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

[Tac-Tics]

The Peano axioms only talk about the successor operation.

Yeah, I was mistaken here.

I conflated peano's axioms with the axioms of TNT in Hofstadter's axiomatization of arithmetic (which is an example of an axiomatic system where distributivity is in fact an axiom).

 

[zgozvrm]

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.

First of all, I'm not complaining.

Secondly, is anyone here really disputing that, for instance, multiplication is not repeated addition. Or, that that the distributive property is NOT an axiom? Or, that you cannot write the value of [itex]\pi[/tex] exactly (which is why we use a symbol to represent it)?

 

[zgozvrm]

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.

By "the entire decimal form," you are referring to 3.141592654... , correct?

Can you show me where I can find reference to this fact? It may be an exact number, but as far as I know, the "decimal part" (0.141592654...) has never been shown to repeat at any point. And therefore, cannot even be shown as a fractional number (an exact ratio). So, until that happens, how can we say we know the exact decimal value?

 

[Hurkyl]

So, until that happens, how can we say we know the exact decimal value?

By proof. A decimal number is a function from N to the set {0,1,2,3,4,5,6,7,8,9,.} with exactly one occurrence of '.'; while somewhat cumbersome to write and to compute, it's quite straightforward from the Taylor series for arctan(1) centered at 0 together with the Taylor remainder theorem. Much more efficiently computable forms are known these days.

 

[zgozvrm]

By proof. A decimal number is a function from N to the set {0,1,2,3,4,5,6,7,8,9,.} with exactly one occurrence of '.'; while somewhat cumbersome to write and to compute, it's quite straightforward from the Taylor series for arctan(1) centered at 0 together with the Taylor remainder theorem. Much more efficiently computable forms are known these days.

Yes, that is what a decimal number is. But that doesn't mean that we know the exact value of [itex]\pi[/tex]. If we did, we'd be able to express it in terms of the ratio of 2 whole numbers.

You said it yourself... "Much more efficiently computable forms are known these days."
The value of [itex]\pi[/tex] has been computed to many more places (and more efficiently), but it has yet to be exactly determined.

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

Can we just accept that the OP has received the answer he was looking for, and stop nit-picking?

 

[Hurkyl]

Yes, that is what a decimal number is. But that doesn't mean that we know the exact value of [itex]\pi[/tex]. If we did, we'd be able to express it in terms of the ratio of 2 whole numbers.

This is only true for your narrow view of the word "value".

You said it yourself... "Much more efficiently computable forms are known these days."

Huh? What does the ability to compute with a function have to do with whether or not we know the function itself?

Can we just accept that the OP has received the answer he was looking for, and stop nit-picking?

The "nitpicking" has continued because you have dug in your heels and defend an indefensible position. The original "nitpick" about your original proof (it was by no means clear that the OP was interested specifically in how to prove the distributive law for the natural numbers from the specific premises you opted to use) comes from the striking resemblance of your argument to one of the standard crackpot stances that only multiplication by an integer is meaningful.

 

[zgozvrm]

Can we just accept that the OP has received the answer he was looking for, and stop nit-picking?

Apparently not.

 

[zgozvrm]

Done.

 

[mathwonk]

one of the earliest treatments of this as a theorem is in euclid, where the quantities are undefined measures of size of rectangles, and he then proves this by decomposing the rectangle with height a, and length (b+c), into two rectangles, of lengths b and c.

 

[Mentallic]

This seems to have gotten a little out of hand. It's exactly what I was looking for, a proof to show that a(b+c)=ab+ac and the proof that zgozvrm gave was enough to satisfy this for me. I didn't really mind about what restrictions were placed on the constants but mathwonk's little geometric proof extends this to all real numbers. That was nice of you as well.

This is what I wanted to see, not pages of bickering. Thank you.

And by the way, the answer has been kind of lost in the mess, but if the distributive property can be proven by more simple definitions (or axioms) of multiplication and addition, is the distributive property itself an axiom?

 

[l'Hôpital]

I think it really depends on what you define axioms. I mean, do we take "+" for granted? What are the axioms?

If we work with, say, the Peano Axioms (and the elements are natural numbers), then it's a theorem. In particular, it can be proven rigorously using induction/the successor function.

 

[Mentallic]

Yes you're right. It all just depends on how you define axioms and since I have a primitive understanding of them - I assumed they mean "starting off points" in maths.

