到数论，望高手请教。我看到一个国外的文献貌似不错，但是不是数学系的看不太懂啊。还望高手指教。
Why is multiplication commutative?
The main point of this page is that the above question can be interpreted in many different ways and that different interpretations lead to different answers. A secondary goal is to discuss whether the fact that we can `just see' that multiplication is commutative means that we have some ability that computers could never have (as, for example, Penrose has suggested).


mn=nm for any pair of positive integers m and n.
Why does this statement seem obvious? After all, if you decide to multiply
together two numbers such as 395 and 428 using long multiplication, then the
calculations you do will depend very much on whether you work out 395 428s or
428 395s. In the first case you will find yourself adding 128400, 38520 and 2140
while in the second you will add 158000, 7900 and 3160. Why isn't it a miracle
that both triples of numbers add to 169060?
Most people would probably say that 395x428 and 428x395 both count the number
of points in a 395-by-428 rectangular grid. If you rotate such a grid through 90
degrees then you turn it into a 428-by-395 rectangular grid but you clearly
leave the number of points unchanged. (They might not express the argument in
exactly this form, but this would be the rough line of reasoning.)
This argument, compelling as it is, doesn't quite qualify as a mathematical
proof. Can we turn it into one? Here is an attempt

Atempted Proof
If A is a set of cardinality m and B is a set of cardinality n, then the
Cartesian product AxB has cardinality mn. But the map (a,b)-->(b,a) is easily
seen to be a bijection between AxB and BxA, from which it follows that BxA has
cardinality mn. But we already know that it has cardinality nm, so mn=nm.

Was that a proof?
It was certainly a valid argument, as long as one was prepared to accept the
initial assertion about the sizes of Cartesian products. But what is the status
of that assertion? Is it a theorem of set theory? Is it the definition of
multiplication? In fact, what is the definition of multiplication?
I have discussed definitions of familiar concepts elsewhere . Even if you have not read that, you are
probably aware that there is often a considerable arbitrariness in what one
regards as a `basic' definition and what one thinks of as an easy theorem. So
instead of asking about "the definition" of multiplication, it may be better to
ask, "Could one, in principle, define mn to be the cardinality of AxB when A has
cardinality m and B has cardinality n?"
If we want to adopt this as a definition, then we must prove that it is
well-defined - that is, that the cardinality of AxB depends only on the
cardinalities of A and B and not on the sets themselves. This is easily done. If
C has the same cardinality as A and D has the same cardinality as B, then there
are (by the definition of `has the same cardinality as') bijections f:A-->C
and g:B-->D. It is then easy to check that (a,b)-->(f(a),g(b)) is a
bijection between AxB and CxD, which shows that AxB and CxD have the same
cardinality.
We now have a serviceable definition of multiplication, at least for positive
integers, and a 99% rigorous proof that multiplication, thus defined, is
commutative. The extra 1% consists in observing that, for any positive integer
m, there is some set A of cardinality m: the set {1,2,...,m} will do,
for example.
Is there any possible objection to the above definition and argument? As I
have said, it is completely rigorous, but one might choose to object on the
aesthetic grounds that multiplication is a very basic arithmetical operation and
it should not be necessary to use set theory, however simple, to define it. What
about our intuition that multiplication is "repeated addition", an idea that can
be understood by a child who has never heard of sets? (One indication that
children do not initially think of multiplication as the size of something like
a Cartesian product is that both my sons went through a phase of understanding
it reasonably well without seeing that it was obviously commutative.
Now they do. I don't yet know about my daughter.)
A rough definition of mxn that does justice to the repeated-addition idea is
that mn=n+n+....+n (m times). The roughness comes in the not entirely standard
notation, but there is a standard way of making such definitions precise, which
is to use induction. A precise definition of multiplication as repeated addition
is that 1n=n and mn=(m-1)n+n when m> 1. Hence, for example,
5n=4n+n=3n+n+n=2n+n+n+n=n+n+n+n+n, a series of equalities that corresponds very
closely to our intuitive idea, especially when we look at them in the reverse
order.
Notice that with this definition it is no longer quite so obvious that mn=nm.
Here is a proof.

