Demonstration of a property of potentiation

[normanlff]

Can someone explain to me how exactly this demonstration works?

[math]\frac{A^n}{A^m} = A^{n-m}[/math]
Demonstration:

Note that [math]A^m \cdot A^{n-m} = A^{m+(n-m)} = A^m[/math]
and since then[math]A \neq 0[/math] passing [math]A^m[/math] dividing the other side of the equality we find that [math]\frac{A^n}{A^m} = A^{n-m}[/math]

 

[mario99]

What do you mean by potentiation?

Can someone explain to me how exactly this demonstration works?

At where exactly are you stuck?

Note that [math]A^m \cdot A^{n-m} = A^{m+(n-m)} = A^m[/math]

This line should be:

[imath]A^m \cdot A^{n-m} = A^{m+(n-m)} = A^{m+n-m} = A^n[/imath]

 

[Dr.Peterson]

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?

[math]\frac{A^n}{A^m} = A^{n-m}[/math]
Demonstration:

Note that [math]A^m \cdot A^{n-m} = A^{m+(n-m)} = A^m[/math]
and since then[math]A \neq 0[/math] passing [math]A^m[/math] dividing the other side of the equality we find that [math]\frac{A^n}{A^m} = A^{n-m}[/math]

This assumes you have already proved that [imath]A^m\cdot A^n=A^{m+n}[/imath], and uses the fact that you can divide both sides of an equation by a non-zero number.

 

[mario99]

It means (in some dialect other than mine) "powers" (exponentiation).

Ahh, first time to read the word "potentiation"!

 

[Dr.Peterson]

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.