Proving Identities with 0 < a in R

[Nusc]

How do you prove these?
Let 0 < a, an element of R, and x, y an element of R. Then:
(a^x)(a^y) = (a^(x+y))
a^(-x) = 1/(a^x)
Thanks.

 

[Nusc]

A hint before midnight would be great!

 

[Nusc]

Or maybe they are not so simple?

 

[1800bigk]

what class is this for? There is prolly many ways to prove this but some of the techniques might not be appropriate for all classes.

how about write
(aaaaa...a)(aaaaaaa...a) so we have a times itself x+y times
...x times... y times

=a^(x+y)

 

[2k5 yzf-r1]

for the second part start with a^(-x) and multiply by a well placed 1.
Since 1(anything)=anything
so try multiplying by 1 where your 1= (a^x)/(a^x)

 

[lurflurf]

How do you prove these?
Let 0 < a, an element of R, and x, y an element of R. Then:
(a^x)(a^y) = (a^(x+y))
a^(-x) = 1/(a^x)
Thanks.

The second follows easily from the first by the case y=-x
To handle the first you need to consider the definition you are using for a^x
for instance one possible definition is
a^x:=exp(x*log(a))
in which case the problem is reduced to showing
exp(x+y)=exp(x)*exp(y)
for this one must considerthe definition of exp(x)
one possible definition being
exp(x) is the unique function for which
exp(x) is a real number if x is a real numberjk
exp(x)*exp(y)=exp(x+y) if x and y are real numbers
limit x->0 [exp(x)-1]/x=1
This is a nice definition for this problem, of course if other definitions are used you will need to prove your statement other ways.
The other common definition of a^x is to let a^x be defined for rational numbers r then
a^x:=lim n->infinity a^r_n
where r_n is a rational sequence for which
lim n->infinity r_n=x
it is also possible to stay with
a^x:=exp(x*log(a))
and use other definitions for exp and log

 

[Nusc]

(a^x)(a^y) = exp(x*ln(a))exp(y*ln(a)) = exp((x+y)*ln(a)) = exp(ln(a)^(x+y)) = a ^ (x+y)

Okay there. Now the second one?

exp(-x*ln(a)) = exp(ln(a)^-x) = a ^-x = 1/a^x

I'm not too sure about the second one...

 

[lurflurf]

(a^x)(a^y) = exp(x*ln(a))exp(y*ln(a)) = exp((x+y)*ln(a)) = exp(ln(a)^(x+y)) = a ^ (x+y)
Okay there. Now the second one?
exp(-x*ln(a)) = exp(ln(a)^-x) = a ^-x = 1/a^x
I'm not too sure about the second one...

The second is a special case of the first
a^x*a^-x=a^(x+(-x))=a^(x-x)=a^0=1
hence
a^-x-1/a^x
just remember what is needed to rigorize this
1)take as a definition
a^x:=exp(x*log(a))
2)show that definition is equivalent other definitions if needed
3)define exp and log as desired
4)show these definitions are equivalent other definitions if needed
5)show that exp(x+y)=exp(x)*exp(y)
6)show that 5 and 1 together can be used to prove the desired result