1 to the 0 power
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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.
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@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 \(1.\)
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Well, that's just my point of view on this. I was also about to ask why the code was weird
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@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!
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Interesting discussion! What do you guys think \(0^0\) is?
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@thomas 0
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@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.
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@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?
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@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
