How can x^0 be equal to 1?

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In summary, the conversation discusses the concept of any number raised to the power of 0 and how it can be equal to 1. Different proofs and arguments are presented, with some discussing the definition of exponentials and others using algebraic manipulations. The conversation also touches upon the definition of 0^0, with some arguing that it is arbitrary and others stating that it is typically defined as 1 for convenience. Overall, the conversation explores the mathematical reasoning behind the concept of 0^0 and its various interpretations.
  • #1
Joza
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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
 
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  • #2
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.
 
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  • #3
Kurdt said:
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.

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 ex and stuff without respect to where e came from in the first place.
 
  • #4
[tex]1 = \frac{a^n}{a^n} = a^{n-n} = a^0[/tex]What could be more simpler than that?
 
  • #5
Zurtex said:
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 ex 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.
 
  • #6
neutrino said:
[tex]1 = \frac{a^n}{a^n} = a^{n-n} = a^0[/tex]What could be more simpler than that?
I like this proof best. And I also think it is the formal proof.
 
  • #7
Kurdt said:
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.
I don't really get it, would you explain more please?
 
  • #8
Shing said:
I don't really get it, would you explain more please?

If you set n = 0 you get the following:

[tex] n^0 = e^{0\ln{n}} = e^0 = 1[/tex]
 
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  • #9
Shing,
[tex]\forall a \ne 0,\;a^n = a^{n - 1} a \Leftrightarrow a^n = \frac{{a^{n + 1} }}{a}[/tex]
Given that a1=a,
[tex]a^0 = a^{ - 1} a \Leftrightarrow a^0 = \frac{{a^1 }}{a} = 1[/tex]
 
  • #10
Joza said:
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)
 
  • #11
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
 
  • #12
This is not a proof, it's a pattern. x^0 = 1 is defined.
 
  • #13
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...
 
  • #14
neutrino said:
[tex]1 = \frac{a^n}{a^n} = a^{n-n} = a^0[/tex]What could be more simpler than that?

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].
 
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  • #15
gunch said:
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].

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.
 
  • #16
prasannapakkiam said:
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.
 
  • #17
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.
 
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  • #18
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 =)
 
  • #19
Gib Z said:
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.
 
  • #20
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.
 
  • #21
Gib Z said:
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)?
 
  • #22
gunch said:
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:
[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]
 
  • #23
VietDao29 said:
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.
 
  • #24
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.
 
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  • #25
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.
 
  • #26
Kurdt said:
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.

Even with technicalities about proving a^0 = 1 versus defining it, your argument there is circular. You want to "prove" that [itex]a^0 = 1[/itex], and then proceed to do so by writing it as [itex]e^{x\ln a}[/itex], 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!
 
  • #27
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.
 
  • #28
gunch said:
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].

perfect example of what will happen if people study more than actually required
 
  • #29
For a decreasing sequence of integers as exponents, we might use the following definition of exponentiation to prove the statement:
[tex]a^{n-1}=\frac{a^{n}}{a}, a\neq{0}, a^{1}=a, n\leq{1}[/tex]

This definition is sufficient to prove that several properties we would like exponentiation to have actually hold.
 
  • #30
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.

:)
 
  • #31
neutrino said:
[tex]1 = \frac{a^n}{a^n} = a^{n-n} = a^0[/tex]


What could be more simpler than that?

I think this actually proves it the previous once doesn't really prove it, because you'd need the information above for it. Besides this is such a simple proof and anyone can understand it.
 

1. What is the significance of proving x^0 = 1?

The significance of this proof lies in its application in various mathematical equations and calculations. It allows for simplification and ease of solving problems involving exponents.

2. How does the proof of x^0 = 1 work?

The proof is based on the fundamental property of exponents, where any number raised to the power of 0 is equal to 1. Therefore, x^0 can be rewritten as x^(1-1), which using the exponent rule becomes x^1/x^1. Since any number divided by itself is equal to 1, x^1/x^1 simplifies to 1.

3. Can this proof be applied to any number or only specific ones?

This proof is applicable to any real number, as long as it follows the rules of exponents. This includes positive, negative, and fractional numbers.

4. What are some real-life examples where this proof is useful?

This proof can be used in various fields such as physics, engineering, and economics. For example, in physics, it can be used to calculate the velocity of an object at rest (v = u + at), where the initial velocity (u) is equal to 0.

5. Are there any exceptions to this proof?

The only exception to this proof is when x = 0. In this case, x^0 is undefined as any number raised to the power of 0 would result in a division by 0, which is undefined in mathematics.

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