Question

Is 0 a polynomial ?
what is the degree ?

I said YES, '0' is a polynomial and f(x)=0 has degree \(-\infty\)

Answer 1

am i correct?

Answer 2

it has a degree of 0

Answer 3

sure??
just because its constant ?

Answer 4

only \(e^{-\infty}\) equals 0..the rest of the variables don't

Answer 5

well the degree is also infinite

Answer 6

0x^2 = 0

Answer 7

so whats the final degree?
i have read it somewhere that its \(-\infty\) or -1

Answer 8

undefined ..?

Answer 9

http://mathworld.wolfram.com/ZeroPolynomial.html

Answer 10

personally i think it's indeterminate :|

Answer 11

yeah, i have read there only, wolfram

Answer 12

some crazy authors do say that given in the link .. but it is basically undefined

Answer 13

The degree of the zero polynomial is either left undefined, or is defined to be negative (usually −1 or −∞).

--> wikipedia
http://en.wikipedia.org/wiki/Degree_of_a_polynomial

Answer 14

http://planetmath.org/ZeroPolynomial2.html

Answer 15

hmm...thanks for the replies,though.

Answer 16

but everyone is sure about 0 being a polynomial, right ?

Answer 17

Sorry, but, you should not define
\[
\deg(0)=0
\]Typically, it's a good idea to define it as:
\[
\deg(0)=-\infty
\]But, it is mainly used for well-ordering principles in rings \(R\), such that it is \(R[x]\). It can also be left undefined, but either definition causes problems at some point or another in a proof.

Answer 18

And I say "well-ordering principles" with a grain of salt. I mean it to be something similar to the applied well-ordering of \(\mathbb{Z}\), or absolute ordering of \(\mathbb{Q}\) or \(\mathbb{R}\).

Answer 19

Last random, separated post... hopefully. So, we take, for example:
\[
P(x), Q(x)\in R[x]
\]Where \(R[x]\) is a polynomial ring, then, we wish to maintain the following desirable properties (required for any absolute measure):
\[
\deg(P(x)+Q(x))\leq\max(\deg(P(x)), \deg(Q(x)))\\
\deg(P(x)Q(x))\leq \deg(P(x))+\deg(Q(x))
\]That's why it leads us to the conclusion of \(\deg(0)=-\infty\) (where \(0\in R\) is the additive identity). The point is, though, it *could* screw you over, either way, in a proof (as it has me, more than once in number theory), unless you define that specific case as something different, or treat it as a separate case.

Answer 20

really appreciate that ^^ thanks :)

Answer 21

Sure thing... great question, by the way.