# Homework Help: Simple Proofs

1. Nov 1, 2005

### 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.

2. Nov 1, 2005

### Nusc

A hint before midnight would be great!

3. Nov 1, 2005

### Nusc

Or maybe they are not so simple?

4. Nov 1, 2005

### 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.

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

=a^(x+y)

Last edited: Nov 1, 2005
5. Nov 1, 2005

### 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)

6. Nov 1, 2005

### lurflurf

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

7. Nov 1, 2005

### 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...

Last edited: Nov 1, 2005
8. Nov 1, 2005

### lurflurf

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 equivelent other definitions if needed
3)define exp and log as desired
4)show these definitions are equivelent 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