 

[Gerenuk]

To believe zgozvrm's "proof" is naive, because he showed it for integer numbers only.
There is a reason why millions of mathematicians in the world will tell you that the distributive law is an axiom. And the majority of people here agree.
zgozvrm should first read how to write down rigorous proofs, before claiming he can generalize his toy example to irrational numbers.

EDIT: but reading all the lasts posts now I see that there are enough clever people here who can explain it better than me.

 

[Mentallic]

There is a reason why millions of mathematicians in the world will tell you that the distributive law is an axiom. And the majority of people here agree.

Right, so what is it? I'm not disputing this but it just seems logical for the moment that since proofs were given for this rule, it isn't an axiom.

 

[Gerenuk]

Right, so what is it? I'm not disputing this but it just seems logical for the moment that since proofs were given for this rule, it isn't an axiom.

You have think in the "pedantic" way a mathematician does and there every little bit is important: Strictly speaking the proof that was given *presupposes* a definition of multiplication. Basically it _assumes_ a*b=(a-1)*b+b. That's the hidden assumption there. It's reasonable from the real world perspective.
But it cannot be generalized to irrational numbers, so you have not proven the distributive law for general numbers! Therefore mathematicians take another approach, which incidently includes the simple integer case.

 

[deluks917]

You might want to consider reading a "Construction of the Real numbers." I'd recommend chapters 28-30 of Spivak's Calculus, he proves the distributive law after defining the real numbers as certain sets of rational numbers. I'm sure there are other similar treatments.

 

[Mentallic]

You have think in the "pedantic" way a mathematician does and there every little bit is important: Strictly speaking the proof that was given *presupposes* a definition of multiplication. Basically it _assumes_ a*b=(a-1)*b+b. That's the hidden assumption there. It's reasonable from the real world perspective.
But it cannot be generalized to irrational numbers, so you have not proven the distributive law for general numbers! Therefore mathematicians take another approach, which incidently includes the simple integer case.

Oh yes you're right. But you must admit that while the distributive rule isn't self intuitive at first, this "proof" by the definition of multiplication which is self-intuitive makes it much more understandable. It has in my eyes at least.
And mathwonk's geometric proof extends this to all real numbers.

 

[Gerenuk]

I think the following is a good wording:
Distributivity is a property that can in some case apply to the physical world.
That's why mathematicians prepared a mathematical tool (our algebra), that handles distributivity. This tool is distributive by definition and cannot be proven.

Whenever you encounter some distributive property in the real world you can luckily use the mathematical tool. That's the case for geometric squares and so on.

And if some real world case is not distributive then you couldn't use the algebra.

But the algebra itself is just distributive by definition.

 

[zgozvrm]

To believe zgozvrm's "proof" is naive, because he showed it for integer numbers only.
There is a reason why millions of mathematicians in the world will tell you that the distributive law is an axiom. And the majority of people here agree.
zgozvrm should first read how to write down rigorous proofs, before claiming he can generalize his toy example to irrational numbers.

EDIT: but reading all the lasts posts now I see that there are enough clever people here who can explain it better than me.

Okay, so I only showed that the distributive law works for integers. Big deal. I simply gave "a proof" not "the proof." When we learn the distributive law in grade school, we are shown how it works by a method similar to the one I gave. At that point in our education, we are hardly expected to understand complex numbers, for instance, and therefore there is no need to prove whether or not the law works with complex numbers. Rather (as I've stated before), as we're introduced to other types of numbers, we then have to prove whether or not laws such as the distributive law still hold true.

Mentallic asked a very general question, I gave him a general answer. An answer which can be expanded upon for other numbers as well.

Let it go. Not everyone is a career mathematician. Sometimes a simple answer is all that is needed.

 

[Gerenuk]

Actually that you didn't give a general answer, but the other way round. You gave a very specific answer which does not address the full question.

And in school no claimed they gave you proofs. Instead they gave an example how it worked. So don't use the word proof, if you're not sure what it is. And don't claim it's easily generalized to real numbers, if you don't know how.

But some people in this thread told you already. It's fine not being a mathematician, but then at least don't claim to have given an appropriate proof.

 

[zgozvrm]

Actually that you didn't give a general answer, but the other way round. You gave a very specific answer which does not address the full question.

And in school no claimed they gave you proofs. Instead they gave an example how it worked. So don't use the word proof, if you're not sure what it is. And don't claim it's easily generalized to real numbers, if you don't know how.

But some people in this thread told you already. It's fine not being a mathematician, but then at least don't claim to have given an appropriate proof.

