# What is x^0 – Detailed Explanation & Examples

The answer to the question of what is x to the power of 0 is very simple and easy as $x^{0} = 1$.

It looks very simple, but the question of how x^{0} = 1 now arises, and how true it is for all the values of “$x$”.

What is $x^{0}$, when $x = 0$ itself?

In this complete guide, we will study the expression $x^{0}$ and what it means. Does the answer to $x^{0}$ always equal to “$1$” or are there some exceptions?

## What Is x^0 Equal To?

X to the power of 0 is always equal to 1, which results in this formula: $x^{0} = 1$. This is an interesting question and there are various ways to answer this question. Let us discuss some of the answers that explain why $x^{0} = 1$.

If any variable has power, we basically multiply the same variable by itself depending upon the power value on it. For example, $2^{2} = 2 \times 2 = 4$, $8^{4}= 8\times 8 \times 8 \times 8 = 4096$. So, if a variable has a power of “$0$”, then it means that we are multiplying the variable to itself zero times.

What does it mean that a variable is multiplying to itself zero times? Well, to explain this let us review the concepts of additive identity and multiplicative identity.

Additive identity states that when a number is added to “$0$”, the answer is the number itself. For example, when “$x$” is added to “$0$”, the answer is “$x$”: $x + 0 = x$. So basically, we can say that if we add no numbers to “$x$”, the answer will always be “$x$”. Adding no numbers is basically an additive identity.

Similarly, multiplying no numbers gives us a multiplicative identity that is equal to “$1$”. In the case of multiplicative identity, if we multiply any number by “$1$” it gives us the same number. For example, if a variable “$x$” is multiplied by “$1$”, the answer is “$x$”.

Our main question, “How is $x^{0} = 1$, $x^{0}$?” means that any number having zeroth power and any number to power zero means that no numbers are being multiplied with each other, and that is a multiplicative identity which is equal to “$1$”.

Hence, we can conclude that when no numbers are being multiplied, it gives us the multiplicative identity that is equal to “$1$”.

Any number or variable having a power means that we multiply that number or variable to that power. For example, if we are given $5^6$, we can write it as $5^{6}= 5\times 5\times \times 5 \times 5 \times 5 \times 5$. Now let us draw a pattern by decreasing the power by $”1”$.

$5^{6} = 5\times 5\times 5\times 5 \times 5 \times 5 \times 5 = 15,625$

$5^{5} = 5\times 5\times \times 5 \times 5 \times 5 = 3125$

$5^{4} = 5\times 5\times \times 5 \times 5 = 625$

$5^{3} = 5\times 5\times \times 5 = 125$

$5^{2} = 5\times 5 = 25$

$5^{1} = 5$

So if you look at the pattern closely, what is basically happening here? We are decreasing the power of “$5$” in each step and whenever we decrease one power, we divide the above expression by “$5$”. For example, $5^{6} = 15,625$, and if we divide it by “$5$” we will get $3125$, which is the next answer to $5^{5}$.

So what will happen when we divide $5^{1} = 5$ by “$5$”? The answer would be equal to “$1$”. Hence, any number to the power $0$” will always be equal to “$1$.

Any number to the power zero is always “$1$” and there is a quick method to prove it. For example, let us look at the sequence from $4^{1}$ to $4^{4}$.

$4^{1} = 4$

$4^{2} = 4\times 4\times = 16$

$4^{3} = 4\times 4\times 4 = 64$

$4^{4} = 4\times 4\times 4\times 4 = 216$

From the above sequences and patterns, we can deduce that:

$4^{3} = \dfrac{4^{4}}{4}$

$4^{2} = \dfrac{4^{3}}{4}$

$4^{1} = \dfrac{4^{2}}{4}$

## x^0 = 1 proof

So we can form the formula for the power for any variable “$x$”

$x^{n-1}= \dfrac{x^n}{x}$.

$x^{0}$ will happen when the value of “$n$” is equal to “$1$. Plugging in the value of “$n$” in above equation:

$x^{1-1} = \dfrac{x^1}{x}$

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

Hence, $x^{0} = 1$

Let us prove that any number to the power zero is always “$1$” by using the exponential rule of mathematics. When two numbers having the same base are multiplied by each other, we add their powers or exponents.

