The Degree of the Zero Polynomial: Why is it Defined as -∞?

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Discussion Overview

The discussion revolves around the definition of the degree of the zero polynomial, specifically why it is defined as -∞. Participants explore the implications of this definition in relation to polynomial properties and the extended real numbers.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Mathematical reasoning

Main Points Raised

  • Some participants note that the zero polynomial is defined as the additive identity in the context of polynomials, leading to questions about its degree.
  • One participant suggests that the zero polynomial can be viewed as an infinite degree polynomial with all coefficients equal to zero, drawing a parallel to the dual nature of zero as both a real and imaginary number.
  • Another participant questions the relevance of the extended real numbers to this definition and points out that polynomials typically have non-negative degrees.
  • A participant references a precalculus textbook that states the zero polynomial has no degree, prompting a discussion about the correctness of this assertion.
  • Some participants argue that the choice of defining the degree of the zero polynomial as -∞ is somewhat arbitrary, noting that it allows for the formula deg(P) + deg(Q) = deg(PQ) to hold true when Q is the zero polynomial.
  • There is acknowledgment that definitions can vary, with some sources defining the degree of the zero polynomial as -1 or stating it has no degree at all.

Areas of Agreement / Disagreement

Participants express differing views on the definition of the degree of the zero polynomial, with no consensus reached on whether it should be defined as -∞, -1, or as having no degree.

Contextual Notes

Participants highlight the dependency of the degree definition on the context of polynomial operations, particularly in relation to the formula involving the degrees of products of polynomials.

marellasunny
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I understand that mathematicians have had to define the number '0' also as a polynomial because it acts as the additive identity for the additive group of poly's.What I do not understand is why they define the degree of the zero polynomial as [ tex ]-\infty[ /tex ].

An explanation on planetMath wasn't that helpful,at the end they point-out to refer to the extended real numbers(don't they mean 'projectively extended real numbers??)
 
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Well, I guess it's similar to how one sometimes regards zero as both a real and an imaginary number because you can write 0 = 0 + i0. Similarly, you can write

0 = 0 + 0x + 0x2 + 0x3 + ...

i.e., you can write '0' as an infinite degree polynomial with all coefficients zero.

(There may be a more rigorous reason, but that's an intuitive one).
 
Thanks!Could you explain what extended real numbers have got to do with this?
But,polynomials always have non-negative degrees.
deg[P(x)]=+n
Why would mathematicians define a polynomial with a negative degree?
 
Huh, I'm looking at a precalculus textbook (Larson, 8th Ed.), and it states that the zero polynomial has no degree. Is that wrong? (Note that no degree ≠ zero degree -- a polynomial that consists of a single non-zero number has a degree of zero.)
 
The choice is pretty arbitrary. Sometimes it defined as having no degree, sometimes it's -1, sometimes it's -\infty.

A handy formula for polynomials is

deg(P)+deg(Q)=deg(PQ)

If we want this formula to hold for the zero polynomial, then we see (by taking Q=0) that

deg(P)+deg(0)=deg(0)

must hold for all P. This is only satisfied with deg(0)=-\infty. This is the reason why they defined it this way. But again, it's pretty arbitrary.
 
micromass said:
This is only satisfied with deg(0)=-\infty. This is the reason why they defined it this way. But again, it's pretty arbitrary.
I see now. Thank you.
 
micromass said:
The choice is pretty arbitrary. Sometimes it defined as having no degree, sometimes it's -1, sometimes it's -\infty.

A handy formula for polynomials is

deg(P)+deg(Q)=deg(PQ)

If we want this formula to hold for the zero polynomial, then we see (by taking Q=0) that

deg(P)+deg(0)=deg(0)

must hold for all P. This is only satisfied with deg(0)=-\infty. This is the reason why they defined it this way. But again, it's pretty arbitrary.

Awesome!Thanks a ton.
 

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