0th root
Phil Oberforcher
What's the 0th root of a number?
lwa...@lausd.net
> Help settle a bet.
> What's the 0th root of a number?
There is no zeroth root of a number. It's undefined.
There are two reasons to see why:
1) Every nonzero complex number z has exactly 2 square
roots, 3 cube roots, 4 fourth roots ... n nth roots. So
z must have exactly 0 zeroth roots.
2) The nth root of a number z is z^(1/n). So the 0th
root of a number z is z^(1/0), which is undefined
since division by zero is undefined.
David W. Cantrell
> Help settle a bet.
>
> What's the 0th root of a number?
I'd suppose that some would prefer to leave the 0th root undefined.
But it would be reasonable to define the 0th root of x to be
{ 0 if 0 <= x < 1
{ 1 if x = 1
{ infinity if x > 1.
Note: By "infinity", I mean the positive infinity of the two-point
extension of the reals.
David
cbr...@cbrownsystems.com
> On Oct 16, 7:12 pm, Phil Oberforcher <ph...@feeberfoo.net> wrote:
>
> > Help settle a bet.
> > What's the 0th root of a number?
>
> There is no zeroth root of a number. It's undefined.
>
Well, that depends on how "nth root" is defined (one assumes here that
n is an integer).
> There are two reasons to see why:
>
Yes but...!
> 1) Every nonzero complex number z has exactly 2 square
> roots, 3 cube roots, 4 fourth roots ... n nth roots. So
> z must have exactly 0 zeroth roots.
>
Your observation follows for n > 0 from solving x^n - 1 = 0 in the
field of complex numbers. But how does that proof apply to solving x^0
- 1 = 0?
> 2) The nth root of a number z is z^(1/n).
Don't you mean "an nth root" if we are talking about the complex
numbers?
> So the 0th
> root of a number z is z^(1/0), which is undefined
> since division by zero is undefined.
Seems to be a bit of an abuse of notation there; and you have moved
from one situation (the complex numbers) to quite a different one (the
reals).
"The" nth root of a number isn't necessarily defined in general unless
we are talking about a set of numbers which can be ordered. For
example, in the reals, x is /the/ nth root of y iff x^n = y and x >=
0. That's why /the/ square root of 4 is 2, and not -2.
Instead we might more generally say that in a ring R (ordered or not),
w is "an" nth root of z iff w^n = z.
So for example, in the ring of integers mod 6, there are 1 square
roots of 0 (i.e., 0), 2 square roots of 1 (i.e., 1 and 5), /no/ square
roots of 2, 1 (not "exactly 2"!) square roots of 3 (i.e., 3), 2 square
roots of 4 (i.e. 2 and 4), and /no/ square roots of 5.
From that view point, it is still consistent to say that every element
r of a ring R is /a/ 0th root of 1 (assuming R has a 1) because r^0
denotes the empty product; and the /only/ element of R that has /a/
zeroth root is the 1 of R.
And so if R is the field of real numbers, /every/ real number is "a"
0th root of 1; but the usual definition of "nth root" in the reals
does not allow us to select any particular x as /the/ 0th root of 1.
So in the end, I would concur that the answer the the OP's question is
probably: even though in the reals, every number is /a/ 0th root of
1, /the/ 0th root of a real number is undefined, even when that number
is 1.
Cheers - Chas
William Elliot
> Phil Oberforcher <ph...@feeberfoo.net> wrote:
> >
> > What's the 0th root of a number?
>
> I'd suppose that some would prefer to leave the 0th root undefined.
> But it would be reasonable to define the 0th root of x to be
>
> { 0 if 0 <= x < 1
> { 1 if x = 1
> { infinity if x > 1.
>
When -1 < x < 1, the root is 0.
G. A. Edgar
there is no solution ... *unless* ... we declare that 0^0 is an
indeterminate form, and can have every value, and hence the 0th
root of a is 0 for any a ...
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
beew...@hotmail.com
> Help settle a bet.
>
> What's the 0th root of a number?
The discussions in this thread brings to mind that every operation has
its mechanics (how to crunch the operands) and its applicable range
(the set of operands for which the crunching is defined). For the "n-
th root of x" operation there are two operands: n and x. The
mechanics is (can be) defined by x^(1/n). The usual range on n is any
non-zero complex number. The usual range on x is any complex number.
There is also the special case that the n-th root is not defined by
x^(1/n) for x = 0 and for n being any purely real and non-positive
number. To extend the range of the n-th root operation beyond the
ususal range just stated, one must also carefully define the mechanics
(number crunching) to be applied. In the present example, the n-th
root of x cannot be defined by x^(1/n) for n = 0, but must be defined
by some ancilary mechanics, several of which have been given in this
thread.
- MO
Igor
But then, technically, the zeroth root of one could be any arbitary
number.
David W. Cantrell
Maybe you say that because the 0th power of any number is 1. Nonetheless,
it would still seem reasonable to take _the_ (i.e., the principal) 0th root
of 1 to be 1.
David
David R Tribble
>> roots, 3 cube roots, 4 fourth roots ... n nth roots. So
>> z must have exactly 0 zeroth roots.
>
Chas Brown wrote:
> Your observation follows for n > 0 from solving x^n - 1 = 0 in the
> field of complex numbers. But how does that proof apply to solving
> x^0 - 1 = 0?
Indeed, does that also imply that there are 3/2 roots for
z^(3/2) - 1 = 0? Or -1 roots for z^(-1) - 1 = 0?
I'm not completely sure of this (the caffeine hasn't kicked
in yet), but is it not the case that there are n roots for
z^n for all positive integers n, but only one root/solution
for non-positive integer powers?