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mathslover
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Hey guys,
I know that the degree of zero "polynomial" is "undefined" ...
But .... can't it be "any whole number" ?
 one year ago
 one year ago
mathslover Group Title
Hey guys, I know that the degree of zero "polynomial" is "undefined" ... But .... can't it be "any whole number" ?
 one year ago
 one year ago

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mathslover Group TitleBest ResponseYou've already chosen the best response.0
\[\LARGE{0\times x^{3}=0}\] and similarly any whole number can stay there at the place of 3
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@KingGeorge
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
\[\LARGE{0\times x^{1}=0}\] But this is not considered as a polynomial since the power of x is in negative and hence this is not a polynomial
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
So why not "any whole number" ? Why is it undefined?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.2
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 wellordering 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 "wellordering principles" with a grain of salt. I mean it to be something similar to the applied wellordering 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.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
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).
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.2
Oh, thread is here: http://openstudy.com/study#/updates/504acbf6e4b0b72c4ea8acf7
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
@LolWolf My question is again that : How can there be a polynomial having degree as negative ? \[\large{\sqrt{2x} \space \textbf{Not belongs to} \space \frak{Polynomial}}\]
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
here x has the power of 1/2 .... and hence it is not a polynomial and similarly \[\large{x^{1}\space \textbf{Is also not a polynomial}}\]
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
If i am wrong please correct me..
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.2
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?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
^^basically what he said. The negative exponent just makes some properties true, that we wanted to be true.
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.2
Yeah, typically leaving it undefined also helps quite a bit in some proofs, since the case can be ignored or treated separately.
 one year ago

mathslover Group TitleBest ResponseYou've already chosen the best response.0
good explanation folks... thanks ;)
 one year ago
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