# Anything to the zero power equals one?

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I understand that anything other than zero to the zero power equals one. But this doesn't seem to make sence to me. Could someone explain how this is, rather than that it is? I'd like as much feedback on this as possible. Thank you.

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I like to think of it this way:

xn=x(n+1)/x

Although the above equation is pretty obvious, it helps to illustrate the idea. Let x be 4 and n be 0.

40=4(0+1)/4=1

You could think of it another way. When multiplying or dividing by indices, you add or subtract the indices. So:

41/41=1

Which could be rewritten as:

40=1

I hope it makes more sense now.

Edited by Samm
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I like to think of it this way:

xn=x(n+1)/x

Although the above equation is pretty obvious, it helps to illustrate the idea. Let x be 4 and n be 0.

40=4(0+1)/4=1

You could think of it another way. When multiplying or dividing by indices, you add or subtract the indices. So:

41/41=1

Which could be rewritten as:

40=1

I hope it makes more sense now.

not yet

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not yet

So let's start with the basics, the integers. Exponentiation extends multiplication just as multiplication extends addition. Multiplication in the integers is defined as repeated addition. For example, 4*3 = 4+4+4. Recursively, the base case is$a*1 = a$ and $a*n= a*(n-1) + a$ is the reduction rule. Exponentiation is similarly defined as repeated multiplication. The base case is $a^1=a$, but now the reduction rule uses multiplication rather than addition: $a^n=a^{n-1}*a$.

This definition of raising a number to the power of a positive integer still applies when one is working in the rationals or the reals:

$x^n = x^{n-1}*x$

So, let's naively substitute n=1:

$x^1 = x^0*x$

Now solve for $x^0$, and use the fact that by definition $x^1=x$:

$x^0 = \frac{x^1}{x} = \frac {x}{x} = 1$

This is valid for all x not equal to zero. It is not valid for x=0 because that would involve computing 0/0, which isn't defined.

Edited by D H
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So let's start with the basics, the integers. Exponentiation extends multiplication just as multiplication extends addition. Multiplication in the integers is defined as repeated addition. For example, 4*3 = 4+4+4. Recursively, the base case is$a*1 = a$ and $a*n= a*(n-1) + a$ is the reduction rule. Exponentiation is similarly defined as repeated multiplication. The base case is $a^1=a$, but now the reduction rule uses multiplication rather than addition: $a^n=a^{n-1}*a$.

This definition of raising a number to the power of a positive integer still applies when one is working in the rationals or the reals:

$x^n = x^{n-1}*x$

So, let's naively substitute n=1:

$x^1 = x^0*x$

Now solve for $x^0$, and use the fact that by definition $x^1=x$:

$x^0 = \frac{x^1}{x} = \frac {x}{x} = 1$

This is valid for all x not equal to zero. It is not valid for x=0 because that would involve computing 0/0, which isn't defined.

Mabye im looking for a simplified answer that just aint there. Does this fact need to be taken on fath? Is it just true by definition?

Edited by Anura
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My answer wasn't simple enough? Sigh.

Try this one more time. Suppose x is not zero. That means we can divide x raised to some power by x. For example, $x^4/x = x^3$, $x^3/x = x^2$ and $x^2/x = x^1=x$. Note that $x^n/x = x^{n-1}$. So what is $x^1/x$, both expressed as a numeric value and as a power of x?

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This is valid for all x not equal to zero. It is not valid for x=0 because that would involve computing 0/0, which isn't defined.

Put that into a diagram.

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My answer wasn't simple enough? Sigh.

Try this one more time. Suppose x is not zero. That means we can divide x raised to some power by x. For example, $x^4/x = x^3$, $x^3/x = x^2$ and $x^2/x = x^1=x$. Note that $x^n/x = x^{n-1}$. So what is $x^1/x$, both expressed as a numeric value and as a power of x?

I understand that all of what your saying is true. D H. Is it possible im learning in the wrong order

? I dont have all of first year algabra figured out yet. And if thats the case would someone like to tutor me? Im not doing this for school but my own personal gain.

