Find Arthimetic Proofs: Series Factors & Rules

[Werg22]

I was asked to find a paper with proofs on the basis of arthimetics. For example proof that the product of a serie of factors is the same no matter what is the order the're read, and other arthimetic rules.

 

[fourier jr]

what you're talking about is associativity. i think any book on abstract algebra would have what you're looking for.

 

[matt grime]

And they're axioms not theorems

 

[shmoe]

what you're talking about is associativity. i think any book on abstract algebra would have what you're looking for.

You know, I don't think I could name one. You'll usually get something like "using induction and the associativity law, we can unambiguously write multiplications without brackets," but I can't recall ever seeing it worked out in full gory detail.

 

[JasonRox]

You know, I don't think I could name one. You'll usually get something like "using induction and the associativity law, we can unambiguously write multiplications without brackets," but I can't recall ever seeing it worked out in full gory detail.

Find a book that starts with the Peano Axioms and that should do it.

You even have to prove that 1 + x = x + 1!

It can be done.

 

[shmoe]

I know it can be done, I just can't recall ever seeing it in an abstract algebra book. It's usually waffled over though and with good reason-it's not really difficult or enlightening, just messy.

 

[JasonRox]

I know it can be done, I just can't recall ever seeing it in an abstract algebra book. It's usually waffled over though and with good reason-it's not really difficult or enlightening, just messy.

Yeah, an Abstract Algebra text certainly would never have it.

I don't see why it should though. You talk about associativity, but that's if groups have that property, which is what you must show yourself.

 

[shmoe]

I don't see why it should though. You talk about associativity, but that's if groups have that property, which is what you must show yourself.

groups are all associative, it's part of the definition of a group.

 

[DeadWolfe]

I might be remembering wrong, but doesn't Spivak's Calculus start with some of this stuff?

 

[JasonRox]

groups are all associative, it's part of the definition of a group.

DAMN! Big mistake.

I meant to say something else. We check if sets under a given binary operation is associative, which is one of the steps to showing that a set is a group under a given binary operation.

That makes me laugh because it's so so so wrong what I said.

 

[JasonRox]

I might be remembering wrong, but doesn't Spivak's Calculus start with some of this stuff?

He does, but I think his proof of associativity is just the same as any Abstract Algebra text.

Well, you can prove that (a + b) + c = a + ( b + c), then it is a direct consequence that it will be true for n elements.

 

[Werg22]

Can someone simply proove it right here?

 

[JasonRox]

When you ask a common question, just GOOGLE IT!

http://www.answers.com/topic/addition-of-natural-numbers

 

[Werg22]

As a matter of fact I did but I didn't find anything relevant. Thanks alot, but this is only concerning sums, not multiplications/divisions, not what I was looking for.

 

[JasonRox]

As a matter of fact I did but I was never skilled in the area of research. Thanks alot!

You are welcome.

 

[HallsofIvy]

Here's a link to a monograph I wrote myself several years ago. It details proofs of basic properties of: Natural numbers, Integers, Rational numbers, and Real numbers. The natural numbers are the first chapter:
http://academic.gallaudet.edu/courses/MAT/MAT000Ivew.nsf/ID/918f9bc4dda7eb1c8525688700561c74/$file/NUMBERS.pdf

 

[Werg22]

HallsofIvy, I lack the proper knowledge to understand your paper. Is the proof of these axioms absolutly require basis in Number Theory?

 

[JasonRox]

HallsofIvy, I lack the proper knowledge to understand your paper. Is the proof of these axioms absolutly require basis in Number Theory?

Axioms don't require proofs, only Theorems.

That is the basics for Number Theory, and you don't need much mathematics for it.

I had a link to a free basic Number Theory book, but I guess I lost it. It's easier to read, and doesn't use quantifiers, which is probably what you don't understand in HallsofIvy's text.

Regardless, HallsofIvy wrote a nice monograph, and maybe one day I'll one too. Maybe a different topic though.

 

[HallsofIvy]

HallsofIvy, I lack the proper knowledge to understand your paper. Is the proof of these axioms absolutly require basis in Number Theory?

Well, you asked about basic properties of numbers, didn't you? There are other ways of approaching the natural numbers but they essentially boil down to the same proofs.

The essential property of the natural numbers, also called the "counting numbers" is- counting! Which, in technical terms is just "induction": the fact that given any natural number there is a "next" number and that if we start at 1 and to each "next" number in turn, we get all natural numbers.

All proofs of basic properties of natural numbers reduce to proofs by induction. You must remember that anytime you ask for proofs of the fundamentals of anything, you are getting into deep water!

Here is, briefly, the proof of the associative law for addition:
The set of natural numbers, N, is first defined as a set of objects together with a function f (the "successor function) such that:
a) There exist a member of N, called "1" such that f is a one-to-one, onto, function from N to N\{1}. That is, for every natural number, n, f(n), the "successor" of n, is a natural number and every natural number except 1 is the successor of some other unique natural number.
b) If U is a set of natural numbers such that
(i) 1 is in U
(ii) Whenever n is in U, f(n) is also in U
then U is the set of all natural numbers: U= N.
(The principal of induction- the natural numbers are precisely those numbers we get by starting at 1 and going to the "next" number (the successor).)

Addition is defined by: For any natural number, n, n+ 1= f(n). If m is not 1, then m is the successor of sum natural number, i: m= f(i). We define n+ m= n+ f(i)= f(n+i).
One would use induction to show that this is "well defined" (that the sum of two natural numbers is a unique natural number)- that was "theorem 1".

