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Distributive rule

by Mentallic
Tags: distributive, rule
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Mentallic
#1
Sep18-10, 06:04 AM
HW Helper
P: 3,540
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.
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csopi
#2
Sep18-10, 06:39 AM
P: 79
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
#3
Sep18-10, 06:57 AM
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P: 3,540
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
#4
Sep18-10, 07:30 AM
P: 79
Distributive rule

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
#5
Sep18-10, 08:31 AM
P: 1,060
Quote Quote by Mentallic View Post
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
#6
Sep18-10, 09:55 AM
P: 754
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
#7
Sep18-10, 10:12 AM
HW Helper
P: 3,540
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
#8
Sep18-10, 10:26 AM
P: 754
Quote Quote by Mentallic View Post
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
#9
Sep18-10, 10:35 AM
HW Helper
P: 3,540
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
#10
Sep18-10, 10:35 AM
P: 754
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
#11
Sep18-10, 10:36 AM
P: 754
Quote Quote by Mentallic View Post
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
#12
Sep18-10, 11:33 AM
HW Helper
P: 3,540
Yes, thank you And that proof was very elegant, I appreciate it.
jgens
#13
Sep18-10, 12:07 PM
P: 1,622
Quote Quote by zgozvrm View Post
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, 21/2 times?
zgozvrm
#14
Sep18-10, 12:12 PM
P: 754
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
#15
Sep18-10, 12:27 PM
P: 754
Quote Quote by zgozvrm View Post
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
#16
Sep18-10, 12:50 PM
P: 898
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
#17
Sep18-10, 01:09 PM
P: 754
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
#18
Sep18-10, 01:30 PM
P: 754
Quote Quote by slider142 View Post
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.


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