Edited by Anura
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$x^0 = \frac{x^1}{x} = \frac {x}{x} = 1$

This is valid for all x not equal to zero. It is not valid for x=0 because that would involve computing 0/0, which isn't defined.

Put that into a diagram.

What are you complaining about? My use of "this"? My use of "for all x not equal to zero"? Is this better:

$x^0 = \frac{x^1}{x} = \frac {x}{x} = 1$

This result, x0=1, is valid for all x except for x=0. It is not valid for x=0 because that would involve computing 0/0, which isn't defined.

I understand that all of what your saying is true. D H. Is it possible im learning in the wrong order

? I dont have all of first year algabra figured out yet. And if thats the case would someone like to tutor me? Im not doing this for school but my own personal gain.

First year algebra in high school? If that is the case, perhaps you are learning things a bit out of order.

Edited by D H
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(...)This result, x0=1, is valid for all x except for x=0. It is not valid for x=0 because that would involve computing 0/0, which isn't defined.

The horizontal line represent the result of x^0 for all numbers. For an infinitesimal close to zero on the positive side, the result is 1. The same occurs on the negative side. The line is like the trajectory of a bullet. Do we have to wonder what happen to this line at point zero?

Does a notion of continuity exist in math?

Edited by michel123456
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The horizontal line represent the result of x^0 for all numbers. For an infinitesimal close to zero on the positive side, the result is 1. The same occurs on the negative side. The line is like the trajectory of a bullet. Do we have to wonder what happen to this line at point zero?

Does a notion of continuity exist in math?

That is but one view of determining the value of 0^0. Let's try a couple more.

1. For all positive values of x, 0^x=0. Thus $\lim_{x\to 0} 0^x = 0$.

So perhaps we should have 0^0 be zero?

2. Even worse, given any number a one can find a pair of functions f(x) and g(x) such that both f(x) and g(x) approach zero as x approaches some value x0 and such that $\lim_{x\to x_0} f(x)^{g(x)} = a$.

Bottom line: 0^0 can only be defined in terms of limits, but different limits can be found that let 0^0 take on any value whatsoever. It is an indeterminate form.

That said, it is convenient to define 0^0 to be 1 as an abuse of notation. This lets us write power series such as

$f(x) = \sum_{n=0}^{\infty} a_n x^n$

Without this convenient abuse of notation, this series would not make sense at x=0.

Edited by D H
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It helps to know how to do math with exponents. When you multiply the same number with an exponent by the same number with a different exponent, you add up the exponents. When you divide by a number with an exponent, it is the same as having a negative exponent. Feel free to test this out on as many real numbers as you want.

$3^1 = 3$, $3^2 = 3*3=9$, $3^3 = 3*3*3 = 27$, $3^3*3^4 = (3*3*3)*(3*3*3*3) = 3^7$

Use that knowledge to solve for division; for example $3^5/3^2 = (3*3*3*3*3)/(3*3) = 3^3 = 3^{(5-2)}$, which if you test works for all numbers. But when the exponents are equal, $3^n/3^n = 3^{(n-n)} = 3^0$, but also , $3^n/3^n = 1 = 3^0$. This is not a formal proof since it uses specific numbers, but if you test it with any number you like (other than zero, because division by zero is undefined), then you get the same.

And yes, you're learning things out of order but that can be fun too.

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Here's what i figured. When you multiply any number (n) by itself zero times, Whatever that number is then becomes one unit (n/n), thus becoming one. Is this correct?

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Here's what i figured. When you multiply any number (n) by itself zero times, Whatever that number is then becomes one unit (n/n), thus becoming one. Is this correct?

That's the reasoning I used. But then, I'm some silly high school student.

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That is but one view of determining the value of 0^0. Let's try a couple more.

1. For all positive values of x, 0^x=0. Thus $\lim_{x\to 0} 0^x = 0$.

So perhaps we should have 0^0 be zero?

You mean this:

I see no problem.

IMHO it is an algebraic notation issue: the horizontal axis is the input, the vertical axis is the result, they are different. So IMHO there are 2 zero's, with a different meaning. If you had written the formula's with red & green numbers, you would have seen that red zero is not the same as green zero. Just as if the graph was a projection of a 3D situation where the horizontal axis is 100 meters above the vertical one.

