# 0^0 ?= 0

#### Iconoclast

##### New member
Any nonzero real number to the power of zero is postive one. Common theoretical evaluations of 0^0 are zero and positive one. It is not evaluated by software, but it was not called an undefined term.

For a time the correct response was zero. I first agreed with this after thinking about fundamental physics of dynamic base arithmetic, but I noticed Google states otherwise.

After I asked around it seems an older text book described a contradiction to the evaluation of +1.
Code:
``````Simplification:  x^a / x^b = x^(a - b)
Instantiation:   0^3 / 0^3 = 0^(3 - 3) = 0^0``````

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#### Iconoclast

##### New member
Code:
``````Simplification:  x^a / x^b = x^(a - b)
Instantiation:   0^3 / 0^3 = 0^(3 - 3) = 0^0

0^3 / 0^3 = 0 / 0 =
0^0 =``````

The expression may have been evaluated to zero and postive one, but the quotient of zero and zero has not been said to have been either of those in particular.

#### hcs

##### Active member
Gah, somehow I kept misreading that as 3^0 / 3^0.

#### Iconoclast

##### New member
That is well enough; I admit you have me scared either way. I thought this was a proof of 0^0 != 1, but after your post I see I completely skimmed something the first time I read it. I read it deeper afterwards, but I completely missed something. Either the text book was incorrectly recalled or it was just a personal question of the informer.

As it turns out this is instead a contradiction to the property of equality:
x^a / x^b = x^(a - b)

The property must now be:
x^a / x^b = x^(a -b); x != 0

If this property is not revised the below statement is not true. Focus on just the first two sides of the equation.

0^3 / 0^3 = 0^(3 - 3) = 0^0

Now I read something about a few physical justifications for 0^0 = 1, but that guy made all these statements violating my favorite fundamentals. He confused the hell out of me with material I don't even know yet as a Pre-calculus student for the time.
http://www.mathforum.org/library/drmath/view/55764.html

">1) Why do we say things like 1/0 are undefined? Can't you call 1/0
>infinity and -1/0 negative infinity? Why not?

1/0 is said to be undefined because division is defined in terms of
multiplication. a/b = x is defined to mean that b*x = a. There is
no x such that 0*x = 1, since 0*x = 0 for all x. Thus 1/0 does not
exist, or is not defined, or is undefined." *end quote*

a/b = x (Given)
a / b * b = x * b (Multiplication Property of Equality)
a = bx (simplified)

However the Multiplication Property of Equality specifically mentions dealing with only real numbers. If that is not the circumstance this *proof* is invalid. If b is 0 the value of a is irrelevant, so I don't fully trust this guy is my point. Fundamental stuff like that is always skipped over I feel.

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#### Ugly Joe

##### New member
Consider the value of base 2 as the value of the exponent approaches zero. It may help to know that raising a base to (1/2) is the same as taking the square root, (1/3) is the same as the cube root, and so on:

2^2 = 4
2^1 = 2
2^(1/2) ~= 1.4142
2^(1/4) ~= 1.1892
2^(1/16) ~= 1.0442
2^(1/64000) ~= 1.00001

As you can see, as the exponent approaches zero, the value approaches 1, not zero. Bases between zero and one approach 1 from the other direction. That is why x^0 = 1.

#### Iconoclast

##### New member
The geometric asymptote is one justification I remember from there.

However the base is +2 instead of 0.

So I read in ancient mathematics negative numbers didn't exist.

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#### Ugly Joe

##### New member
Make the base as close to zero as you want, it's still holds true. I don't see how making it exactly zero would result in an anomaly.

So I read that in present mathematics cinnamon numbers don't exist.

#### Iconoclast

##### New member
Zero to the power of any nonzero real number is zero. No asymptote concept can apply to this abstract exponent expression.
f(x) = 0^x

Have you graphed the function modeling Euler's number?
f(x) = [(x + 1) / x]^x
graphed a few negative and positive inputs with their outputs...pretty fun

#### Garrot

##### Member
Code:
``````Simplification:  x^a / x^b = x^(a - b)
Instantiation:   0^3 / 0^3 = 0^(3 - 3) = 0^0``````

assuming 0^3 is zero --and I don't see any reason not to-- aren't you dividing by zero on the left side of the instantiation there making this bullshit anyway? I mean you could take the limit as x->0 of (0^x/0^x) which you could use to maybe justify 0^0=1, but good luck with that.

edit: welp, so much for reading comprehension

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#### hcs

##### Active member
Yeah, he pointed that out here
The property must now be:
x^a / x^b = x^(a -b); x != 0

And yeah I think the limit interpretation is as good as I can think of.

#### Ugly Joe

##### New member
Not that I think it matters, but couldn't you just rewrite it:
x^a * x^b = x^(a+b)
0^3 * 0^-3 = 0^(3-3)

That would avoid the division by zero (although still contradict the equality property).

#### Iconoclast

##### New member
"I mean you could take the limit as x->0 of (0^x/0^x) which you could use to maybe justify 0^0=1, but good luck with that."

Nonzero real numbers divide by themselves to return positive one, but 0^x is not known to ever return positive one.

So this indeed only justifies an under-restricted property definition. I just copied it out of another text book...one that usually specified the input != 0 part but not this time.

Ugly Joe
...You might have found my error in recall if there was one. I will have to look at this later.

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#### hcs

##### Active member
0^3 * 0^-3 = 0^(3-3)

That would avoid the division by zero (although still contradict the equality property).

I don't see how it does avoid division by zero, 0^-3 is still 1/(0^3).

Which means that I now see Garrot's point, it should be x^b != 0 rather than x != 0.

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#### Ugly Joe

##### New member
D'oh. I forgot about that rule (0^x = undefined; x < 0). I knew something was off when I posted that, but I couldn't quite think of it...

#### Iconoclast

##### New member Conversion of 25A from Base Eleven to Decimal

I thought of my own questions when trying to teach for my own repertoire. I completely forgot my attempted imagination on base zero arithmetic. I had to consider this very abstract exponent. 0 digits available means the power is 0 as well.

I remember the question, "How many times does the first digit go into 0^0 = 1," but I remember pure nothing for the time. Such a question seems rather insane and meaningless.

#### Garrot

##### Member
maybe understanding x^0 <-> 0^x is the first step in better understanding of quantum states.

#### Cornellius

##### Active member
I'm with UglyJoe on this one.

Also, some say:

1/3 = .3333

3 x 1/3 = .9999

while others say

3 x 1/3 = 1.000

#### Iconoclast

##### New member
I read in a theory writing online that the number 1.99999... is called "pseudo-one".

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