## Degree of the Zero polynomial

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??)
 Recognitions: Homework Help 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?

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## Degree of the Zero polynomial

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.)
 Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus 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.

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 Quote by micromass 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.

 Quote by micromass 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.

 Tags abstract algebra, algebra