numbers.

All I want is a simple, intuitive proof of commutativity. :tongue2:
 
  • #12
micromass said:
Yes, it is quite complicated. There are actually various layers that you need to go through and Rudin only proves the final layer (which is the most complicated one). The first layer is the construction of the natural numbers and proving the distributive law for that. You can find this in many standard set theory books such as Hrbacek and Jech. The next layer is the construction of the integer and proving the distributive law there. Then you construct the rationals and prove the distributive law there. These two things are rather straightforward.
The most complicated layer is constructing the real numbers from the rational numbers and proving the distributive law for that. That is done in Rudin.

Lol, no mention of commutativity?
micromass said:
This is however just one approach to the real numbers. Many people also prefer to accept the real numbers axiomatically. The axioms that govern the real numbers are called the field axioms. One of the field axioms is the distributive law. In that case, the distributive law becomes an axiom and doesn't need a proof.
The problem with an axiomatic approach is that there doesn't need to be anything that actually satisfies the axiom. To prove that the real numbers actually exist, you really do need to construct them. But many books don't really bother with that.

I see. But real numbers only exist in our minds, don't they?

[IMG codes don't seem to be working]
 
  • #13
greswd said:
Lol, no mention of commutativity?





I see. But real numbers only exist in our minds, don't they?

Yes, but you still have to prove that their properties are consistent.

I mentioned before that there are several different ways to define the real numbers.
One is "Dedekind cuts"- a real number is a set of rational numbers such that:
The set is not empty.
If x is in the set and y< x then x is also in the set.
There exist at least one rational number not in the set.
There is no "largest member" of the set.

Another way is as an equivalence relation on the set of all increasing sequences of rational numbers having an upper bound. We say that two such sequences [itex]\{a_n\}[/itex] and [itex]\{b_n\}[/itex] are "equivalent" if and ony if the sequence [itex]\{a_n- b_n\}[/itex] converges to 0. Then the "real numbers" are equivalence classes defined by that equivalence relation.

A third way is to define rational numbers as that same equivalence relation defined on the set of all Cauchy sequences of rational numbers.

We can then prove that xy= yx for rational numbers by using that property for rational numbers.

Of course, all of those depend upon the rational numbers. We can define rational numbers as equivalence classes using an equivalence relation on the set of pairs (x, y) where x is an integer and y is a positive integer, defining (x, y) to be equivalent to (x', y') if and only if xy'= x'y.

We can prove that xy= yx for rational numbers by using the same property for integers.

And we define integers similarly: integers are equivalence classes using an equivalence relation on the natural numbers (positive integers) where (x, y) is equivalent to (x', y') if and only if x+y'= x'+y.

We can prove that xy= yx for integers by using the same property for natural numbers.

Finally, we can define the natural numbers using the "Peano axioms":
The natural numbers consist of a set of objects, N, called "numbers", and a function, s(x), called the "succesor" function such that:
There exist a unique number, "1", such that s(x) maps N one to one and onto N-{1}.
If a set X, of numbers, contains 1 and, whenever it contains x it also contains s(x), then X= N.

[IMG codes don't seem to be working]
 
  • #14
greswd said:
Lol, no mention of commutativity?

You can replace "distributivity" by "commutativity" everywhere in my post. It will still be valid.

I see. But real numbers only exist in our minds, don't they?

Sure. This is a philosophical topic though, so I guess people might disagree. But I certainly agree. But that does not mean that they don't have a precise definition and construction!
 
  • #15
micromass said:
But that does not mean that they don't have a precise definition and construction!

They do, but only in our minds :tongue2:


In the Maths section, are IMG codes disabled?
 
  • #16
HallsofIvy said:
Yes, but you still have to prove that their properties are consistent.

I mentioned before that there are several different ways to define the real numbers.
One is "Dedekind cuts"- a real number is a set of rational numbers such that:
The set is not empty.
If x is in the set and y< x then x is also in the set.
There exist at least one rational number not in the set.
There is no "largest member" of the set.

