# Looking for a paper

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.

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.

Last edited:
JasonRox
Homework Helper
Gold Member
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.

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).