$x^{m}\times x^{n} = x^{m + n}$

When two numbers are having the same base and are divided by each other, their powers are subtracted from one another.

$\dfrac{x^{m}}{x^{n}} = x^{m – n}$

Now let us assume that the powers and bases are both the same. Consider two numbers, $x^{m}$ and $x^{n}$ while $m = n$, if both these numbers are divided with each other we will get

$\dfrac{x^{n}}{x^{n}} = x^{n – n} =x^{0}$

We know from the properties of rational and integer exponents that $x^{-n}= \dfrac{1}{x^{n}}$. So, any number having a negative exponent is basically denominator of number “$1$”.

With this, we can write:

$\dfrac{x^{n}}{x^{n}} = x^{n}. x^{-n} = x^{n}. \dfrac{1}{x^{n}}$

$\dfrac{x^{n}}{x^{n}} = x^{0} = 1$.

So if any number is divided by itself, the answer will always be zero and any number with the power zero is basically divided by itself. For example, $5^{0}$ can be written as $\dfrac{5}{5}$, $\dfrac{5^{2}}{5^{2}}$etc. Hence, any number with a zero exponent will always be zero.

Now that you have studied detailed reasoning why $x^{0}$ is always equal to “$1$”, you would be able to explain it to someone else, but what if someone asks you what is $0^{0}$ equal to? That means “What is $x^{0}$ when $x = 0$?” and the answer to this question is presented below.

### What Is 0^0 Equal To?

This is a tricky question and to date, there are differences of opinions on this matter, as some mathematicians say that $0^{0} = 1$, while others say that it cannot be determined or it is an indeterminate form. What does $x^0 = 1$ actually mean and what happens if $x = 0$ when $x = 0$? We get $0^0$, so is $0^0 = 1$? We will discuss the justifications for both cases here.

### Why 0^0 Is Equal To 1

Most of the mathematicians in the 1800s and initial 1900”s believed that $0^{0} = 1$ and there was a general consensus that $0^{0} = 1$. This holds for all basic algebra and polynomial series.

We know that a polynomial expression is written in the form $a_ox^{0} + a_1x^{1}……+ a_nx^{n}$ here “$x$” is the variable while “$a$” is the co-efficient. Polynomial addition is done termwise while their multiplication is done through multiplication property of distribution and exponents.

We can say that “$x$” in the polynomial expression are the indeterminates while “$a$” values are the coefficient and together they form a polynomial ring. A polynomial ring is a set of indeterminates with coefficients and it is represented as R[x].

In a polynomial ring $x^{0}$ is treated as the multiplicative identity of the polynomial expression (it is same point we discussed in answer 1). Thus, $x^{0}$ if multiplied by any polynomial function p(x) will always give us the result p(x). Let us look at an example of a binomial theorem $(1+ x)^{i} = \sum_{n=0}^{i}\binom{i}{n} x^{n}$ is only validated for $x = 0$ when the condition $0^{0} = 1$ exists.

Similarly, different power series identities like $\dfrac{1}{1 – x} = \sum_{k=0}^{\infty}x^{k}$ are only valid when $0^{0} = 1$. Likewise, in differentiation $\dfrac{d}{dx}x^{k}= kx^{k – 1}$ is also only valid for $k = 1$ when $x = 0$ only and only if $0^{0} = 1$.

### Why 0^0 is Indeterminate or Undefined

We have made the case for $0^0 = 1$ and it is mostly used in algebra and basic mathematics. We have discussed why $x^{0}$ through examples of exponentials.

$5^{3} = 5\times 5\times \times 5 = 125$

$5^{2} = 5\times 5 = 25$

$5^{1} = 5$

$5^{0}= 1$

We know that each time we decrease the value of power, we are basically dividing the term with “$5$”. Let us take the case of negative powers of $5$.

$5^{-1} = \dfrac{1}{5}$

$5^{-2} = \dfrac{1}{25}$

Keeping in the view of the above example even when we have a negative base e.g. -5, its power to zero will always be 1 and when you plot the graph for $y = x^{0}$, you will see that when $x = 0$, the value of $y = 1$.

