Proof of Distributive Rule: Why a(b+c)=ab+ac

[Mentallic]

I am curious to see a proof as to why a(b+c)=ab+ac.
Also is this a relatively new rule in the world of mathematics because I remember hearing somewhere that expressing quadratics and a product of two factors wasn't done till quite recently (how recent this was I am not sure of).

Thanks.

 

[csopi]

I suppose you mean that a,b,c are elements of a ring. If so, then a(b+c)=ab+ac by the definition of a ring.

 

[Mentallic]

Is there no other reason? Because I don't know what rings are and the fact that they work for all real numbers seems like it should have a decent proof behind it - not a definition in some field of maths.

 

[csopi]

Real numbers are not present in nature, they are abstract creations of mankind. In fact, only positve integers (and maybe the zero) are there "as is". If you want to create negative numbers, not to mention rational numbers and the real numbers in a rigorous mathematic way, you have to make a lot of axioms and algebraic definitions (such as additive and multiplicative inverse, equivalence class, quotient space, limit point ...). Of course, they are choosed "properly", so that they satisfy our intuitions we acquired by multiplying and adding pos. integers, but you have to see, that real numbers are our creations. The only reason, they seem so natural is that you've got used to them (since you've been calculating with them since your early childhood).

The equation is trivial for positive integers, because for them the multiplication is just a short notation for multiple additions. But when you go further, the only reason it remains true is that you choosed the definitions properly...

 

[Gerenuk]

Is there no other reason? Because I don't know what rings are and the fact that they work for all real numbers seems like it should have a decent proof behind it - not a definition in some field of maths.

It's the other way round. Number products are defined by the distributive rule!
Here is a simple reasoning:
We start with 1; We define that 1*1=1
Now we need a new symbol for 1+1; so we invent 2=1+1
Next we need a symbol for 1+1+1 and we invent 3=1+1+1
4=1+1+1+1
5=1+1+1+1+1

OK, but what's 2*2 now?
We can *proof* that it has to be 4, so we *dont have to invent* a new symbol for 2*2:
2*2=(1+1)*(1+1)=(1+1)*1+(1+1)*1=1*1+1*1+1*1+1*1=1+1+1+1=4
You see that basically the distributive law enables us to find what products should be equal to.

 

[zgozvrm]

Try this:

We know that ab = b + b + b + ... + b ("b" added to itself "a" times) by definition of multiplication

To make things a bit easier to follow, let's enumerate each instance of "b" so we can keep track. We'll say that [itex]ab = b_1 + b_2 + ... + b_{a-1} + b_a[/tex]

Therefore, it is also true that [itex]a(b + c) = (b + c)_1 + (b + c)_2 + ... + (b + c)_{a-1} + (b + c)_a[/tex]

This is really [itex](b_1 + c_1) + (b_2 + c_2) + ... + (b_{a-1} + c_{a-1}) + (b_a + c_a)[/tex]

Which can be rearranged with the associative rule into [itex](b_1 + b_2 + ... + b_{a-1} + b_a) + (c_1 + c_2 + ... + c_{a-1} + c_a)[/tex]

By definition of multiplication, this is ab + ac

 

[Mentallic]

While some have said it is a definition, what zgozvrm has posted seems like a valid proof of the distributive law by only using the definitions of what multiplication really is. A definition can't be proven by other definitions can it? I mean, isn't that what the idea of an axiom is?

 

[zgozvrm]

A definition can't be proven by other definitions can it? I mean, isn't that what the idea of an axiom is?

Yes, you have to start somewhere...
At some point multiplication was defined as repeated addition.
Addition was defined prior to that. (We had to have some way of describing what it means to have "a" items and increasing that amount by "b" items).

Multiplication made it easier to deal with larger numbers, arrays, etc.

If everyone agrees on the rules, we can use those rule to create/prove new ones, as long as the original rules aren't broken.

In any formal proof, each step is justified by a definition, theorem, postulate, etc. that has been previously shown to be true. So each proof becomes a building block for other proofs.