Another way is as an equivalence relation on the set of all increasing sequences of rational numbers having an upper bound. We say that two such sequences [itex]\{a_n\}[/itex] and [itex]\{b_n\}[/itex] are "equivalent" if and ony if the sequence [itex]\{a_n- b_n\}[/itex] converges to 0. Then the "real numbers" are equivalence classes defined by that equivalence relation.

A third way is to define rational numbers as that same equivalence relation defined on the set of all Cauchy sequences of rational numbers.

We can then prove that xy= yx for rational numbers by using that property for rational numbers.

Of course, all of those depend upon the rational numbers. We can define rational numbers as equivalence classes using an equivalence relation on the set of pairs (x, y) where x is an integer and y is a positive integer, defining (x, y) to be equivalent to (x', y') if and only if xy'= x'y.

We can prove that xy= yx for rational numbers by using the same property for integers.

And we define integers similarly: integers are equivalence classes using an equivalence relation on the natural numbers (positive integers) where (x, y) is equivalent to (x', y') if and only if x+y'= x'+y.

We can prove that xy= yx for integers by using the same property for natural numbers.

Finally, we can define the natural numbers using the "Peano axioms":
The natural numbers consist of a set of objects, N, called "numbers", and a function, s(x), called the "succesor" function such that:
There exist a unique number, "1", such that s(x) maps N one to one and onto N-{1}.
If a set X, of numbers, contains 1 and, whenever it contains x it also contains s(x), then X= N.


http://en.wikipedia.org/wiki/Construction_of_the_real_numbers#Construction_by_Dedekind_cuts

Hmm, I don't understand the notation.:confused:
 
  • #17
If you want to prove that real numbers have a certain property, then you must use a definition of the real numbers. If you don't, you can't possibly claim to have proved anything about real numbers.

There are many different definitions of the real numbers. Unfortunately, it's impossible to understand any of them without knowing the basics of set theory. So the question is, do you know the basics of set theory?

If you don't, you will have to study it and then come back to this thing. The best answer I can give to a person who doesn't know set theory is that the definition of the set of real numbers was chosen to ensure that multiplication will be commutative. So you may be more interested in knowing why that choice was made than to know how exactly commutativity follows from a construction of the real numbers from rational numbers.

The reason for that, at least if we focus on positive real numbers, is simply that we want the area of a rectangle to be "length of a horizontal side" × "length of a vertical side", and to remain the same if we rotate it 90 degrees. Consider a rectangle with a horizontal edge length of x and a vertical edge length of y. The area is xy. If we rotate it 90 degrees in either direction, the same formula for the area now says that the area is yx. So we need a definition of "real number" that ensures that xy=yx for all positive real numbers x and y.

Similar arguments can be used to motivate all of the properties of the real numbers. For example, if I walk a distance x and then walk a distance y in the same direction, I will have walked a distance x+y. This should be the same distance as if I first walk a distance y, and then a distance x in the same direction. Since these distances are the same, and we want to use real numbers to represent distances*, we require that x+y=y+x for all real numbers x and y.

*) In the 20th century, it was discovered that theories of physics that use a more sophisticated notion of distance are more accurate. This has not changed what we mean by a real number. The definition of the real numbers is motivated by how humans who don't know modern physics think about distances.

If you write down all the properties that we want the real numbers to have for similar reasons, we end up with the definition of a Dedekind complete ordered field. (Those are two separate links). It can be proved that all Dedekind complete ordered fields are isomorphic. This means that for all practical purposes, we can think of them all as the same ordered field. This allows us to just pick any Dedekind complete ordered field and call it "the set of real numbers".

So how do we even know that Dedekind complete ordered fields exist? By explicitly constructing one from the ordered field of rational numbers. How do we know that there exists an ordered field with the properties we want the rational numbers to have? By explicitly constructing one from the ring of integers. How do we know that there's a ring with the properties we want the integers to have? By explicitly constructing one from a set with the properties that we want the non-negative integers to have. How do we know that such a set exists? By explicitly constructing one from the empty set and the standard rules for how to construct new sets from the ones we already have.