A second proof.
First I shall show, by induction, that 1n=n1 for all positive integers n.
This is certainly true when n=1. For larger n we know, using the inductive
hypothesis, that n1=(n-1)1+1=1(n-1)+1=n-1+1=n.
Now I shall show, this time by induction on m+n, that mn=nm for all positive
integers m and n. We have proved the result when either m or n is 1. For m,n>
1 we have
nm=(n-1)m+m=m(n-1)+m=(m-1)(n-1)+(n-1)+m=
(m-1)(n-1)+(m-1)+n=(n-1)(m-1)+(m-1)+n=n(m-1)+n=(m-1)n+n=mn

In what sense was that a proof?
In the course of the above argument, I assumed various "obvious" facts, such
as the commutativity of addition and the principle of induction. Given that the
commutativity of multiplication is itself very basic, why was it legitimate to
make these assumptions?
One possible answer to this question is that addition is more basic than
multiplication since multiplication is defined using addition rather than the
other way round. Another is that the commutativity of addition can itself be
proved, again in terms of more basic concepts. (If you want to see this done,
look here. See below for what is meant by s(m) on
that page.) Of course, one must eventually stop asking "Yes, but why is that
true?" and take certain statements as axioms. The most standard system of
axioms for the natural numbers is Peano's:
1 is a natural number.
For every natural number x there is another natural number s(x), called the
successor of x.
For no natural number x is 1 equal to s(x).
If s(x)=s(y) then x = y.
If A is a set of natural numbers such that 1 belongs to A and s(x) belongs
to A whenever x belongs to A, then all natural numbers belong to A.
The last axiom is a form of the principle of induction, and it allows one to
define addition as follows: if y=1 then x+y=s(x). If y> 1 then y=s(z) for
some z (this is easy to prove by induction) and x+y=s(x+z). One can prove
inductively that addition, thus defined, is commutative, and this proof
naturally appears well before a proof that multiplication is commutative.

A third proof.
Somehow, although the inductive definition of multiplication was more basic
than the Cartesian-products definition, the inductive proof of commutativity was
much less natural than the proof using Cartesian products. Here is a way to get
the best of both worlds.
I would like to prove, by induction, that if A and B are sets of cardinality
m and n respectively, then AxB has cardinality mn. Here, I am defining mn in the
inductive way. Since I have already remarked that the cardinality of AxB depends
only on the cardinalities of A and B, I am free to let A={1,2,...,m} and
B={1,2,...,n}. If m=1 then the result is obvious, since the map (1,x)-->x is
a bijection between AxB and B, B has cardinality n, and 1n is defined to be n.
If m> 1, then let C={1,2,...,m-1}. By induction, CxB has cardinality (m-1)n.
We also know that CxB is a subset of AxB, and that the set-theoretic difference
AxB-CxB consists of all ordered pairs (m,x) such that x is in B. The map
(m,x)-->x is therefore a bijection between AxB-CxB and B, from which it
follows that AxB-CxB has cardinality n. Therefore, AxB has cardinality
(m-1)n+n=mn. It follows that the inductive definition and the Cartesian-products
definition are equivalent, and hence that multiplication (defined inductively)
is commutative.

One way to think of the above argument.
The Cartesian-product definition of mn does justice to our idea that mn
counts the number of points in an m-by-n rectangular grid. The picture suggested
by the inductive definition of mn is of a long line consisting of n points, then
another n, then another n, and so on. The above proof makes rigorous the idea
that we could take the points n at a time from the long line and rearrange them
into an m-by-n rectangle, the first n points forming the bottom row, the next
ones the second row etc.