Fine, but don't claim to know how my school taught me and the many other students who went there. You cannot generalize by saying that no school gave proofs; you don't know that, since you didn't go to my school. In fact, they didn't just give an example, as you say. That would mean that they gave us a(b + c) and showed us how it worked with real numbers and expect us to go on using that law for the rest of our lives. (By "real numbers," I mean "actual numbers" in this case, so don't get all uptight). They didn't just give us something like 4(5+3) and show us that 4 * 8 is the same as (4*5)+(4*3). Sure, that's how it was introduced, but later they proved to us that it works for all integers. Later, when we got to real numbers (mathematically real), they showed us that it still worked via another proof (one that was based on the original proof, but took reals into account). So, at that point in our education we knew, absolutely, that the distributive law holds true for all integers and reals. This continued on through our education as we learned about different kinds of numbers.

Also, don't tell me not to claim that my example is easily generalized to real numbers. It is, and I can. I just didn't do it, and I'm not going to. That wasn't the point. It can also be shown (quite easily) to work with Boolean algebra, but I didn't do that either. I also don't have to justify that multiplication works for all numbers. When introducing the distributive law, it is assumed that the person already knows that.

There are always assumptions made when a person asks a question. My assumption was that the OP wanted a very basic way to show that the distributive law works. If I gave someone a recipe for a cake and asked them to make it, I'm making several assumptions. For instance:
- they speak my language
- they understand (comprehend) what I'm asking
- they will do what I've asked
- they know how to read
- they know how to measure the ingredients
etc.

Obviously, if I'm going to get them to make that cake, they will have to know certain things first. Just as if someone is going to learn the distributive law, they should know how to add and multiply (as well as subtract and divide) first. And, they should have no reason to doubt that those functions work with all of the numbers that they are aware of, thus far (it is not important that they know that the functions work with say, irrational numbers if they have yet to learn what irrational numbers are, and currently have no need to use them). While irrational numbers exist and are very important, a child doesn't need them if he is counting apples, or for that matter slices of (fractions of) apples.

"The full question?" None of us knows what exactly what the OP was asking. Was he looking for an elementary school level answer or a post-graduate school level answer? Judging by his replies, I was right on the money as far as what he was looking for, so why don't you just leave it alone?

Not every mathematical question needs to be answered with a full, formal proof. While the OP asked for a proof, I simply gave him some information that would lead to a proof; information that would hopefully steer him in the right direction.

And, by the way, I never claimed that my method was a "proof." Mentallic asked for a proof, I gave an example of how to show it. I later referred to it as a proof, since that was what the OP was calling it. (My bad).

I would hate to have to talk to you in person. I can see it now:
Me: I'm going to the store.
You: How are you getting there?
Me: I'm driving my car.
You: Are you sure that the car is not carrying you, rather than you driving the car?
You: Can you prove that you are driving it?
You: How do you know that you're not just sitting in the car and the Earth is rotating beneath you to the point that when you get out of the car, you are at the store?
Me: I don't really care. I'm just going to the store.

Yes, you're right. I didn't prove it for all numbers. You win. I said, "uncle," now please stop "beating me up!"

 

[Hurkyl]

Yes, you're right. I didn't prove it for all numbers. You win. I said, "uncle," now please stop "beating me up!"

I can just imagine running into you on the street:
Me: (something that you disagree with)
You: *punch* *punch* *right hook*
You: You win, I'm not fighting. Please stop beating me up.

If you want a subject to drop, then you have to do it first, you can't demand others do it for you.


You got lucky that you were answering the specific variation of the question the opening poster intended to ask. But it's still not clear if you gave the answer that the opening poster should have asked. For example, for a great many purposes, distributivity is not proven.

I'm studying vector spaces, so multiplication distributes over vector addition! Proof? Why would I prove it? I told you I'm studying vector spaces!​

And the proof that a particular multiplication operation distributes over a particular addition operation is not brought up, unless you happen to have a specific structure and decide you want to prove it's a vector space.


The axioms of the real numbers are another example. Who proves the real numbers are distributive?? Pretty much the only time you ever see it are in formal model theoretic contexts -- e.g. proving that the consistency of set theory implies the consistency of real analysis, or maybe you are doing Euclidean geometry synthetically and want to demonstrate the usual construction let's you define a real number line that is a model of the real numbers.


A lot of people get the idea that everything is proven, and don't get the idea that work often proceeds by starting with convenient hypotheses and working from there. Maybe those hypotheses are postulates, maybe they are theorems proven by other people, or even yourself, but that detail is mostly irrelevant.

