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So, I came up with this! ( * = multiplication sign)

1 * 4 * 4 * 4 = 4^3

1* 4 * 4 = 4^2

1 * 4 = 4^1

Therefore, 1 = 4^0

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So, I came up with this! ( * = multiplication sign)

1 * 4 * 4 * 4 = 4^3

1* 4 * 4 = 4^2

1 * 4 = 4^1

Therefore, 1 = 4^0

- #2

Kurdt

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One can define exponentials in terms of the family of standard functions and it falls naturally that anything to the power zero is one. Exponentials are defined as follows:

[tex] n^x = e^{x\ln{n}} [/tex]

Now put x = 0 and see what happens.

[tex] n^x = e^{x\ln{n}} [/tex]

Now put x = 0 and see what happens.

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- #3

Zurtex

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Not such a convincing proof when I saw that, given thanks to my wonderful school system I'd learnt of exponentials as a continuation of powers and got taught how to get the taylor series of eOne can define exponentials in terms of the family of standard functions and it falls naturally that anything to the power zero is one. Exponentials are defined as follows:

[tex] x^n = e^{n\ln{x}} [/tex]

Now put n = 0 and see what happens.

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[tex]1 = \frac{a^n}{a^n} = a^{n-n} = a^0[/tex]

What could be more simpler than that?

What could be more simpler than that?

- #5

Kurdt

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I hear you! I was exactly the same until I started university and finally learned things in the "proper" order. I am still very opposed to the way they teach A-level maths in the UK and that is one of the reasons.Not such a convincing proof when I saw that, given thanks to my wonderful school system I'd learnt of exponentials as a continuation of powers and got taught how to get the taylor series of e^{x}and stuff without respect to where e came from in the first place.

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I like this proof best. And I also think it is the formal proof.[tex]1 = \frac{a^n}{a^n} = a^{n-n} = a^0[/tex]

What could be more simpler than that?

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I don't really get it, would you explain more please?One can define exponentials in terms of the family of standard functions and it falls naturally that anything to the power zero is one. Exponentials are defined as follows:

[tex] x^n = e^{n\ln{x}} [/tex]

Now put n = 0 and see what happens.

- #8

Kurdt

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If you set n = 0 you get the following:I don't really get it, would you explain more please?

[tex] n^0 = e^{0\ln{n}} = e^0 = 1[/tex]

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[tex]\forall a \ne 0,\;a^n = a^{n - 1} a \Leftrightarrow a^n = \frac{{a^{n + 1} }}{a}[/tex]

Given that a

[tex]a^0 = a^{ - 1} a \Leftrightarrow a^0 = \frac{{a^1 }}{a} = 1[/tex]

- #10

StatusX

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Here's my proof that 4^0=0:

So, I came up with this! ( * = multiplication sign)

1 * 4 * 4 * 4 = 4^3

1* 4 * 4 = 4^2

1 * 4 = 4^1

Therefore, 1 = 4^0

4^3=4*4*4+0

4^2=4*4+0

4^1=4+0

4^0=0

Can you explain why your proof is better than mine? (it is, but you haven't shown why)

- #11

mathman

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This is a proof by analogy, but not by logic. You can just as well say:Here's my proof that 4^0=0:

4^3=4*4*4+0

4^2=4*4+0

4^1=4+0

4^0=0

Can you explain why your proof is better than mine? (it is, but you haven't shown why)

4^3=1*4*4*4

4^2=1*4*4

4^1=1*4

4^0=1

- #12

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This is not a proof, it's a pattern. x^0 = 1 is defined.

- #13

prasannapakkiam

For example: x^x = y. Techniqually the domain has integers below 0. But when converting it to y=e^(xln(x)), it has immediately taken those integers away...

Anyway, I think that here it is valid. But how can this be proven for sure? Or is it completely unneccessary to consider here...

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Can you prove that [tex]\frac{a^n}{a^m} = a^{n-m}[/tex] when [tex]n = m[/tex]? Usually the proof is only valid when they aren't equal and then people define [tex]x^0=1[/tex] such that the property remains valid for [tex]n = m[/tex].[tex]1 = \frac{a^n}{a^n} = a^{n-n} = a^0[/tex]

What could be more simpler than that?

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- #15

Gib Z

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Excuse me? Why do we have to work on the case n=m specifically when the proof works for all n and m? n=m is a nice case that lets us prove this nicely.Can you prove that [tex]\frac{a^n}{a^m} = a^{n-m}[/tex] when [tex]n = m[/tex]? Usually the proof is only valid when they aren't equal and then people define [tex]x^0=1[/tex] such that the property remains valid for [tex]n = m[/tex].

- #16

Kurdt

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Thats a good point. My example is the opposite way round which I shall correct. But of course it doesn't solve the problem of something like the function you have stated.However, with Kurdt's method; I accept it. However one must be careful when taking logarithms, as this restricts the limit to which a function may take, as I found the hard way...

For example: x^x = y. Techniqually the domain has integers below 0. But when converting it to y=e^(xln(x)), it has immediately taken those integers away...

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0^0 is either undefined or arbitrarily defined to be some value, typically one. It is clear when examining a plot of z=x^y that the value of z at (x,y)=(0,0) depends on the direction via which one approaches this (x,y) pair. All of the previous posts have approached this point in a direction that makes the value appear to be one. I could just as easily make it appear to be zero by looking at [tex]\lim_{x\to0}0^x[/tex] which is obviously zero. I could just as easily make it any complex number!