Bob `takes away' his portion of the debt by writing it off, and that Charles
then does the same. Then Alice has had -3 pounds of credit removed twice. In
other words, her credit has had -3 subtracted from it twice, or, to put it
another way, has had -3 added to it -2 times. In other words, her credit has
increased by (-2)x(-3) pounds. But we already know that her credit has increased
by 6 pounds (since she was 6 pounds in debt and now isn't) so (-2)x(-3) must
equal 6.
1x(-3)=-3 since one times anything is that thing. 2x(-3)=(-3)+(-3)=-6 since
two times anything is that thing plus itself. By this sort of reasoning we can
form the following pattern: 4x(-3)=-12, 3x(-3)=-9, 2x(-3)=-6, 1x(-3)=-3.
Obviously the pattern should continue with 0x(-3). What will this equal? Well,
the right hand sides have been increasing by 3 every time, so it ought to be
(-3)
+3
=0. After that will come (-1)x(-3) and that should then be 0+3=3. Next is
(-2)x(-3) and that will be 3+3=6.
0=2x0=2x(3+(-3))=2x3+2x(-3)=6+2x(-3). Hence, 2x(-3)=-6. Now,
0=(2+(-2))x(-3)=2x(-3)+(-2)x(-3)=-6+(-2)x(-3). Hence, (-2)x(-3)=6.
Notice that the third of these proofs depends on the distributivity of
multiplication over addition. How do we know that that holds for
negative integers? It won't do to say that they can be constructed, and addition
and multiplication defined, in such a way that the distributive law can be
checked to hold, because we were trying to examine the "direct intuition" that
this construction is meant to capture. The second argument suffers from a
similar defect: how do we know that the pattern continues? Obviously it will be
very tidy if it does, but how do we know that we have the tidiness? It follows
from the distributive law, or even from the special case (x+1)y=xy+y and
induction (assuming we know how to add), but this gets us back to our earlier
difficulty.
Is the first argument any better? Let us try to write it out more
mathematically. Then the middle of the argument claims that
0=(-6)-(-3)-(-3)=(-6)-2x(-3)=(-6)+(-2)x(-3). Adding 6 to both sides then gives
that 6=(-2)x(-3), but the steps of the calculation are still not really based on
a prior intuition about negative numbers, but more on familiar rules for
manipulating them.
These difficulties arise from not thinking about negative numbers in the
correct way. Multiplication of negative numbers is not something we understand
directly - we do not ask how many apples there are if -2 people have -3 apples
each - but is rather a definition that is forced on us if we want it to obey
certain rules. Instead of being grateful that multiplication of negative numbers
happens to be commutative, we are grateful that there happens to be a unique
binary operation on the negative numbers that is commutative, associative,
distributive over addition, with 1 as an identity element. Even this way of
explaining it assumes that the negative numbers themselves are just there,
waiting to be manipulated. Better still is to say this: we are grateful that the
natural numbers, with their familiar operations, can be embedded into a
commutative ring. Since this can be done in many ways, we can go further and say
that one such ring, which we call Z, is the smallest, in the sense that all it
contains is N (or strictly speaking the embedded copy of N), 0 and additive
inverses of every n in N and that these can all be shown to be necessarily
distinct.
A more algebraic way to say something similar is that if R is a commutative
ring and f:N --> R is an embedding (that is, an injection that preserves
addition and multiplication) then there are unique embeddings g:N --> Z and
h:Z --> R such that hg=f. (Exercise: this "universal property" of Z makes it
unique up to isomorphism.)

To be continued ...
This page has been sitting around for some time, waiting for me to go on to
discuss larger number systems. However, it is so closely relevant to what I am
lecturing that I have decided to put it up in unfinished form.

望高手看懂后讲解一下 谢谢了

归纳法~~~

它说本文第一个目的是用不同的方法证明乘法交换律……5楼是用基数，7楼是归纳法……后面就捉鸡了……

楼主你好，请问你能不能将原论文发给我，或者将链接给我，我的邮箱[email protected]
谢谢楼主了

入门水平的文章，要说数论，代数什么的还算不上。

对于自然数，先证明有乘法左右分配率a(n+m)=an+am,(n+m)a=na+ma。然后用归纳法。
a*1=1*a,假设a*n=n*a.那么a(n+1)=an+a*1=na+1*a=(n+1)a.得证。
乘法分配律的证明：
以上就证明了自然数的加法交换律，结合律，乘法交换律，结合律，分配律
有个自然数的运算律后，可以证明有理数，实数，复数形成一个域，乘法交换律自然得到。
数学系的一定要多读点数，不要老是什么戴德金切割定理。

截图取自向武义的《数学分析讲义》头4页