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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

I have a younger cousin who wants to know how to prove addition is commutative and associative for whole numbers .but i don't know how to do that,can anyone help?plus it'd be good if anyone can tell me how the proof is extended to other number sets.

*Last edited by {7/3} (2013-02-16 17:38:06)*

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,919

Hi

It cannot be prpved. It is an axiom of natural numbers and of higher number sets, too.

Here lies the reader who will never open this book. He is forever dead.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi!

And here is a little different way to look at commutativity and associativity.

Commutativity can be viewed more as a language problem than an axiom problem. Let me

explain. Take the example of adding four and five. We typically write and speak in a linear array

of letters or syllables. This forces us to say one of the numbers first and the other second. So

we say "four plus five" or "five plus four" and write 4+5 or 5+4. But we want these two

expressions to mean the same thing; that is, we want them to get us to "9."

As a function (binary operator in this case) we have a function "+" which maps (4,5) to 9 and

also (5,4) to 9. If we had a way to "say" the four and five SIMULTANEOUSLY, then we would

have but one way to "speak" or "write" about the addition of four and five. Thus we wouldn't

have to "invoke" a commutativity rule.

Actually we could define addition of 4 and 5 as a mapping from the SET {4,5} to 9. Since the

set is unordered this gives us only one way to express the combination of the 4 and 5 to get

the 9. Then "4+5" and "5+4" could be viewed as ways of saying that {4,5} maps to 9 or that

these are ways of expressing the result of mapping {4,5} to 9. But then the notation +{4,5}=9

is a bit awkward compared to 4+5=9 or 5+4=9. On the other hand as the sets get larger, for

example {3,4,5}, +{3,4,5}=12 is not so bad. And here with the three numbers using this

approach we don't have to get into associativity either.

If we define 3, 4, and 5 by 3={ooo}, 4={oooo} and 5={ooooo} (multisets) then adding these

three sets together is just in essence dumping them altogether to get {oooooooooooo}. How

we dump them together (all at once, one at a time, etc.) is immaterial. We get the same result.

So commutativity and associativity can be view more as a language problem than an axiom

problem. On the other hand, the additive and multiplicative identity and inverse axioms are

axiom problems in the sense that they do not just involve ways of "saying the same thing".

They inject the notion of "existence" into the system.

All this being said, viewing the way our systems of numbers have evolved, it is perhaps easier

to just introduce commutativity and associativity as axioms and be done with it.

Have a very blessed day!

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LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

Never mind i found ways to prove commutative and associative law of addition of natural numbers in internet[though i'd appreciate it if anyone tells me how this is extended to Z,Q,R,..]

There are 10 kinds of people in the world,people who understand binary and people who don't.

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**scientia****Member**- Registered: 2009-11-13
- Posts: 222

anonimnystefy wrote:

It cannot be prpved. It is an axiom of natural numbers and of higher number sets, too.

It depends on how you are defining your sets of numbers. If you are defining them by axioms, then there is nothing to prove since the axioms will include commutativity and associativity of addition. But if you are constructing them from smaller sets of numbers, then commutativity and associativity need to be proved.

For example, here is briefly how is constructed from . Define a relation on by iff . (Hint: Think of as the "difference" .) Then is an equivalence relation and is defined as the set of all equivalence classes under . Let the equivalence class containing be denoted . Then addition in is defined as

After checking that the operation is well defined, one can proceed to verify that

is commutative and associative.Moreover, zero

in is the equivalence class and multiplication in is defined as .Similarly

can be constructed from as equivalence classes of the equivalence relation on defined by iff . Letting the equivalence class containing be denoted addition in is definied as*Last edited by scientia (2013-02-18 00:19:54)*

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**{7/3}****Member**- Registered: 2013-02-11
- Posts: 210

Thanks,by the way is R constructed the same way?

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**scientia****Member**- Registered: 2009-11-13
- Posts: 222

No, is constructed by means of either Dedekind cuts or equivalence classes of Cauchy sequences of rationals. The process is very different and much more complicated because you are now constructing an uncountable set from a countable one.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,919

I think I've seen that before, but, then it seems that N is not a subset of Z.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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