How can x^0 be equal to 1?

[Joza]

I always wondered, how can any number raised to the power of 0 be 1.

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

 

[Kurdt]

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:

n^x = e^{x\ln{n}}

Now put x = 0 and see what happens.

 

[Zurtex]

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:

x^n = e^{n\ln{x}}

Now put n = 0 and see what happens.

Not such a convincing proof when I saw that, given thanks to my wonderful school system I'd learned 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.

 

[neutrino]

1 = \frac{a^n}{a^n} = a^{n-n} = a^0What could be more simpler than that?

 

[Kurdt]

Not such a convincing proof when I saw that, given thanks to my wonderful school system I'd learned 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.

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.

 

[Shing]

1 = \frac{a^n}{a^n} = a^{n-n} = a^0What could be more simpler than that?

I like this proof best. And I also think it is the formal proof.

 

[Shing]

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:

x^n = e^{n\ln{x}}

Now put n = 0 and see what happens.

I don't really get it, would you explain more please?

 

[Kurdt]

I don't really get it, would you explain more please?

If you set n = 0 you get the following:

n^0 = e^{0\ln{n}} = e^0 = 1

 

[bomba923]

Shing,
\forall a \ne 0,\;a^n = a^{n - 1} a \Leftrightarrow a^n = \frac{{a^{n + 1} }}{a}
Given that a¹=a,
a^0 = a^{ - 1} a \Leftrightarrow a^0 = \frac{{a^1 }}{a} = 1

 

[StatusX]

I always wondered, how can any number raised to the power of 0 be 1.

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

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)

 

[mathman]

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)

This is a proof by analogy, but not by logic. You can just as well say:

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

 

[Werg22]

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

 

[prasannapakkiam]

Neutrino's proof is the basic elementary method. 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...

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

 

[gunch]

1 = \frac{a^n}{a^n} = a^{n-n} = a^0What could be more simpler than that?

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

 

[Gib Z]

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

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 let's us prove this nicely.

 

[Kurdt]

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...

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.

 

[D H]

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 \lim_{x\to0}0^x which is obviously zero. I could just as easily make it any complex number!

Given any complex number a, it is easy to come up with a form that reaches 0^0[/tex] as some limit and evaluates to a. Suppose 0<||a||<1. Using 1=n/n[/tex], a=(a^{n/n})=(a^n)^{(1/n)}\equiv x^y where x\equiv a^n, y\equiv 1/n<br /> <br /> Note that both x and y approach zero as n\to\infty. Thus 0^0=\lim_{n\to\infty}(a^n)^{(1/n)} = a<br /> <br /> ==================<br /> <br /> 0^0 is typically <b>defined</b> 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.

 

[Gib Z]

DH I think you made the mistake I was about to, for this thread. This thread isn't about the commonly talked about 0^0 which one expects to find, its just about the exponent of zero in general. His original post said he proved that 4^0 = 1, which we know on solid ground =)

 

[gunch]

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 let's us prove this nicely.

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.

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.

 

[Gib Z]

I'm talking about that proof that you see in your Elementary Algebra textbook...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.

 

[gunch]

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.

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)?

 

[VietDao29]

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)?

Well, it's just the way we define things. How would you go about computing: 2^-3, and stuff like that?
We notice a nice property of exponential:
\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}, so we define:

x ^ {m - n} = \frac{x ^ m}{x ^ n}, 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:
x ^ {\alpha - \beta} = \frac{x ^ \alpha}{x ^ \beta}, \ \ \ \ \ \alpha , \ \beta \in \mathbb{R}

 

[gunch]

Well, it's just the way we define things.

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.

 

[Werg22]

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 \frac{a^{m}}{a^{n}}, 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

a/b &gt; c/d &gt; 0 then

x^{a/b} &gt; x^{c/d} &gt; 1

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.

 

[Hurkyl]

Not to be a spoilsport, but none of these are proofs; they are simply demonstrations why we might desire things to behave in a certain way.

A proof would invoke a definition of exponentiation. Once a definition is chosen, the issue is quite clear-cut.

 

[Mute]

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:

n^x = e^{x\ln{n}}

Now put x = 0 and see what happens.

Even with technicalities about proving a^0 = 1 versus defining it, your argument there is circular. You want to "prove" that a^0 = 1, and then proceed to do so by writing it as e^{x\ln a}, and then set x = 0 and say, "Oh, e^0 = 1, so a^0 = 1", but you didn't prove that "e^0 = 1"! You used the result you were trying to prove in order to prove it!

 

[Kurdt]

I didn't say I was trying to prove anything. Its merely the standard definition of an exponential function and shows how numbers to the power zero are defined to be 1. And yes I did not explain other definitions such as that of the exponential function, but I was taking it as a well established result.

 

[ank_gl]

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

perfect example of what will happen if people study more than actually required

 

[arildno]

For a decreasing sequence of integers as exponents, we might use the following definition of exponentiation to prove the statement:
a^{n-1}=\frac{a^{n}}{a}, a\neq{0}, a^{1}=a, n\leq{1}

This definition is sufficient to prove that several properties we would like exponentiation to have actually hold.

 

[bLooTuetH]

a^0 = 1

FIRST: We use a one of the laws of exponents which is (a^m/a^n) = 1.

Let m=n,then (a^n/a^n) = a^(n-n) = a^0, where a is not equal to zero.

In short, (a^n/a^n) = a^0.

SECOND: There's a theorem that any nonzero number divided by itself is equal to 1.

That is, (a^n)/(a^n) = 1.


Combining the first and second by transitive property, we have a^0 = 1.

:)