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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Yes. I see nothing wrong with that equality even if x=0. It's the step

x^(1-1) = x^1 * x^(-1) that I have trouble with when x=0. It appears to be applying the law of exponents x^(a+b) = x^a * x^b. When x=0 and b=-1 the second factor is 0^(-1) which is what I have trouble with. To me this is indicating the reciprocal of 0 (equivalent notationally to 1/0) which the multiplicative inverse axiom precludes.

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Yes, it is applying that identity, which should be true for all x,a and b. But because this leads to division by zero, we rather leave 0^0 as undefined.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Yes. I agree that x^0 = x^(1-1) even if x=0. It's the x^(1-1) = x^1 * x^(-1) I have trouble with.

It appears to me that this is an application of the exponent rule x^(n+m) = x^n * x^m. But when x=0 the right hand side has the factor 0^(-1) which to me indicates the reciprocal of zero which does not exist. So it seems a law of exponents is being applied to an expression that has in it an undefined quantity. That appears to me to be an invalid step.

Oops! I thought post #26 got lost in hyperspace, so I redid it as #28.

*Last edited by noelevans (2012-07-22 12:08:43)*

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

It leads to an undetermined expression for x=0 so that means the starting expression is also undetermined for x=0.

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**cmowla****Member**- Registered: 2012-06-14
- Posts: 57

anonimnystefy wrote:

[...], so x^0=x^(1-1)=x/x. Now, when you want to see what 0^0 is, yo would get that it is the same as 0/0, which is indeterminate i.e. undefined.

JaneFairfax wrote:

Not that one. Here is the flawed step:

bossk171 wrote:Compare that with this:

Also, http://mathworld.wolfram.com/ExponentLaws.html

anonimnystefy wrote:

Yes, it is applying that identity, which should be true for all x,a and b. But because this leads to division by zero, we rather leave 0^0 as undefined.

But by my argument (post #21), 0^0 can be defined to be either 0 or 1. If it was undefined, then we couldn't add integers!

And I don't know about the argument

Since

is discontinuous at the origin, is undefined

because what about

, which never goes through the origin. And if you graph , it has strange behavior not just at the origin.Offline

**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

0/0 is symbolism in calculus for an indeterminate form, not for a number in the Reals.

Since 0 has no multiplicative inverse 0/0 which would come from the definition of division and the product of 0 with its reciprocal is NOT an ALLOWABLE construct in the field; that is, we shouldn't even be writing it and viewing it as if it might be a real number.

Also when defining fractions starting with the whole numbers W={0,1,2,3,...} they are typically defined by the set F = { p/q | p and q are whole numbers and q is not zero }. There is no member of this set F that has a zero in the denominator. To write 2/0 for example is to write something that is not a fraction.

But 0^0 as argued earlier is not equivalent to 0/0. We are not then constrained to leave it out of our definition. We can leave it out of our definition of exponentiation or make our defininition of

exponentiation include it. If left out of the definition it remains undefined. If we include it in our definition then one would hope that it would make sense. Defining ... x^3=1*x*x*x, x^2=1*x*x , x^1=1*x and following this "pattern" sensibly produces x^0 = 1 (multiplied by no x's).

I like 0^0 = 1 because as stated several times already it sure is CONVENIENT. One convenience not mentioned before is that for expressions in algebra like (2x-1)^0, this equals 1 when x=1/2. So we don't have to make an exception and say that (2x-1)^0 =1 EXCEPT when x=1/2.

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

My argument from the other topic, Don Blazys', applies here, too. If we remove the removable singularity, we won't get the same function.

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**mathaholic****Member**- From: Juliania
- Registered: 2012-11-29
- Posts: 2,836
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"Double the fun, double the thrill, double the coolness" - Julianthewiki

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

There is no reason to put that into hide tags...

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**n872yt3r****Member**- Registered: 2013-01-21
- Posts: 392

I consider -0 infinity, because it is beyond definition and 0^0 is the same.

- n872yt3r

Math Is Fun Rocks!

By the power of the exponent, I square and cube you!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

The "cleverest" student is also erring, because, just because the limit is 1 at that point, doesn't mean that the value of that function at that point is 1.

Here is a great explanation of why 0^0 is undefined... Problems with zero by numberphile.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

I have to disagree, it is not undefined. It is defined as 1 for convenience.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Who says so?

Did you watch that video I just posted?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Hi;

The mathematician said so, you obviously did not read the link I gave. Now why would I go and confuse myself by hearing what some amateur does not like about the definition?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

Hi Stefy

I say so.

http://www.mathisfunforum.com/viewtopic … 62#p252062

Post 29.

I wrote:

It is occasionally useful to define an expression as having a certain value because the formulas work better that way.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Hm, okay then. But what are the privileges of defining 0.999... to "exist" as a mathematical concept and be equal to 1?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

For one thing we would have to replace the geometric sum with some other answer. Pretty messy!

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Like what?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Hi;

Please evealuate this sum using the rule for geometric sums.

Hint

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

*Last edited by anonimnystefy (2013-02-06 22:14:13)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Yes, so we would have do away with this:

Want to see a reason why we need 0^0 = 1?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

No, the 0^0 is okay.

0.999... isn't. Making 0.999... barred from mathematics makes as much sense as making it equal to 1

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

You did not understand.

Which you have already said was 1 because of the sum of a geometric series.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

No, that sum is 1.

Here lies the reader who will never open this book. He is forever dead.

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