# What Is e^0 (E To The Power Of 0)?

What is e^{0}? If you remember your exponents, the answer to this question is easy. For all numbers, raising that number to the 0th power is equal to one. So we know that:

**e ^{0}=1**

This answer relies on a intrinsic property of the way exponentiation is defined. Exponentiation is defined as iterative multiplication, so the expression x^{n} means you multiply x by itself n times. Therefore, any expression in the form x^{0} has a meaning of multiplying x by itself 0 times. Multiplying a number by itself 0 times should return the same element, which is the same as multiplying the element by 1, the multiplicative identity. So, for all numbers x, x^{0} should give you the multiplicative identity, which is equal to 1 (except when x=0, which is a special case we will consider later).

The above reasoning is based on the definition of the exponentiation operation. Let’s take a look at the general definition of exponentiation and how we can reason to the claim that x^{0}=1.

## What Is Exponentiation?

Exponentiation is a mathematical operation that involves two numbers, a *base* *b* and an *exponent **n.* If *n* is a positive number, than exponentiation correspond to iterative multiplication of the base. So an exponential expression means you multiply *n* copies of the base *b* together. In other words, the expression

*b ^{n}*

means the product of multiplying *n *bases *b* together. We can see how this works by plugging in actual numbers instead of variables. For instance, 2^{3} can be rewritten as (2×2×2)=8. In other words, 2^{3} is just equal to the number two multiplied by itself 3 times. Likewise, 4^{4} is equal to (4×4×4×4)=256.

Conversely, if *n *is a negative number, then exponentiation defined as:

*b ^{-n}* = 1/

*b*

^{n}Exponentiation with a negative exponent *-n *corresponds to the reciprocal of *b ^{n}. *This particular definition of negative exponentiation is a consequence of a useful rule for combining exponents that we will look at in a bit. For now though, just remember the two definitions of exponentiation by positive and negative integers:

*b ^{n }= *(

*b*×

_{1}*b*×

_{2}*b*×….×

_{3}*b*)

_{n}*b ^{-n} = *1/

*b*

^{n}Exponentiation can be seen as the inverse of the logarithm operation

**Concept Check!**

Take a look at these problems involving exponentiation to make sure you understand the basic concept. For each problem, try to rewrite the problem as an extended series of iterative multiplication:

- 5
^{6} - 7
^{-2} - 1
^{-1}

**Solutions**

- 5
^{6 }= (5×5×5×5×5×5) - 7
^{-2 }= 1/(7×7) - 1
^{-1}= 1/1^{1}

Focusing on problem 3 shows us an interesting fact regarding the number 1 and exponentiation. Namely, for any exponent *n*, 1^{n}=1. This is easy to prove. Since exponentiation is defined as iterative multiplication, 1^{n} just means “multiply *n* copies of 1.” However the product of 1 and 1 is always equal to 1, no matter how many 1s you multiply. So it does not matter what *n *is; 1^{n} will always equal 1.

We can prove that 1^{n}=1 holds for negative exponents too. Remember that *b ^{-n} *is defined as 1/

*b*Therefore, 1

^{n}.^{-n}must be equal to 1/1

^{n}. Since we just proved that 1

^{n}=1, 1

*will always equal 1/1, which is just equal to 1. Therefore, for all*

^{-n}*n,*1

^{-n}=1.

**Extra practice: **See if you can prove that for all bases *b*, *b*^{1}=*b*. You should be able to do this simply by reasoning about the meaning of the exponent operation.

## Arithmetic Operations Involving Exponents

The definition of the exponentiation operation gives us an algorithmic way to deal with problems involving the multiplication or division of terms with exponents. We will first look at the example of multiplying exponent terms.

### Multiplication

In general, the rule for multiplying terms with exponents is:

*b ^{n}×b^{m }= b^{n+m}*

We can show that this rule must be true with a simple numerical example. Say we have 2^{4} and 2^{3}. 2^{4} is the same as (2×2×2×2) and 2^{3} is (2×2×2), so 2^{4}×2^{3} = (2×2×2×2)(2×2×2). Since multiplication is associative (it does not matter where we put the parentheses) we can combine all the terms on the right side of this equation into one big parenthetical expression:

**2 ^{4}×2^{3}=(2×2×2×2×2×2×2)=2^{3+4}=2^{7}**

We can generalize this reasoning for any base *b,* and any exponents *m *and *n*. By definition of the exponentiation operation we know that:

*b*^{1} = *b, *

and that

*b ^{n+1} = b^{n}×b*

^{1}

Generalizing this case by setting *b*^{1}=*b ^{m}* gives us our final expression:

*b*^{n}×b^{m }= b^{n+m}### Division

There are also rules for dividing exponentiated terms. In general, the rule is:

*b ^{n}÷b^{m}= b^{n−m}*

Once again, we can show this rule to be true for a specific numerical example, say 2^{5} and 2^{2}. 2^{5 }is the same as (2×2×2×2×2) and 2^{2} is (2×2). So 2^{5}÷2^{2} is the same as (2×2×2×2×2)/(2×2). Since we are dividing, we can cancel out pairs of like terms and get rid of them, which just leaves (2×2×2).

