Re: how to deal with the algebra problem?


SU(2) wrote:
> assume a1*a2*...*an=1
> does there exist b1,b2,...,bn such that:
> a1(b1/b2)=1
> a2(b2/b3)=1
> .
> .
> .
> an(bn/b1)=1
>
> if exist, how to find these bi (i=1~n) ?
>
> thanks in advance.

Not only do such bi exist, there are an infinite number of sets.

proof of "if there is one solution, there are infinitely many":
assume b1,b2,...,bn is a solution. Since the constraining equations
only have quotients of bi by bj, i<>j, if all bi are multiplied by a
(non-zero) value, x, then you are simply multiplying the numerator and
denominator of a fraction by the same value, leaving it unchanged.
Thus, x*b1,x*b2,...,x*bn is another solution.


Proof that a solution exists:
a1 (b1/b2)=1, so b1=b2/a1

similarly, bk=b(k+1)/ak, k<n and bn=b1/an

Chose b1 arbitary.
Then, b(n-1)=bn/a(n-1)=b1/(an*a(n-1))
b(n-2)=b(n-1)/a(n-2)=b1/(an*a(n-1)*a(n-2))
..
..
..
b2=b3/a2=b1/(an*a(n-1)*...*a3*a2
b1=b2/b1=b1/(an*a(n-1)*...*a3*a2*a1)=b1

Thus, any b1 you choose (except, of course, b1=0) will work, and that
will force the values of all the other bi. The fact that b1=b1 in the
last equation shows that the equations are all consistant.

example, with n=4:
let a1=1/4, a2 = 1/3, a3=2, a4=6
choose b1 = x
b1=b2/a1, so b2=b1*a1=x/4
b2=b3/a2, so b3= b2*a2=x/12
b3=b4/a3, so b4=b3*a3=x/6

b1/b2=x/(x/4)=4=1/a1
b2/b3=(x/4)/(x/12)=3=1/a2
b3/b4=(x/12)/(a/6)=1/2=1/a3
b4/b1=(x/6)/x=1/6=1/a4

So, everything checks out.

.

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