# arithmétic

• February 16th 2008, 09:22 AM
j-r
arithmétic
hello

I am 16. I am a new french member and I hope me to improve my standart in english.

Quote:

a, m and n are three natural entires.
I must compare :

$(a^m-1\wedge a^n-1)$ and $a^{m\wedge n}-1$
what that I do :

I have posed: $d=m\wedge n$

$m=dq$ et $n=dk$ with $q\wedge k=1$.

$\begin{cases} a^n-1= (a^d-1)\sum_{i=0}^{k-1}(a^d)^{k-1-i} \\ a^m-1= (a^d-1)\sum_{i=0}^{q-1}(a^d)^{q-1-i}\end{cases}$

d'où:

$a^n-1\wedge a^m-1 = (a^d-1)\times (\sum_{i=0}^{q-1}(a^d)^{q-1-i} \wedge \sum_{i=0}^{k-1}(a^d)^{k-1-i} )$

so here I suppose that: $a^n-1\wedge a^m-1= a^{m\wedge n}-1$

and now I try to show:

$(\sum_{i=0}^{q-1}(a^d)^{q-1-i} \wedge \sum_{i=0}^{k-1}(a^d)^{k-1-i})=1$

but I am not finding

thanks