There is no need to waste time "reinventing the wheel."


What you can't do, is use a definition to define itself like:
"Addition is the process of adding 2 or more numbers together, resulting in a single sum."

 

[Mentallic]

Right, so isn't the distributive rule a consequence of the multiplication and addition definitions as you've shown, rather than a definition in itself?

I mean when they were creating the foundations of mathematics, they would start with defining addition and such, and then defining multiplication from addition, but rather than defining the distributive rule, they proved the distributive rule using the already known definitions.

 

[zgozvrm]

By the way, axioms are the building blocks of mathematical theorems.
They can't be proven, they are just accepted to be true; they are self-evident.

The fact x = x is an axiom. There is no mathematical proof of this, yet we all know it to be true. (Given that we know and accept what the concept of equality is).

 

[zgozvrm]

Right, so isn't the distributive rule a consequence of the multiplication and addition definitions as you've shown, rather than a definition in itself?

I mean when they were creating the foundations of mathematics, they would start with defining addition and such, and then defining multiplication from addition, but rather than defining the distributive rule, they proved the distributive rule using the already known definitions.

Absolutely!

 

[Mentallic]

Yes, thank you And that proof was very elegant, I appreciate it.

 

[jgens]

Try this:

We know that ab = b + b + b + ... + b ("b" added to itself "a" times) by definition of multiplication

To make things a bit easier to follow, let's enumerate each instance of "b" so we can keep track. We'll say that [itex]ab = b_1 + b_2 + ... + b_{a-1} + b_a[/tex]

Therefore, it is also true that [itex]a(b + c) = (b + c)_1 + (b + c)_2 + ... + (b + c)_{a-1} + (b + c)_a[/tex]

This is really [itex](b_1 + c_1) + (b_2 + c_2) + ... + (b_{a-1} + c_{a-1}) + (b_a + c_a)[/tex]

Which can be rearranged with the associative rule into [itex](b_1 + b_2 + ... + b_{a-1} + b_a) + (c_1 + c_2 + ... + c_{a-1} + c_a)[/tex]

By definition of multiplication, this is ab + ac

What would it mean to add a number to itself, say, 2^1/2 times?

 

[zgozvrm]

I assume you meant 2.5 (rather than [itex]2^{1/2} = \sqrt{2}[/tex]) times?

You would first have to introduce/define fractions and division.

 

[zgozvrm]

I assume you meant 2.5 (rather than [itex]2^{1/2} = \sqrt{2}[/tex]) times?

You would first have to introduce/define fractions and division.

Then you could show that [itex]2 \frac{1}{2} = \frac{5}{2}[/tex]

Then you would multiply the numerators (using the definition of multiplication)
and multiply the denominators (again, using the definition of multiplication)

The result would be the product of the numerators over the product of the denominators.


... but I'm sure you already knew this (just trying to be "smart")!

 

[slider142]

This is a good motivation for defining the distributive property for real numbers, but it does not work as a theorem in general. For example, how would your example work for Euler's number multiplied by the constant pi or irrationals in general?

 

[zgozvrm]

Multiplication is still repeated addition.

e.g. 4 * [itex]\pi[/tex] can never be exactly determined. But, we can deternime that value to a certain number of places.

Let's use [itex]\pi[/tex] to 2 places: 3.14
We can show this as 3 + 0.1 + 0.04
If we were to multiply this by 4, we can show that 4 * 3.14 = 4 * (3 + 0.1 + 0.04) = 4*3 + 4*0.1 + 4*0.04

You can continue [itex]\pi[/tex] out as far as you like and the distributive property still holds true. The only problem is that we have to stop at some point because we have yet to find an exact value of [itex]\pi[/tex]

This can be shown for all types of numbers, so, yes it does work as a theorem in general.

 

[zgozvrm]

it does not work as a theorem in general

In fact, it is a theorem, in general!

I may not have shown it to work "in general" but I also didn't define fractional, irrational or complex numbers and how to multiply them.