On the contrary, what happens if we take the equation $y = 0^{x}$? Here the base is constant while we are changing the exponent, so let us see if we decrease the value of “$x$” from $3$ to $1$.

$y = 0^{3} = 0$

$y = 0^{2}= 0$

$y = 0^{1}= 0$

Let us assume that $0^{0}= 1$, then

$0^{-1}$ should be $= \dfrac{0}{0}$ as $5^{-1}$ was $\dfrac{1}{5}$.

We know that anything divided by zero is infinity. So for $0^{x}$, what does $x=0$ on a graph look like? For expression $0^{x}$, what is $x=0$ called?

Well, the answer is simple as the answer is undefined in this case because $0^{x}$ is “1” for all positive values and infinity for all the negative values of “$x$”.

So is $x=0$ no solution in this case? The answer is yes and the graph will look like this:

From the graph, we can draw the contradiction to $0^{0}$ being equal to $1$. So we can draw an interesting conclusion here, when we are dealing with the formula $x^{0}$ then $0^{0}$ will always be $1$.

But on the other hand, when are dealing with the formula $0^{x} then 0^{0}$ is undefined. This in itself creates ambiguity and this point has been raised by many mathematicians.

$0^{0}$ is also taken as an undefined term when you are studying calculus, specifically when you are studying the topics of limits, you will find out that $0^0$ is undefined or indeterminate.

When you are solving the problem of the limits and you are asked to evaluate the limit of $0^{0}$, then the limit of such form is always called the limits of indeterminate. We use special techniques like L’Hopital’s rule to solve such limits evaluating a limit of form $0^0$, and limits of that form are called “indeterminate forms.” You will need to use a special technique such as L’Hopital’s rule to evaluate them.

Let us take a simple limit $\lim_{x\to 0^{+}}f(x)$, what would happen if the function is of the form $[f(x)]^{g(x)}$, while $f(x) = 0$, $g(x) = 0$ and $x$ is approaching to 0 this gives us an indeterminant answer.

If we are given a two variable function, say $t^{n}$, and it is continuous on ${(t,n): t > 0}$ but it will not be continuous on ${(t,n): t > 0} U {(0,0)}$ no matter what is the value of $0^{0}$. Hence, while solving the limits and calculus problems it is desired that $0^{0}$ is taken as undefined term.

So, $x^{0} = 1$ is the general consensus while questions are made whether or not $0^0 =1$ or not. You now have an in-depth idea about the topic, but if you really want to dig deep into the debate of whether or not $0^0 = 1$, you can study the work of the mathematicians listed below.

1.  George Baron
2.  Augustin-Louis Cauchy
3.  Leonhard Euler

## The Difference Between $(-1)^{0}$ And $-1^{0}$

Yes, there is a difference in $(-1)^{0}$ and $-1^{0}$. In the expression $(-1)^{0}$, we are taking “$0$” as the power for number “$-1$” so in short, the base is “$-1$” and answer for $(-1)^{0} = 1$. While for $-1^{0}$, the base is “$1$” as $-1$ is basically “$-1 \times 1$”, $1^{0 }= 1$ while the negative signs make it “$-1$”. Hence, $-1^{0} = -1$.

### Is There Any Difference Between Exponent and Power?

Yes, there is a major difference between exponent and power, as Power is considered as a whole expression or answer. Any base to an exponent or its answer is considered power. For example, 81 is considered as the power of 3, since $3^{4} = 81$. In this example, “$3$” is the base while “$4$” is the exponent, and the expression $3^{4}$ is considered as power.

## Conclusion

Let us summarize the whole article through the list of points below.

• In simple mathematics and generally speaking, x^0 will always be equal to 1.
•  x^0 = 1, and x = 0 when we are dealing with simple algebra, polynomials, and power series, while 0^0 is undefined in several topics of calculus, most prominently when dealing with limits or L’hopital’s rule.
• When the base is not zero, for example, when we are given x^0, then it will always be equal to 1. But when we are given zero as a base and the exponent is variable 0^x, then 0^0 will be undefined as “0” to power negative values, giving us undefined values or infinity as an answer.
Through this guide, we can finally make a conclusion about what is the value of $x^{0}$.