0^0 ?= 0

I

Iconoclast

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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.
http://www.google.com/search?q=0^0

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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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.
 
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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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.
 
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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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.
 
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
 
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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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).
 
"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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Ugly Joe said:
0^3 * 0^-3 = 0^(3-3)

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

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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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...
 
convertme0.gif

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.
 
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