How do we prove the distributive property of multiplication?


by greswd
Tags: distributive, multiplication, property, prove
greswd
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Nov12-12, 06:59 AM
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How to prove that 3 x 2 = 2 x 3?
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Erland
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Nov12-12, 08:40 AM
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Quote Quote by greswd View Post
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.
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Mar31-13, 10:54 PM
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Quote Quote by Erland View Post
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.

micromass
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Apr1-13, 12:02 AM
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How do we prove the distributive property of multiplication?


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.
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Apr6-13, 07:52 AM
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Quote Quote by micromass View Post
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?
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Apr6-13, 12:43 PM
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Quote Quote by greswd View Post
alright, I'll check it out. Is it complicated?
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.
42Physics
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Apr6-13, 01:04 PM
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See, in mathematics distributing different numbers/factors, EVENLY, doesnt 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.
AnTiFreeze3
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Apr6-13, 01:20 PM
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Quote Quote by 42Physics View Post
See, in mathematics distributing different numbers/factors, EVENLY, doesnt 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?!?
HallsofIvy
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Apr7-13, 08:27 PM
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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.
HallsofIvy
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Apr7-13, 08:29 PM
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Quote Quote by greswd View Post
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.
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Apr8-13, 12:15 PM
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Quote Quote by HallsofIvy View Post
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.
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Apr8-13, 12:24 PM
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Quote Quote by micromass View Post
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?



Quote Quote by micromass View Post
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?

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HallsofIvy
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Quote Quote by greswd View Post
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]
micromass
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Apr8-13, 02:27 PM
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Quote Quote by greswd View Post
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!!
greswd
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Apr8-13, 09:36 PM
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Quote Quote by micromass View Post
But that does not mean that they don't have a precise definition and construction!!
They do, but only in our minds


In the Maths section, are IMG codes disabled?
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Apr8-13, 09:44 PM
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Quote Quote by HallsofIvy View Post
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/Constru..._Dedekind_cuts

Hmm, I don't understand the notation.
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Apr9-13, 05:12 AM
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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".
greswd
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Apr9-13, 05:35 AM
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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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