Is 0^0 Equal to 1? An Explanation of Mathematical Concepts and Terminology

[3.1415926535]

I am very pleased with your translation which appears quite perfect.
I contact you by private message for acknowlegments.

Excellent work JJacquelin!

 

[JJacquelin]

Hi !
The paper, now entitled "Zéro Puissance Zero" & "Zero To Zero^th Power" has been updated with the spamiam's English translation. Pubished on Scribd : http://www.scribd.com/JJacquelin/documents
Thanks again,
JJ.

 

[dimension10]

Did you prove my equation for all 0 < j < k already? So that you can now complain about notational issue with 0=j and j=k cases?

But seriously... I just said that I define 0^0 to be 1. Therefore 0^0=1, by definition. OK? What do you think about dialogue like this:

Person A: "I have defined f so that it is f(x)=x^2"

Person B: "I see. I think I'm going to define f so that f(x)=x^3... What a minute! Did you just say that f(x)=x^2? That's wrong! By (my) definition f(x)=x^3, and x^2\neq x^3".

Person A: ""

Well, but actually you 0^0 is not 1 or 0. I mean, I thought it was interdeterminate?

 

[Hurkyl]

Well, but actually you 0^0 is not 1 or 0. I mean, I thought it was interdeterminate?

Viewing the expression 0^0 as a limit form, it is indeed indeterminate.

But the math problem that spawned this thread was not about limit forms.

 

[Orion1]

\lim_{x \rightarrow 0} 0^x = 0
\lim_{x \rightarrow 0} x^0 = 1
0^0 = \text{indeterminate}
[/Color]

 

[micromass]

\lim_{x \rightarrow 0} 0^x = 0
\lim_{x \rightarrow 0} x^0 = 1
0^0 = \text{indeterminate}
[/Color]

Indeed, when used as a limit form, it is indeterminate!
But when working with integer exponents only, then 0⁰ is a convention that is often made!

 

[SteveL27]

\lim_{x \rightarrow 0} 0^x = 0
\lim_{x \rightarrow 0} x^0 = 1
0^0 = \text{indeterminate}
[/Color]

True.

But:

\lim_{x \rightarrow 0} x^x = 1

and

| \emptyset^ \emptyset| =1 in set theory.

So in some contexts it makes perfect sense to take 0^0 = 1 as a convention.

I believe these points have already been made several times in this thread.

 

[dimension10]

Let a be a number such that a^a=a/a

a^a=a/a
a^(a-1)=1/a
a^(a-1)=a^-1
a-1=-1
a=0.

Thus, 0^0=0/0. Since 0/0 is intedeterminate, 0^0 is intedeterminate. However, now we have more information. So, we can say:

0^5=1*0*0*0*0*0
0^4=1*0*0*0*0
0^3=1*0*0*0
0^2=1*0*0
0^1=1*0
0^0=1

Thus,

0^0=1.

However, with limits, it is both 0 and 1. But when we have more information, can't we claim it to be 1?

 

[disregardthat]

Since this discussion will never end, we should settle on some compromise. So let's declare 0^0 = 1/2.

 

[Hurkyl]

0^5=1*0*0*0*0*0
0^4=1*0*0*0*0
0^3=1*0*0*0
0^2=1*0*0
0^1=1*0
0^0=1

This argument only makes sense if exponentiation is repeated multiplication. This is why the convention 0^0=1 works very well when plugging 0 into a polynomial or Taylor series, because the exponentiation there is repeated multiplication. (and for the same reason it is often used in algebra and combinatorics)

However, the exponentiation operator on real numbers is not repeated multiplication, and continuity is rather important for calculus and real analysis, which is why real exponentiation leaves 0^0 undefined.

Let a be a number such that a^a=a/a

a^a=a/a
a^(a-1)=1/a
a^(a-1)=a^-1
a-1=-1
a=0.

Ah, the classic fallacies of forgetting your hypotheses, and reversing the flow of logic.