This probably isn't what the opening poster asked. But did he even know that he might want to?

 

[zgozvrm]

For example, for a great many purposes, distributivity is not proven.

I'm studying vector spaces, so multiplication distributes over vector addition! Proof? Why would I prove it? I told you I'm studying vector spaces!​

I'm assuming that the indented statement is sarcasm. Maybe I'm wrong. But either way, you can't use a new mathematical concept with an existing system without first showing that it works in all cases (without breaking any of the rules we have already defined) and vice-versa ... you can't use an old (previously understood) mathematical concept with a new system without first showing that it works (by proof, theorem, postulate, etc.).

Vector spaces are no exception. First (after defining what a vector is, of course), a definition of vector addition had to be made (which is based upon our "normal" concept of addition). The same is true for vector multiplication. Now, just because we've accepted the concepts of vector addition and vector multiplication, that doesn't mean that we should automatically expect (or accept) all other mathematical concepts (like the distributive law) to hold true, as well. We have to somehow show (prove?) that they do, in fact, work. Do you think any mathematicians would have gotten where they are (with knowledge of math) if they just took every piece of (mathematical) information given to them, as being true? No, that would be ignorant. We can use mathematical concepts without proofs, theorems and postulates, but that's not the same as understanding mathematical concepts. We understand them and accept those concepts because they were somehow proven to us to work.

It so happens that not only does multiplication distribute over addition in Boolean algebra (yet, another system we deal with mathematically), addition also distributes over multiplication. In other words, A*(B+C) = (A*B)+(A*C) and (A*B)+C = (A*C)+(B*C).

At some point, both forms of the distributive law had to be proven to work for Boolean algebra in all cases. Should we have assumed that distribution of addition over multiplication doesn't work in Boolean algebra, since it doesn't work with integer, reals, complex numbers, vectors, irrationals, etc.? No. We checked to see if it did work in all cases by proving it somehow and then it became one of the laws of Boolean algebra.

 

[Mentallic]

Guys, I'm just going to say it bluntly - quit arguing. Many have now made it clear that the distributive property is not proven, but an axiom in itself. That's fine, but that answer alone would not have satisfied my question because all I wanted was an intellectual understanding of why it is true. I have been using it for many years now and just accepted it because it worked for all cases. It wasn't intuitive to me at first but now because of zgozvrm's and mathwonk's proofs it is. This is what I wanted, so thank you.

I seem to stand in the same position as zgozvrm here because it seems logical to me that the distributive property is proven, but since the consensus seems to sway towards it being an axiom, fine, I will accept that.

 

[Hurkyl]

But either way, you can't use a new mathematical concept with an existing system without first showing that it works in all cases (without breaking any of the rules we have already defined) and vice-versa ... you can't use an old (previously understood) mathematical concept with a new system without first showing that it works (by proof, theorem, postulate, etc.).

Yes you can, it's not all that difficult. Occasionally you wind up working with the degenerate theory (the one containing a contradiction), but on the whole it's very effective.

You can do it even if you want to be strictly rigorous -- you just start out by postulating all of the facts you want to assume and work from there. This is rather common practice even in mathematics, at least in certain fields. For example:

• A lot of work in number theory that is contingent upon the Riemann hypothesis and its generalizations.
• Real analysis tends to begin with "Let R be a complete ordered field". The proof of relative consistency to set theory is relegated to an off-hand remark, or maybe an appendix.

Proofs do have pedagogical use. Some proofs are useful to convince us of facts we might otherwise be skeptical of. Some proofs serve as references to be modified to suit new circumstances. Other proofs provide much needed practice using the tools and concepts the student is supposed to be learning.

A great many proofs, if you understand the subject, are quite trivial. For example, I assert -- at least as far as real analysis is concerned -- that there is nothing interesting or illuminating about Cantor's proof that the Cauchy real numbers are a model of the real numbers. The interesting part of Cantor's work is the idea of systematically naming points in a space via Cauchy sequences in a dense subspace, which is of great utility. But once you have the idea, the proof is a straightforward, tedious, and unenlightening exercise in basic set theory or higher-order logic.

None of this is meant to discount the work great mathematicians did in working out the right foundations for analysis -- but today we get to take full advantage of the fact that we have good foundations to work with.

Vector spaces are no exception.

Vector spaces, by definition, are distributive and so forth. If they weren't, we wouldn't call them a vector space.