## mathslover 3 years ago Hey guys, I know that the degree of zero "polynomial" is "undefined" ... But .... can't it be "any whole number" ?

1. mathslover

$\LARGE{0\times x^{3}=0}$ and similarly any whole number can stay there at the place of 3

2. mathslover

@KingGeorge

3. mathslover

$\LARGE{0\times x^{-1}=0}$ But this is not considered as a polynomial since the power of x is in negative and hence this is not a polynomial

4. mathslover

So why not "any whole number" ? Why is it undefined?

5. KingGeorge

That's exactly why it's undefined. Since we can fill in any value for $$k$$, rather than pick out a value, people usually say it's undefined. However, some people do call it degree "-1" Just to distinguish it.

6. KingGeorge

I should also point out that when things are left undefined, it usually means that if we defined them to be something, it would cause problems. For example, take $$y_1=x$$ which is a degree 1 polynomial. If we let $$y_2=0$$ be a degree $$k$$ polynomial, then we have$y_1\cdot y_2=x\cdot0=0$However, it has degree $$1+k\neq k$$. So the degree of the zero polynomial can't be defined as a single integer.

7. mathslover

uh.. How can the degree of any polynomial be in "negative" ? ? ? Ok you mean like this? : $\large{0\times x^{3} \times 0\times x^{5}=0}$ $\large{x^{8}=0\times x^{0}}$ $\large{8\ne 0}$ right ?

8. LolWolf

Here's the response I gave to @hartnn , some time ago. Post 1: 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. Post 2: 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 $$\mathbb{R}$$. Post 3: 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.

9. KingGeorge

When mathematicians use a degree of -1 (or $$-\infty$$), it's merely signifying that the degree of the zero polynomial is not a natural degree to use. If you use $$-\infty$$, the problem I was just mentioning sort of goes away, and you can do $$k-\infty=-\infty$$ (sort of, infinity is weird).

10. LolWolf

11. mathslover

@LolWolf My question is again that : How can there be a polynomial having degree as negative ? $\large{\sqrt{2x} \space \textbf{Not belongs to} \space \frak{Polynomial}}$

12. mathslover

here x has the power of 1/2 .... and hence it is not a polynomial and similarly $\large{x^{-1}\space \textbf{Is also not a polynomial}}$

13. mathslover

If i am wrong please correct me..

14. LolWolf

There's not, directly. The only reason why we wish to have $$\deg(0)=-\infty$$ is so that any absolute comparison (like, for example for some $$x, y$$ we have $$|x|>|y|$$) follows its definition. So, think about it this way, how do we know if a polynomial is bigger than another? And, if we use this measure, what properties does it have?

15. KingGeorge

^^basically what he said. The negative exponent just makes some properties true, that we wanted to be true.

16. KingGeorge

And that's a whole different reason some people leave it undefined. They don't like to call it as a negative degree, so they leave it undefined.

17. LolWolf

Yeah, typically leaving it undefined also helps quite a bit in some proofs, since the case can be ignored or treated separately.

18. mathslover

good explanation folks... thanks ;)