# 1 to the 0 power

• So I was on Khan Academy math, and they told me that any number to the 0 power is 1 because normal exponents like 2 to the second power can also be expressed as 1 x (2 x 2), and so any number to the zero power would just be that 1 left over.

But personally, I feel like something is wrong with this. Exponents are how many times a number is multiplied by, right? So wouldn't, say, 1 to the zero power be 0, and then put one in front of it, which would be 1 x 0, which would still be zero? It feels like with exponents of 0, we're not being consistent.

On a material level, it's the same thing. I have 0 copies of some amount of some thing, so I have one copy of it? Isn't mathematics created so we can use it in the real world?

• @The-Rogue-Blade
I’m not an expert myself, but I’ll try to answer your question.
First, let’s look at $$1 \times 0$$. $$0$$, right?
Why is the answer $$0$$?
Multiplication can be thought repeated addition, and 1 added to itself $$0$$ times is 0.
But is there anything interesting about $$0$$?
Anything $$+0$$ is always going to be equal to itself. I think this is called the “addition identity”.
Now, let’s look at $$1^0$$. The answer is $$1$$.
But why $$1$$?
First, exponents or powers can be thought as repeated multiplication.
$$1$$ is what we call the “multiplication identity”, as anything $$\times 1$$ is always equal to itself.
So my assumption is that mathematicians continued this pattern of “anything to the operation of $$0$$ is equal to the identity of the operation in question’s definition, which is another operation repeated.”
This is a bit long, so I hope it doesn’t waste too much time reading this post.
Hope it helps!

• @RZ923 That is an excellent answer! I'd like to add another comment here which comes from a different angle:

@The-Rogue-Blade said in 1 to the 0 power:

But personally, I feel like something is wrong with this. Exponents are how many times a number is multiplied by, right? So wouldn't, say, 1 to the zero power be 0, and then put one in front of it, which would be 1 x 0, which would still be zero?

As an example of exponent multiplication, when we times together $$2^3$$ and $$2^4,$$ for example, then we get

$$2^3 \times 2^4 = 2^7$$

since

$$\textcolor{blue}{2 \times 2 \times 2} \times \textcolor{red}{2 \times 2 \times 2 \times 2} = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$$

We add the powers together.

$$\textcolor{blue}{2^3} \times \textcolor{red}{2^4} = 2^{3 + 4}$$

If we multiply something to the zero power, like $$2^0,$$ and another power of $$2,$$

$$2^0 \times 2^4 = 2^{0 + 4}$$

we will expect the power on the exponent $$2^4$$ to remain the same, since $$0$$ is the identity under addition. (This relates to what @RZ923 said, about additive identities.)

Thus $$2^0$$ is the multiplicative identity, and it's because when the powers sum together, $$0$$ is summed as the additive identity.

$$2^0 \times 2^{\textcolor{red}{\text{anything}}} = 2^{0 + \textcolor{red}{\text{anything}}}$$

$$2^0 \times 2^{\textcolor{red}{\text{anything}}} = 2^{\textcolor{red}{\text{anything}}}$$

$$1 \times 2^{\textcolor{red}{\text{anything}}} = 2^{\textcolor{red}{\text{anything}}}$$

$$\boxed{1 = 2^0}$$

• I mean, there's also like continuing a pattern, but shouldn't math be applied and not just working in theory?

• We already have a ton of exceptions in math, and including one here wouldn't be a problem. Besides, it would be possible to derive this exception easily.

• @The-Rogue-Blade Good point! In practice, $$x^0 = 1$$ works in the sense that it's better than $$x^0 = 0,$$ because if $$x^0 = 0,$$ then we get

$$\textcolor{red}{x^0} \times x^4 = 0$$

$$\textcolor{red}{0} \times x^4 = 0$$

which is a bit strange, since $$\textcolor{red}{x^0}$$ would be able to have this destructive property of obliterating all other exponents.

Additionally, when we prime factorize a number, like $$60,$$

$$60 = 2^2 \times 3^1 \times 5^1$$

we might want to be able to list them out with all primes as a sort of basis, like

$$60 = 2^2 \times 3^1 \times 5^1 \times \textcolor{red}{7^0} \times \textcolor{red}{11^0} \times \textcolor{red}{13^0} \ldots$$

We would only be able to do so if we define $$x^0$$ to equal $$1.$$

• Well, that's just my point of view on this. I was also about to ask why the code was weird

• @The-Rogue-Blade It might be because it's a fresh post, and it takes awhile for the site to render the code...? I'm not sure, but refreshing often fixes the problem!

• Interesting discussion! What do you guys think $$0^0$$ is?

• @thomas It's probably undefined. Since 0^x = 0 and x^0=1 for all nonzero x, if x were to be 0, we would have 0=1, which is very, uh,





let's just not think about that.

• @Potato2017 Interesting! It seems like it's unclear if we can even do it at all...

Would perhaps this graph change your mind?

This is a graph of the function $$x^x$$. What this graph is saying is that if you look at $$1^1$$, $$0.1^{0,1}$$, $$0.01^{0.01}$$, ... these numbers get closer and closer to $$1$$. Do you think that means that $$0^0$$ should be $$1$$? Why or why not?

• @The-Blade-Dancer
I came up with another way of explaining why $$1^0$$ is $$1$$. It had something to do with Place Value.
Let’s just look at base 10.
Any number, say $$1729$$, can be represented with powers of 10.
For example, the $$1$$ represents $$1000$$, or $$1 \times 10^3$$. The $$7$$ represents $$7 \times 10^2$$, and the $$2$$ represents $$2 \times 10^1$$.
Now let’s look at the $$9$$. What does it represent.
Continuing our pattern, it represents $$9 \times 10^0$$.
So $$10^0$$ must be one. If it’s $$0$$, then we don’t have place value.
Hope that helps

• lol you have a point there