Wednesday, September 26, 2007

a^0=1 and negative powers.

In college I was told that a^0=, and I remember why that is the case being explained. However, the finer details escaped me and I quickly forgot 'why' or how we can show this. Today, I once again came across this fact and it is them neat little mathematical things, that make you go 'awe'. ;) [Well it came across as cool!]

We have to know about certain rules before proceeding, but I will only be stating the few used.

(a) \text{If }ab=ac, \text{where } a,b,c \in \mathbb{R} \text{ then } b=c. Also, a must not equal to zero. [Cancellation law for multiplication].

(b)\text{For } a\in \mathbb{R}, a^1=a.

(c)\text{If } a\in \mathbb{R}\text{ and } i,j \in \mathbb{N}, \text{ then } a^{i+j}=a^i \times a^j.

Now there are a few ways to go about doing this, but it depends on what you already take for granted I suppose. I will do it in the way I read it from the book first, because that is needed for negative powers.

From (b), a= a^1 = a^{1 +0} = a^1 \times a^0 = a \times a^0. Hence, (a) [the multiplication cancellation laws] then tell us that no matter what a is, a^0=.

Now for the negative powers, we use the 'infamous trick' of adding in zero. [We actually used it in a lecture today - bringing a wry smile to my face.]

We start from knowing what a^k is equal to, then: 1= a^0 = a^{k+(-k)}= a^k \times a^{-k} . (By using the rules mentioned above). So from this we can see that a^{-k} = \frac{1}{a^k}.


If we know about negative powers, we can always use that fact to show that a^0=.

Consider: \displaystyle 1= \frac{a^l}{a^l} = a^{l+ (-l)} = a^0.

It does seem circular, but I guess you don't have to be 'that' rigorous. It's quite embarrassing, the way I had the 'ah ha' moment when I recalled the above. We take a lot of things for granted in maths, and 'nice' little things like the above are just 'nice'.


beans said...

I feel 'naughty' because I started the post title with a^0=1!!

Anonymous said...

"a" should be assumed nonzero
(check the "cancellation" law
in particular). yrs in the faith.

Henning said...

In your condition (a) you need to require that
a is not equal to zero.

beans said...

I have corrected it now. Many thanks to both of you.