The very premise of your problem requires a to be a positive number. (0 is not a positive number)

Also, what you have proved is

If a is a (positive) number satisfying a^a=a/a, then a = 0​

which is very different from

If a=0, then a^a = a/a.​

 

[Mute]

Let a be a number such that a^a=a/a

a^a=a/a
a^(a-1)=1/a
a^(a-1)=a^-1
a-1=-1
a=0.

Ah, the classic fallacies of forgetting your hypotheses, and reversing the flow of logic.

The very premise of your problem requires a to be a positive number. (0 is not a positive number)

Also, what you have proved is

If a is a (positive) number satisfying a^a=a/a, then a = 0​

which is very different from

If a=0, then a^a = a/a.​

Nevermind the fact that he/she divided by a in the derivation, and took a logarithm of both sides (which also happened to remove the a = 1 solution).

0^5=1*0*0*0*0*0
0^4=1*0*0*0*0
0^3=1*0*0*0
0^2=1*0*0
0^1=1*0
0^0=1

Thus,

0^0=1.

0^5=17*0*0*0*0*0
0^4=17*0*0*0*0
0^3=17*0*0*0
0^2=17*0*0
0^1=17*0
0^0=17

Thus,

0^0=17?​

 

[JJacquelin]

Citation : " We have not seen the last of the Power Less monster ! "

 

[dimension10]

0^5=17*0*0*0*0*0
0^4=17*0*0*0*0
0^3=17*0*0*0
0^2=17*0*0
0^1=17*0
0^0=17

Thus,

0^0=17?

But aren't powers defined that way? I mean,

3^5=1*3*3*3*3*3
3^4=1*3*3*3*3
3^3=1*3*3*3
3^2=1*3*3
3^1=1*3
3^0=1

Isn't that why any number taken up to the power of 0 is 1?

 

[dimension10]

This argument only makes sense if exponentiation is repeated multiplication. This is why the convention 0^0=1 works very well when plugging 0 into a polynomial or Taylor series, because the exponentiation there is repeated multiplication. (and for the same reason it is often used in algebra and combinatorics)

However, the exponentiation operator on real numbers is not repeated multiplication, and continuity is rather important for calculus and real analysis, which is why real exponentiation leaves 0^0 undefined.

Oh, ok. I did not know that.

 

[dimension10]

The very premise of your problem requires a to be a positive number. (0 is not a positive number)

You are right. I think I forgot that 1^1=1/1

1=0

I guess it is a form of zero proof?

 

[dimension10]

So, should we conclude that the answer is different when you look at it from different perspectives? I presume 0 is not a number but just a concept.

 

[Hurkyl]

So, should we conclude that the answer is different when you look at it from different perspectives?

More accurately, the answer is different when you change what the symbols mean.

I presume 0 is not a number but just a concept.

Where did you get that idea?

 

[dimension10]

The evidence micromass pointed out is sufficient. The idea that 0^{0}=0 doesn't make sense.

You cannot say it does not make sense because, viewing it as a limit, it is true in the sense of limit _{x approaches 0} 0^x=0.

In fact, we could claim that it does not make sense to carry out operations with 0 or an extreme number like infinity, since it easily runs into paradoxes.

 

[dimension10]

Where did you get that idea?

I have read quite a few articles that state that it is impossible to consider 0 as a number.

http://answers.yahoo.com/question/index?qid=20090326225006AA1dOJm"

http://answers.yahoo.com/question/index?qid=20110314151806AAvae04"

A number is defined as a mathematical object applied to counting or measurement. 0 cannot be used to count. Thus, it is not a number. Numerical manipulations with 0 often paradox, again suggesting that 0 is not a number.

 

[HallsofIvy]

0 is not a counting number (aka "positive integer") but it is a number. I don't think that "answers.yahoo.com" is a very good source of technical information.

 

[micromass]

Wow, my respect for yahoo answers is suddenly much lower...

 

[Dadface]

The way I like to look at this is that for any equation of the type y=k^x then for all finite values of k,y tends to unity as x tends to zero,even when k tends to zero.I am unable to see any reality in the limiting case where k and x become zero but that's probably just me,I have forgotten most of my maths.

