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mathslover

  • 2 years ago

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

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  1. mathslover
    • 2 years ago
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    \[\LARGE{0\times x^{3}=0}\] and similarly any whole number can stay there at the place of 3

  2. mathslover
    • 2 years ago
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    @KingGeorge

  3. mathslover
    • 2 years ago
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    \[\LARGE{0\times x^{-1}=0}\] 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
    • 2 years ago
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    So why not "any whole number" ? Why is it undefined?

  5. KingGeorge
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    Oh, thread is here: http://openstudy.com/study#/updates/504acbf6e4b0b72c4ea8acf7

  11. mathslover
    • 2 years ago
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    @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
    • 2 years ago
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    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
    • 2 years ago
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    If i am wrong please correct me..

  14. LolWolf
    • 2 years ago
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    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
    • 2 years ago
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    ^^basically what he said. The negative exponent just makes some properties true, that we wanted to be true.

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

  18. mathslover
    • 2 years ago
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    good explanation folks... thanks ;)

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