Given any complex number [itex]a[/itex], it is easy to come up with a form that reaches [itex]0^0[/tex] as some limit and evaluates to [itex]a[/itex]. Suppose [itex]0<||a||<1[/itex]. Using [itex]1=n/n[/tex], [tex]a=(a^{n/n})=(a^n)^{(1/n)}\equiv x^y[/tex] where [itex]x\equiv a^n, y\equiv 1/n[/itex]

Note that both [itex]x[/itex] and [itex]y[/itex] approach zero as [itex]n\to\infty[/itex]. Thus [tex]0^0=\lim_{n\to\infty}(a^n)^{(1/n)} = a[/tex]

==================

0^0 is typically**defined** to be one for the sake of simplicity. This definition eliminates zero as a special case for power series, binomial expansions, etc. It is just a convention however, not a proof.

Given any complex number [itex]a[/itex], it is easy to come up with a form that reaches [itex]0^0[/tex] as some limit and evaluates to [itex]a[/itex]. Suppose [itex]0<||a||<1[/itex]. Using [itex]1=n/n[/tex], [tex]a=(a^{n/n})=(a^n)^{(1/n)}\equiv x^y[/tex] where [itex]x\equiv a^n, y\equiv 1/n[/itex]

Note that both [itex]x[/itex] and [itex]y[/itex] approach zero as [itex]n\to\infty[/itex]. Thus [tex]0^0=\lim_{n\to\infty}(a^n)^{(1/n)} = a[/tex]

==================

0^0 is typically

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- #18

Gib Z

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What proof? Assuming we are dealing with real numbers then the proof usually makes the assumption that a^0=1 when dealing with the case n=m. For such proofs neutrino's proof would use circular reasoning and therefore not be valid.Excuse me? Why do we have to work on the case n=m specifically when the proof works for all n and m? n=m is a nice case that lets us prove this nicely.

In my experience either a^n/a^m=a^(n-m), a^0=1, a^n=a*a^(n-1) or some similar identity is just taken as an axiom and the rest are proved from that (can you show me a proof of one of those without use of the others?) a^n=a*a^(n-1) may seem intuitive, and it definitely is for most value of n, but when n=1 it isn't really clear why it should hold.

Actually the conclusion of the original post is false as others have noted. 0^0 is generally not considered to be 1, but the original post said any number. That means anything which would prove it needs to be incorrect.

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Gib Z

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When m isn't equal to n then we are left with a^(m-n). No doubt about that. The problem is that when n=m and we cancel them out, then we end up with 1/1=1. How do you go from there to concluding that a^n/a^n=a^(n-n)?the one where a^m/a^n is expressed as m terms of a divided by n terms of a, then we cancel them out till we are left with a^(m-n). No assumption that a^0=1 needed there.

- #22

VietDao29

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Well, it's just the way weWhen m isn't equal to n then we are left with a^(m-n). No doubt about that. The problem is that when n=m and we cancel them out, then we end up with 1/1=1. How do you go from there to concluding that a^n/a^n=a^(n-n)?

We notice a nice property of exponential:

[tex]\frac{x ^ 5}{x ^ 3} = \frac{x \times x \times x \times x \times x }{x \times x \times x} = x ^ 2 = x ^ {5 - 3}[/tex], so we define:

[tex]x ^ {m - n} = \frac{x ^ m}{x ^ n}[/tex], i.e, this equation holds true for every m, and n, including m = n, and m < n.

Then, we expand this to the real, and have:

[tex]x ^ {\alpha - \beta} = \frac{x ^ \alpha}{x ^ \beta}, \ \ \ \ \ \alpha , \ \beta \in \mathbb{R}[/tex]

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My point, exactly. Either we need to define a^0=1 for nonzero a and then derive identities like: a^n/a^m=a^(n-m) or we will have to just assume that a^n/a^m=a^(n-m) and then derive a^0=1 from that as neutrino did.Well, it's just the way wedefinethings.

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It has always been my impression that a^0 = 1 in order to make the function a^x continuous at x = 0. Defining a^k/l as *the ***lth** root of **a** raised to the power **k** determines this definition. As for [tex]\frac{a^{m}}{a^{n}} [/tex], it's the same principle. As n, which is kept smaller than m, approaches m, the expression m - n approaches 0 and a^m -n approaches 1 consequently. The only thing to prove here is that, according to our definitions, if

[tex]a/b > c/d > 0 [/tex] then

[tex]x^{a/b} > x^{c/d} > 1[/tex]

Once this is proven, we can show that the limit as the rational number k in x^k approaches 0 is 1. It's clear that as k goes to infinity and hence k goes to 0, a^1/k approaches 1. Hence, we could always chose a/b < 1/k for any k hence showing that as ab approaches 0, x^a/b approaches 1. For negative powers, a similar treatment is needed and we need only to show that positive and negative powers make up a continuous and differentiable function at x = 0.

[tex]a/b > c/d > 0 [/tex] then

[tex]x^{a/b} > x^{c/d} > 1[/tex]

Once this is proven, we can show that the limit as the rational number k in x^k approaches 0 is 1. It's clear that as k goes to infinity and hence k goes to 0, a^1/k approaches 1. Hence, we could always chose a/b < 1/k for any k hence showing that as ab approaches 0, x^a/b approaches 1. For negative powers, a similar treatment is needed and we need only to show that positive and negative powers make up a continuous and differentiable function at x = 0.

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- #25

Hurkyl

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A proof would invoke a

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