 

[dimension10]

0 is not a counting number (aka "positive integer") but it is a number. I don't think that "answers.yahoo.com" is a very good source of technical information.

Its not the only place I saw that. I've also seen it written in the publications "When is Arithmetic meaningful?" and "The story of 0".

 

[micromass]

Its not the only place I saw that. I've also seen it written in the publications "When is Arithmetic meaningful?" and "The story of 0".

Non of which are mathematics texts.

I agree that 0 was a bit of controversial number and it took a long time before it was accepted. However, over 99.99999999999% of the mathematicians today see 0 as a number, just like 1, -1, pi and i are numbers (and not just concepts).

There will of course be people who disagree with 0 being a number, which is a philosophically valid point of view. But it's not the point of view of current mathematics.

Remember, there are also people that believe the Earth is flat. So you can't just accept whatever people say... You must form your own opinion by thorough research.

 

[gb7nash]

Non of which are mathematics texts.

I wish you were right. I've seen some math texts that regard 0 as a natural number and some that don't.

 

[micromass]

I wish you were right. I've seen math texts that regard 0 as a natural number and some that don't.

Yes, yes, of course. There's a bit of a disagreement wether 0 is a natural number. But everybody agrees that 0 is a number (=integer, rational number, real number, complex number) and not just a concept.

 

[dimension10]

Non of which are mathematics texts.

I agree that 0 was a bit of controversial number and it took a long time before it was accepted. However, over 99.99999999999% of the mathematicians today see 0 as a number, just like 1, -1, pi and i are numbers (and not just concepts).

There will of course be people who disagree with 0 being a number, which is a philosophically valid point of view. But it's not the point of view of current mathematics.

Remember, there are also people that believe the Earth is flat. So you can't just accept whatever people say... You must form your own opinion by thorough research.

I wish you were right. I've seen some math texts that regard 0 as a natural number and some that don't.

Why is it not possible to consider 0 the absence of a number, rather than a number itself?

Yes, yes, of course. There's a bit of a disagreement wether 0 is a natural number. But everybody agrees that 0 is a number (=integer, rational number, real number, complex number) and not just a concept.

The product of a transcendental number with another non-transcendental number should give a transcendental number. But pi multiplied by 0 is 0. 0 is certainly not transcendental.

 

[HallsofIvy]

Why is it not possible to consider 0 the absence of a number, rather than a number itself?

What would that even mean?




The product of a transcendental number with another non-transcendental number should give a transcendental number. But pi multiplied by 0 is 0. 0 is certainly not transcendental.[/QUOTE]
Then why even mention it? No one has suggested that 0 was a transcendental number.
(And it it certainly NOT true the the product of a transcendental number and a non-transcendental number is a transcendental number. I have no idea what you mean by "should give".)

 

[micromass]

Why is it not possible to consider 0 the absence of a number, rather than a number itself?

OK, what do you consider to be "a number"? Do you think -1 is a number? What about pi or i?

The product of a transcendental number with another non-transcendental number should give a transcendental number. But pi multiplied by 0 is 0. 0 is certainly not transcendental.

Why "should" this be the case? You don't have the power to tell mathematics what is true and what's not! (this was said to me by my high school teacher who thought I was complaining too much, I've used the phrase ever since )

 

[dimension10]

What would that even mean?

The product of a transcendental number with another non-transcendental number should give a transcendental number. But pi multiplied by 0 is 0. 0 is certainly not transcendental.

Then why even mention it? No one has suggested that 0 was a transcendental number.
(And it it certainly NOT true the the product of a transcendental number and a non-transcendental number is a transcendental number. I have no idea what you mean by "should give".)

Why "should" this be the case? You don't have the power to tell mathematics what is true and what's not! (this was said to me by my high school teacher who thought I was complaining too much, I've used the phrase ever since )

Well, I mean that:

2pi is transcendental.
3pi is transcendental.
4pi is transcendental.
...
22993.341486513pi is transcendental.

Is there any number that does not follow this rule? Isn't a transcendental number defined as an irrational number that is not the multiple of any 2 or more numbers?