1 to the 0 power

The Blade Dancer

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?

RZ923

@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!

debbie

@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.

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

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

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

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

The Blade Dancer

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

The Blade Dancer

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.

debbie

@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 $1.$

The Blade Dancer

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

debbie

@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!

thomas

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

The Blade Dancer

@thomas 0

Potato2017

@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,

$$$$
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let's just not think about that.

thomas

@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?

RZ923

@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

The Blade Dancer

lol you have a point there