Slavakion

Is there an actual mathematical proof as to why a number raised to the 0th power equals one? Today in our physics class, someone said that she asked the math teacher, who said that it just does. The physics teacher couldn't explain it, and made a wild stab at it.

10^3 = 10*10*10
10^2 = 10*10
10^1 = 10*1
10^0 = 10*.1

It was a good effort, but it only works for ten. Any ideas?

irseg

Interesting question. I did a quick search on Google, and found:

http://mathforum.org/dr.math/faq/faq...to.0power.html

spived2

3^4
1 = ----- = 3^(4-4) = 3^0
3^4


This equation pretty much sums it up. Good find irseg

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Yakk

Lets start with x^a

What happens when you take x^a * x^b?

You get x^(a+b), if a and b are positive integers.

Hmm. So, what should x^0 be? Well, that's a nice rule. Can we make it true for x^0 as well as other positive integers?

x^a * x^0 = x^(a+0) for all a and x
x^a * x^0 = x^a for all a and x
x^0 = 1 for all x for which x^a does not equal zero

So, if x is not zero, if you want the x^a*x^b = x^(a+b) rule to work for x^0, x^0 must be zero.

The same lets you work out what happens with negative numbers: you get x^-a = 1/x^a

This is how you get many rules in mathematics: you take the basic definition for positive integers, find a nice property, and extend it to all integers, or even all real numbers, in such a way that the nice property still holds.

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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

Slavakion

awesomeness. I'll probably use yours in class monday, Yakk. Only because it looks more like the traditional proofs.

rsl12

Consider the limit of y=2^x, where x is a number very very close to 0 but not quite, such as one thousandth (1/1000). In other words, y is equal to some number that you would have to multiply a thousand times with itself in order to get the number 2. Obviously y would have to be really close to 1 in order for this to happen. Consider if x = one millionth. one over a googol. you get the idea.

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rsl12

i dunno what level math your are in, but if you want to do it right, the limit argument is only convincing if also consider the limit from the other side as well (ie, negative #s very very close to 0). The same logic applies.

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Yakk

RSL, how was it decided that
x^(1/100) = 100th root of x

They took the arithmetically true statements (x^a*x^b = x^(a+b), (x^a)^b = x^a*b, etc, where x^a = x*x*..*x (a times)), and extended them to the rationals, then the reals.

x^(1/2) = square root of x
because
x^(1/2) * x^(1/2) = x^(1/2+1/2) = x
the non-rational reals are then defined by making the function continuous (which also, independantly, results in x^0 = 1 for all x not equal to zero).

Interestingly, f(x,a) := x^a is not continuous everywhere: f(0,a) either has a discontinuity at a=0, or f(x,0) has a discontinuity at x=0. Hence the 0^0 problem.

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Last edited by JHVH : 10-29-4004 BC at 09:00 PM. Reason: Time for a rest.

rsl12

yup yakk, i wonder if ever there will be a new set of numbers (like imaginary numbers) based on 0^0...

your explanation is more rigid. mine, i'd like to think, is more common sense. I'm an engineer, not a mathematician.

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llama8

I always remember being told by my maths lecturer that it is true because "all the equations work when it is true".

Hardly a rigorous proof but there you go!

Grothendieck

That x^0=1 is a convention which makes formulae nicer. There is no "proof" of it. For instance, sometimes it is more useful to set 0^0=0. The philosophy is that 1 is the neutral element for multiplication, so if we multiply nothing (i.e. ^0), we stay with the neutral element. Much as if we add nothing, we get 0, the neutral element for addition.

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MoJoPokeyBlue

You're absolutely right...there's no good reason why a number raised to the 0th power equals one!

I hereby decree, that from this day forward, all numbers raised to the 0th power equal 3.

Who's with me?

Tempboy

Yikes, those all look overly complicated up there..

If I recall my old high school math, this is how it was taught to me:

Let's take x squared for a simple example: x^2

(x^2) / (x^2)

To solve this, we would subtract the exponents...so 2-2=0
We end up with:
x^0

Also, with basic math we know that any number divided by itself equals 1.

Therefore we end up with:
(x^2) / (x^2) = x^0
(x^2) / (x^2) = 1

And so:
x^0 = 1