(note for the reader, this is not what you will learn in standard math)

Edited by michel123456
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2. Even worse, given any number a one can find a pair of functions f(x) and g(x) such that both f(x) and g(x) approach zero as x approaches some value x0 and such that $\lim_{x\to x_0} f(x)^{g(x)} = a$.

Does this theorem have a name?

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(note for the reader, this is not what you will learn in standard math)

Right. I guess that exemplifies why nobody teaches the non-standard michel123456 math.

What you will learn in standard math is that

• In vicinity of the origin of the half plane x>0, xy takes on all values from 0 to infinity. This means from the perspective of a limit (or continuity), 0^0 is indeterminate.
• However, it is extremely convenient if we define it to be one, at least in certain contexts such as the binomial expansion and power series.

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2. Even worse, given any number a one can find a pair of functions f(x) and g(x) such that both f(x) and g(x) approach zero as x approaches some value x0 and such that $\lim_{x\to x_0} f(x)^{g(x)} = a$.

Does this theorem have a name?

0^0 is indeterminate. http://www.wolframal.../input/?i=0%5E0

As far as finding such a pair of functions, here is one such pair: f(x)=x, g(x)=ln(a)/ln(x). Note that with f(x) and g(x) defined in this manner that f(x)g(x) is identically equal to a for all x>0. Note also that both f(x) and g(x) approach zero as x→0+ as required.

Edited by D H
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• 4 weeks later...

If we take things based on the basic definition of what is the power,

since power is multiplicity of self, its definition was made recursive,

$a^{-p} = \frac{1}{a^p} = { \left( \frac{1}{a} \right) }^{p}$

$a^p = a^{p-1} \times a$ and $a^p = \frac{a^{p+1}}{a}$

with base case is for: $a^{0} = 1$, ${0}^x = 0$,

and $0^{0} =$ undefined OR 1, $0^{+\infty} =$ undefined OR 0, $0^{-\infty} =$ undefined OR 0

where undefined might be 1 for $0^{0}$ and 0 for $-\infty$ $+\infty$ according to some mathematicians

that's all I know ...

Edited by khaled
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If x = 1 any power = 1 so x to the zero power = 1

If x is not equal to 1 any power is not equal to one except the power of zero

2^x doubles so 2 to the power of zero is half of two = 1

5^x is 5, 25 125 so increases five fold so 5to the power of 0 is 5/5 = 1

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i0=1

i0=10

i=1 i= all real number, all imaginary number

Can we use x0as a term in the equation?

Edited by alpha2cen
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i0=1

i0=10

i=1

Can we use x0as a term in the equation?

In what equation?

As I have said many times now, you cannot prove that 00 is one because xy takes on all values for (x,y) in the vicinity of the origin. 00 is indeterminate. However, it is often extremely convenient to define x0 to be 1 for all x, including x=0. This is the interpretation of x0 in the binomial theorem, power series, etc.

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In what equation?

As I have said many times now, you cannot prove that 00 is one because xy takes on all values for (x,y) in the vicinity of the origin. 00 is indeterminate. However, it is often extremely convenient to define x0 to be 1 for all x, including x=0. This is the interpretation of x0 in the binomial theorem, power series, etc.

From #20

i is an imaginary number, i.e., sqrt(-1)

Edited by alpha2cen
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i0=1

i0=10

i=1

Can we use x0as a term in the equation?

You cannot undo a relation that maps multiple numbers to one number. Thus your line of reasoning fails at the third step.

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x0=1

x=all real number, all imaginary number possible

20, 30, -20, -30, -i0, i0,........................= x0

Can we use x0as a term in the equation?

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I understand that anything other than zero to the zero power equals one. But this doesn't seem to make sence to me. Could someone explain how this is, rather than that it is? I'd like as much feedback on this as possible. Thank you.

It is a convention designed to make the usual algebra of exponents work

$1 = x/x = x^1 x^{-1} = x^{1-1} = x^0$

Usually $0^0 =1$ by definition.

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