How do we know that the empty set exists and that those rules are valid? We don't. What we do know is that those assumptions define a branch of mathematics that's rich enough to include all the number systems, all the mathematics of physics, and a lot more. So you might as well take those assumptions as the definition of what we mean by "mathematics".
 
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  • #18
Guess I have a lot to learn. :blushing:
After reading through, it seems like there's no actual proof of commutativity.

By the way, why are IMG codes disabled?
 
  • #19
greswd said:
After reading through, it seems like there's no actual proof of commutativity.
What would be an "actual proof"?

greswd said:
By the way, why are IMG codes disabled?
I don't know. Maybe something went wrong when the forum changed to a new style.
 
  • #20
Fredrik said:
What would be an "actual proof"?

Good question. I really don't know. :rolleyes:

I've always assumed that mathematical proofs are final and irrefutable.
 
  • #21
greswd said:
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.
 
  • #22
lavinia said:
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. :confused:
 
  • #23
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 [itex]x\in X[/itex] then [itex]s(x)\in X[/itex], 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 [itex]1\in X[/itex]
If [itex]x\in X[/itex] then [itex]s(x)*1= s(x)[/itex] while [itex]1*s(x)= s(1*x)= s(x)[/itex] so that [itex]s(x)\in X[/itex]. 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, [itex]1\in X[/itex]. If [itex]y\in X[/itex] then [itex]x*s(y)= x*y+ x[/itex] while [tex]s(y)*x= yx+ x[/tex]. Since [itex]y\in X[/itex], those are equal. That is, if [itex]y\in X[/itex] then [itex]s(y)\in X[/itex] 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 [itex]\frac{m}{n}[/itex]. 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 [itex]\{a_n\}[/itex] equivalent to [itex]\{b_n\}[/itex] if and only if [itex]\lim_{n\to \infty} a_n- b_n= 0[/itex]. If x is the equivalence class containing [itex]\{a_n\}[/itex] and y is the equivalence class containing [itex]\{b_n\}[/itex] then we define x+ y as the equivalence class containing [itex]\{a_n+ b_n\}[/itex] and x*y as the equivalence class containing [itex]\{a_n*b_n\}[/itex]. Again, that is the same class as [itex]\{b_n*a_n\}[/itex].

(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 "[itex]\pi[/itex]".)
 
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  • #24
greswd said:
yeah, the intuitive geometric proof.

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

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.
 
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  • #25
All that just sailed over my head. :blushing:
 
  • #26
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.
 
  • #27
HallsofIvy said:
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.
 
  • #28
greswd said:
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).
 
  • #29
commutativity is a natural way of thinking.
 
  • #30
What should I read as prep before Goldrei?
 
  • #31
greswd said:
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.
 
  • #32
Fredrik said:
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. :smile:

Damn, that thread stretched 8 pages.
 
  • #33
I'm struggling with No.5 in Book of Proof.
 
  • #34
greswd said:
I'm struggling with No.5 in Book of Proof.
Chapter 5? Problem 5? You may need to be more specific. :smile:

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.
 

1. What is the distributive property of multiplication?

The distributive property of multiplication states that when multiplying a number by a sum, you can first multiply each addend by the number and then add the products together to get the same result as multiplying the number by the sum.

2. How do we use the distributive property to solve equations?

To use the distributive property to solve equations, we can break down the equation into smaller parts and apply the property to simplify it. This is particularly useful when dealing with equations involving variables.

3. Can you provide an example of using the distributive property?

Sure, for example, if we have the equation 2(x + 3), we can use the distributive property to rewrite it as 2x + 6. This is because we first multiply 2 by x and then 2 by 3, and then add the two products together.

4. How do we prove the distributive property of multiplication?

The distributive property can be proven using algebraic manipulation, where we show that the left side of the equation is equal to the right side. We can also use visual representations, such as area models or arrays, to demonstrate how the property works.

5. Why is the distributive property important in mathematics?

The distributive property is important in mathematics because it allows us to simplify complex equations and expressions, making them easier to solve. It is also a fundamental concept in algebra and is used in many mathematical operations, such as factoring and expanding expressions.

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