Incidentally, this particular rule of dividing exponentiated terms is what gives us our definitions of exponentiation by 0.

### Exponentiation

There are also rules for raising exponential terms themselves to exponents. In general, raising an exponent term to an exponent is:

**( b^{n})^{m }= b^{n×m}**

This particular definition of nested exponentiation is a consequence of the associativity of multiplication. terms. The expression (*b ^{n}*)

*tells us to multiply together*

^{m}*m*copies of the base

*b*The base,

^{n}.*b*tells us to multiply

^{n}*n*copies of the base

*b*. So all together, we would have

*m*copies of

*b*where each

^{n}*b*has

^{n}*n*copies of

*b*. To figure out the total number of

*b*s multiplied together, we can just multiply

*m*by

*n to*give us our final exponent expression.

Here is a numerical example to illustrate the point. Say we have (2^{2})^{3}. This expression tells us to multiply together 3 copies of the base 2^{2}. This is the same as (2×2)×(2×2)×(2×2). Since multiplication is associative, we can get rid of the parentheses and combine all the terms into one expression (2×2×2×2×2×2) = (2^{2})^{3} = 2^{6}.

## Exponentiation By 0

Now that we have some exponent rules under our belt, we can see exactly why raising any number to the 0th power always equals 1. Recall that when dividing exponentiated terms, the general rule is:

*b ^{n}÷b^{m}= b^{n−m}*

This is equivalent to *b ^{n}/b^{m}. *If we set both

*n*and

*m*equal to 1 we get:

*b ^{1}÷b^{1} = b^{1−1}= b^{1}/b^{1} = b^{0}*

Since *b ^{1 }*is equal to just

*b*and any number divided by itself is equal to 1, it follows necessarily that

*b*^{0 }= *b ^{1}/b^{1}= *

*b/b*= 1

This expression is true for any non-zero number *b*. Another way of saying this is that *b*^{0} corresponds to dividing *b* by itself, which will always equal 1.

## Exponentiation By Negative Numbers

Now that we have proved that exponentiation by 0 always equals 1, we can generate the definition of negative exponentiation hinted at earlier. Recall that for negative exponentiation:

*b ^{-n} = *1/

*b*

^{n}According to the definition of exponentiation, we know that:

*b ^{n} =* (

*b*

^{n+1})/

*b*

The above statement just follows from what it means to apply the exponentiation operator. We also know that this statement is true in the case that *n=*0:

*b ^{0}=*

*b*

^{1}/

*b = 1*

Now, extending this definition to *n = *-1 gives us:

*b*^{-1} = *b*^{0}/*b* = 1/*b*

We can extend this reasoning to all negative *n *to arrive at our identity statement for negative exponentiation:

*b ^{-n} = 1/b^{n}*

## Raising 0 To A Power

All the above rule are defined with a non-zero base *b*. What happens if we allow *b = *0? The operation of raising zero to an exponent has some special rules.

First, when n is positive, 0^{n}=0. This is easy enough to see; the expression 0^{n} means “multiply together *n* copies of 0.” 0×0 is always equal to 0 no matter how many copies of 0 you have, so it follows that for all positive *n, *0* ^{n}* = 0.

When *n *is negative, 0* ^{n}* is undefined. The reason why is that raising 0 to a negative exponent implies division by 0, which is undefined. Think of it this way, we know that for all

*b*and

*n*,

*b*1/

^{-n}=*b*If we let

^{n}.*b*= 0, then we get:

0^{-n} = 1/0^{n} = 1/0.

Division by zero is an undefined operation, and so raising 0 to some negative power is also an undefined operation.

### 0 To The 0th Power

What about when both *b* and *n* are 0, so 0^{0}? Here we start to get to the controversial territory. Currently, there is no universally accepted value for the expression 0^{0}. Depending on the field, mathematicians will give you different answers. In basic algebra, 0^{0} is normally just defined as equal to 1, however, this is often for reasons of simplicity rather than logical rigor.

Some mathematicians argue that 0^{0} is undefined for the same reason that raising 0 to a negative exponent is undefined; they both imply a division by 0, which is an undefined operation in algebra. Following the normal rules for non-zero exponents while substituting 0 seems to indicate that the expression 0^{0} is the same as 0/0.

Other mathematicians argue on this basis that 0^{0} is not undefined, just *indeterminate*. The reason why is that the expression 0/0 does not give us enough information to determine its value. Expressions like 0/0 are commonly seen in the context of calculus, as one takes the ratio of two derivatives as they approach their limits. In these cases, an answer of 0/0 does not mean that the answer is undefined, just that we need more information to determine its value.