Demonstration of a property of potentiation

normanlff

New member
Joined
Jul 24, 2024
Messages
1
Can someone explain to me how exactly this demonstration works?

AnAm=Anm\frac{A^n}{A^m} = A^{n-m}
Demonstration:

Note that AmAnm=Am+(nm)=AmA^m \cdot A^{n-m} = A^{m+(n-m)} = A^m
and since thenA0A \neq 0 passing AmA^m dividing the other side of the equality we find that AnAm=Anm\frac{A^n}{A^m} = A^{n-m}
 
What do you mean by potentiation?

Can someone explain to me how exactly this demonstration works?
At where exactly are you stuck?

Note that AmAnm=Am+(nm)=AmA^m \cdot A^{n-m} = A^{m+(n-m)} = A^m
This line should be:

AmAnm=Am+(nm)=Am+nm=AnA^m \cdot A^{n-m} = A^{m+(n-m)} = A^{m+n-m} = A^n
 
What do you mean by potentiation?
It means (in some dialect other than mine) "powers" (exponentiation).
Can someone explain to me how exactly this demonstration works?

AnAm=Anm\frac{A^n}{A^m} = A^{n-m}
Demonstration:

Note that AmAnm=Am+(nm)=AmA^m \cdot A^{n-m} = A^{m+(n-m)} = A^m
and since thenA0A \neq 0 passing AmA^m dividing the other side of the equality we find that AnAm=Anm\frac{A^n}{A^m} = A^{n-m}
This assumes you have already proved that AmAn=Am+nA^m\cdot A^n=A^{m+n}, and uses the fact that you can divide both sides of an equation by a non-zero number.
 
Ahh, first time to read the word "potentiation"!
Actually, I'm not sure I ever saw it before either; it was just obvious from my knowledge of Latin, and from the context. When I search for it, all I find is a page obviously written by a non-English speaker, and this comment on something similar. It looks like an attempted translation of Spanish potenciación, or something similar.
 
Top