why root 0 of 0 does not return error?

Question

0 to the power of 0 is 0 multiplied 0 times by itself, a very illogical and twisted statement which returns error, but what about root 0 of 0, the calculator says 0. An example might help us:

$2^3=8$ --> what is the result of multiplying $2 \times 2 \times 2$? $8$

$\sqrt[3]{8}=2$ --> which number to the power of $3$ gives $8$? $2$

about $0$:

1)what is the result of multiplying 0, 0 times by itself?

2)which number to the power of 0 gives 0?

if the answer to question 2 was 0 then 0 to the power of 0 would also be 0. It doesn't make sense, it is contradictory. I think that because the root is the inverse operation of the power it should return error instead of 0.

$2^3$ is the number of functions from a set of three elements to a set of two elements. It follows in this interpretation that $x^0=1$ always ($1$ is for the empty set which is a function always). — Henno Brandsma, Apr 10 '21 at 11:53
@5201314 that's weird. I pressed the button which says C (in orange) and Units (in blue) then the second element in the second row and wrote 0 at the base and at the exponent. the result was 0, if you want I can attach a picture — serax, Apr 10 '21 at 12:07
In the end, you should not be blindly trusting calculators to give the correct results, especially in fringe cases like this. All sorts of weird things can happen, like $0.1+0.2 = 0.30000000000000004$ or $11^6/13 = \dfrac{156158413}{3600}\pi$. — JMoravitz, Apr 10 '21 at 13:20

score 2 · Accepted Answer · answered Apr 10 '21 at 12:11

2

What a calculator returns depends on how it's programmed to define the operation. If $\sqrt[0]0$ is defined as $\sqrt[a]b=b^{1/a}$ as $a$ and $b$ both approach $0$ from positive values ($\to 0^+$), then indeed the limit of $0$ is well-defined. But, if we allow $b$ to approach from negative values or from values that alternate in sign, or if we drop all directional constraints and work in the complex domain, then the limit is no longer defined on all paths.

answered Apr 10 '21 at 12:11

Oscar Lanzi

39,403

Ok, I'll bite. Why the -1? – Oscar Lanzi Apr 10 '21 at 13:08
1

"What a calculator returns depends on how it's programmed" This, exactly. Not every calculator is programmed the same way and not every mathematician necessarily uses the same definitions in some of these somewhat ambiguous situations. – JMoravitz Apr 10 '21 at 13:14

score 0 · Answer 2 · answered Apr 10 '21 at 11:50

0

Usually $x^0 = 1$ for any $x$, also $x=0$. This convention has some advantages in general formulas. So taking the $0$-th power is far from being (almost) injective, so has no inverse that we could call a $0$-th root function.

answered Apr 10 '21 at 11:50

Henno Brandsma

242,131

score 0 · Answer 3 · answered Apr 10 '21 at 14:22

0

Consider $$x^x=e^{x \log(x)}=1+x \log (x)+\frac{1}{2} x^2 \log ^2(x)+\cdots$$ If $x\to 0^+$,$\forall n$, $[x\log(x)]^n \to 0$.

So, what is the limit ?

answered Apr 10 '21 at 14:22

Claude Leibovici

260,315

why root 0 of 0 does not return error?

3 Answers3