Theorem 2 (associative law for addition):
For all a,b,c ε N, (a+b)+c = a+ (b+c)

For a,b ε N let U_ab= {c | (a+b)+ c= a+ (b+c)}
1) (a+b)+ 1= f(a+b)= a+f(b)= a+ (b+1)
Therefore 1 ε U_ab.
2) Assume c ε U_ab.
Then (a+b)+ f(c)= f((a+b)+ c)= f(a+ (b+c)) since cε U_ab
f(a+ (b+c))= a+ f(b+c)= a+ (b+ f(c))
Therefore f(c) ε U_ab.
Therefore Uab = N.

 

[fourier jr]

maybe I'm too fast for my own good. iit isn't just associativity??

 

[shmoe]

maybe I'm too fast for my own good. iit isn't just associativity??

I had thought it was associativity for a general number of elements, what let's you write a product unambiguously as abcd without brackets for example. This is what I don't think I've seen in an abstract algebra book, just mentioned or left as an exercise. Which is fair, once you're at that stage you should be capable of muddling through the details.

Werg22, you can prove this sort of thing by induction on the number of elements in the product. I don't know the simplest way to proceed, but you could assume that every product with n terms is equal no matter how the parentheses are arranged. Then for n+1 terms, break it up into smaller products and go from there. Here you'd try to prove that no matter what you start with, you could reduce it to the same arrangement of parantheses, say multiply from left to right e.g. for 4, reduce to ((ab)c)d.

 

[HallsofIvy]

maybe I'm too fast for my own good. iit isn't just associativity??

I can't speak for how fast you are but I did say "Here is, briefly, the proof of the associative law for addition".

Yes, it is the associative law- that's why I called it that!

Of course, once you have (a+ b)+ c= a+ (b+ c) the more general form follows easily.

 

[shmoe]

So I don't seem totally off my rocker, Werg22 had said "For example proof that the product of a serie of factors is the same no matter what is the order the're read", which I understood to be associativity for a general number of parenthesis.

 

[Werg22]

I found a way to proove it somehow by induction.

It is easy to proove those identities:

1. ab=ba

2.abc=(ab)c=a(bc)

Then we can proove that for this special case (3 elements) the order dosen't matter. Thus for 4 elements;

abcd = (abc)d By definition .

Now we can proove tha not matter what the order in the parenthesis, the result remains unchanged. The proove the general case, we have to proove that you can place any of the element to any wished position without chaging the result. By property 1, d can be the first element and the last. It is easy to see that any number can be put first, and since there is 3 remaining elements, and we already prooved it can be arranged a wished, then any combination can be formed with those elements without changing the result. Then we prooved this for 4 elements, and we proove it for 5 the same way.

Q.E.D.?

 

[JasonRox]

I found a way to proove it somehow by induction.

It is easy to proove those identities:

1. ab=ba

2.abc=(ab)c=a(bc)

Then we can proove that for this special case (3 elements) the order dosen't matter. Thus for 4 elements;

abcd = (abc)d By definition .

Now we can proove tha not matter what the order in the parenthesis, the result remains unchanged. The proove the general case, we have to proove that you can place any of the element to any wished position without chaging the result. By property 1, d can be the first element and the last. It is easy to see that any number can be put first, and since there is 3 remaining elements, and we already prooved it can be arranged a wished, then any combination can be formed with those elements without changing the result. Then we prooved this for 4 elements, and we proove it for 5 the same way.

Q.E.D.?

What you described as a proof of associativity for 4 elements isn't a proof for associativity. You basically just described commutativity.

The element d being on the left or right has nothing to do with associativity.

 

[Werg22]

You misunderstoof what I meant. What I meant is that any of the element can be put to be the first number, and since we prooved any combination can be made with three numbers, leaving the same result, thus we proved that any number that start with a certain element gives the same result and every combination gives the same result as well. Now that it is proved for 4 elements, it can be proved for 5 ans so on.

 

[Werg22]

Also note that we need to prove that for any number of elements (it is already proved for 3 elements)

abc... = a(bc...)

It can be proved by induction

a(bcd)

a(b(cd))

ab(cd)

(ab)(cd)

(ab)cd

abcd

Now proved for 4 elements, it can be proved for 5

a(bcde)

a(b(cde))

ab(cde)

(ab)(cde)

(ab)c(de)

(((ab)c)d)e

And by definition,

abcde

...

If you do not understand the previous proof, this is an example:

(abc)d = abcd = (acb)d = (bac)d = (bca)d = d(bca) ...

Any number can put to be the first number and by previous proofs, we know the result remains unchanged. Since any number can be put to be the first number, and the three left can be arranged as wanted it to be, then we can form any combination possible with those numbers. Thus, any order can be formed without changing the order, which is the same as saying the order dosen't influence the result.

 

[JasonRox]

a(b(cd))

ab(cd)

First, who said you can take those brackets off.

Justify every step. You need to do this for every proof. YES, proofs can be tedious.

Second, that's not a proof by induction.

Third, a trick to proving associativity for 4 elements is to let cd = m.

 

[Werg22]

The trick you give me is what I used, ab =(ab). If you want proof of the other transformation here it is:

1.Property 1

ab = ba



...a

Repeated a certain number of time, b

...a
...a
...a
...a
.
.
b

So this is a group represented by a columns and b rows. If flipped 90 degree, you have b columns and a rows. This is by definition, b*a. The same number of points is kept so

ab = ba

Property 2.

abc = (ab)c = a(bc)

abc = (ab)c, this is simply by definition.

Then let's prove abc=a(bc)

abc=
...a
...a
...a
...a
.
.
b
-
...a
...a
...a
...a
.
.
b
_
c

So you have a number of rows,a, and a number c of groups. We firmly know that the number of rows is proportional to the number of groups and b. That said, for every group there is b rows. Thus, the total number of rows, by definition is b*c

Then looking at the figure, the number of points is given by rows*columns;

a(b*c)

Thus

abc=a(b*c).