I believe the distributive property is generally introduced to math students before they learn about numbers other than integers. When they learn about other types of numbers, that is when the elementary laws of mathematics (associative, distributive, commutative) are shown to still work, not the other way around.

 

[Mentallic]

Multiplication is still repeated addition.

e.g. 4 * [itex]\pi[/tex] can never be exactly determined. But, we can deternime that value to a certain number of places.

Let's use [itex]\pi[/tex] to 2 places: 3.14
We can show this as 3 + 0.1 + 0.04
If we were to multiply this by 4, we can show that 4 * 3.14 = 4 * (3 + 0.1 + 0.04) = 4*3 + 4*0.1 + 4*0.04

You can continue [itex]\pi[/tex] out as far as you like and the distributive property still holds true. The only problem is that we have to stop at some point because we have yet to find an exact value of [itex]\pi[/tex]

This can be shown for all types of numbers, so, yes it does work as a theorem in general.

This method wouldn't be able to show that $$\sqrt{2}$$ multiplied by itself is in fact 2.

 

[zgozvrm]

This method wouldn't be able to show that $$\sqrt{2}$$ multiplied by itself is in fact 2.

It's not supposed to!

The fact that [itex]\sqrt{2} \times \sqrt{2} = 2[/tex] has NOTHING to do with the distributive property!


It also doesn't show that 4/2 = 2, or that a rectangle has 4 sides, or that the volume of a sphere is equal to [itex]\frac{4}{3}\pi r^3[/tex]

 

[zgozvrm]

This method wouldn't be able to show that $$\sqrt{2}$$ multiplied by itself is in fact 2.

Although [itex]\sqrt{2} \times \sqrt{2}[/tex] is a multiplication, you are actually dealing with exponents and square roots here:

$$\sqrt{2} \times \sqrt{2} = (\sqrt{2})^2$$

Now, you can multiply them using traditional methods (i.e. repeated addition) but then you'd be dealing with approximations because [itex]\sqrt{2}[/tex] is just another irrational number (approximately equal to 1.414213). So my explanation of how to multiply a number by [itex]\pi[/tex] holds for [itex]\sqrt{2}[/tex], as well!

 

[zgozvrm]

While the proof I gave uses multiplication and division, I don't have to show that multiplication and division work for all types of numbers. Rather, it's the other way around!

No matter WHAT kind of number you're talking about, multiplication comes down to repeated addition (look it up).

I merely offered a way to show that the distributive property is valid.

 

[Tac-Tics]

The distributive property of natural numbers is axiomatic.

All other distributive properties (of integers, rationals, reals) follow from this (and the other axioms of arithmetic).

 

[zgozvrm]

The distributive property of natural numbers is axiomatic.

All other distributive properties (of integers, rationals, reals) follow from this (and the other axioms of arithmetic).

No, the distributive property of natural numbers is not "axiomatic," as you say. It is not an axiom because it can be proven mathematically and it is not self-evident. (Look up the definition of a mathematical axiom.)

 

[Tac-Tics]

You can choose whatever axioms you want as long as they don't lead to contradiction.

And using the standard peano axioms of arithmetic, yeah, distributivity is axiomatic.

You can, of course, be clever and come up with an alternative set of axioms where distributivity is a theorem, but it's such a fundamental property, any proof would be uninteresting to the OP.

 

[zgozvrm]

You can choose whatever axioms you want as long as they don't lead to contradiction.

And using the standard peano axioms of arithmetic, yeah, distributivity is axiomatic.

You can, of course, be clever and come up with an alternative set of axioms where distributivity is a theorem, but it's such a fundamental property, any proof would be uninteresting to the OP.

Again, the distributive property is not an axiom since it is not self-evident. The fact that x = x is an axiom, because it is self-evident and can't be proven formally.
After all it is called the distributive property, not the distributive axiom!



And, I disagree ... the proof is, in fact interesting to the OP, since that is exactly what they were asking for!

I am curious to see a proof as to why a(b+c)=ab+ac.

 

[Tac-Tics]

After all it is called the distributive property, not the distributive axiom!

The property in question is included as one of the ring axioms. It's true by definition, and thus satisfies the criteria of axiom-hood. (Being "self-evident" has no bearing).

 

[Hurkyl]

Multiplication is still repeated addition.

e.g. 4 * [itex]\pi[/tex] can never be exactly determined.

$4\pi$ seems rather exact to me.



Anyways, this thread is getting too silly. You're viewing things far too narrowly, zgozvrm.

 

[zgozvrm]

The property in question is included as one of the ring axioms. It's true by definition, and thus satisfies the criteria of axiom-hood. (Being "self-evident" has no bearing).

The distributive property is hardly true "by definition." The definition of the distributive property is a(b + c) = ab + ac, but that's not immediately evident, and so must be shown to work (i.e. proven) before it can be used.

Axioms are used in proving that the distributive property works, but that doesn't make the distributive property, itself, an axiom. Otherwise every mathematical theorem, property, etc. would be considered an axiom.

An axiom is a building block upon which properties, theorems, etc. are proven.

 

[zgozvrm]

$4\pi$ seems rather exact to me.



Anyways, this thread is getting too silly. You're viewing things far too narrowly, zgozvrm.

Really? So you can exactly determine the value of [itex]\pi[/tex]?
You must publish your results. Mathematicians have been trying to find that value for centuries!

 

[zgozvrm]

$4\pi$ seems rather exact to me.



Anyways, this thread is getting too silly. You're viewing things far too narrowly, zgozvrm.

Besides, [itex]\pi[/tex] is merely a symbol representing a value; a value which cannot be exactly determined.

And, no, I'm not viewing things too narrowly. Rather, others are trying to dispute a known fact, perhaps in an attempt to try and get me to print a "perfect" widely-acceptable definition.

The OP was looking for a way to prove the distributive property, and I gave him one. I gave him an algebraic proof that works. Although I didn't prove that it works for irrational numbers, or complex numbers, or integers, or fractional numbers, etc. specifically, my proof can be used to show that it does work for all those types of numbers (using other known axioms, properties, etc.) But that wasn't the point of the original question, now was it?

By the way the OP was happy with the solution I gave, so why everyone has to pick it apart afterward, is beyond me.

 

[Tac-Tics]

Really? So you can exactly determine the value of [itex]\pi[/tex]?
You must publish your results. Mathematicians have been trying to find that value for centuries!

You're trolling the wrong forum if you think pi doesn't have a exact and well-understood value.

 

[Hurkyl]

Really? So you can exactly determine the value of [itex]\pi[/tex]?
You must publish your results. Mathematicians have been trying to find that value for centuries!

What's to determine? Even the ancient Greeks knew the exact value of pi, and the entire decimal form has been known for centuries, at least.

And using the standard peano axioms of arithmetic, yeah, distributivity is axiomatic.

The Peano axioms only talk about the successor operation.

 

[Hurkyl]

And, no, I'm not viewing things too narrowly. Rather, others are trying to dispute a known fact, perhaps in an attempt to try and get me to print a "perfect" widely-acceptable definition.

Yes, you are. You have your own pet idea about words like "value" or "axiom" or "property", and you complain whenever anybody doesn't conform to your narrow world view.

 

[zgozvrm]

More often than not, this forum seems to be a place where people try and disprove those who attempt to help others rather than offering up additional information to help support the answer. I guess they were too small to stand up for themselves on the playground, and now get their jollies "bullying" people here.

We all know that the distributive property is indeed a property (not an axiom) and that it works. The OP was looking for a way to prove it (perhaps to help explain to someone else that it works in all situations) He wasn't asking me to prove that it is an axiom or that it is a property. I tried to help him out. No need to question that multiplication is repeated addition ("What would it mean to add a number to itself, say, 21/2 times?") or that my example doesn't work "for Euler's number multiplied by the constant pi or irrationals in general" (when, in fact, it does ... I just didn't show it).