20

My teacher says-

The degree of the zero polynomial is undefined.

My book says-

The degree of the zero polynomial is defined to be zero.

Wikipedia says-

The degree of the zero polynomial is $-\infty$.

I am totally confused and want to know which one is true or are all true?

psmears
  • 767
Soham
  • 9,990
  • 11
    Wanna be more confused? I think I have also seen it be defined as $-1$. Haha. – Em. May 23 '16 at 06:04
  • 4
    It's a matter of convention, and there are a few reasons to argue that the best convention is $-\infty$. – Qiaochu Yuan May 23 '16 at 08:29
  • It happens all the time that a Wikipedia page only reflects the page author's thought. Just don't consider it to be the ultimate truth. On the other hand, I very much prefer to consider the degree of the zero polynomial either equal to $-\infty$ or to leave it undefined. When it comes to definitions, none is right or wrong: definitions don't admit “proofs”. – egreg May 23 '16 at 09:36
  • 5
    Yet, I'd argue that the book is wrong, since defining $\deg(0)=0$ thwarts one of the main uses of degree (for univariate polynomials over a field), namely to get the property that Euclidean division of any$~a$ by any nonzero element$~b$ is possible leaving a remainder with strictly lower degree than $\deg(b)$. – Marc van Leeuwen May 23 '16 at 10:17
  • 1
    I agree, the book is wrong. What book is that?! The definition used by your teacher and the definition used by Wikipedia are both useful and common. So you have to be careful what definition the author you read uses. – Jeppe Stig Nielsen Mar 14 '17 at 08:57
  • @JeppeStigNielsen -Would you like to contact the authors?I can give you their email.They say in the preface any criticism or comments on the book contents are welcome... – Soham Mar 14 '17 at 09:03
  • I agree, whatever the degree of the zero polynomial is, it should be strictly less than zero. – azimut Jun 08 '20 at 21:08

3 Answers3

10

Well, it depends.

Mathematical practice shows that sometimes it is useful to define the degree of the zero polynomial to be zero, sometimes to define it to be $-\infty$ and sometimes to leave is undefined. Which option one chooses depends on what one is trying to do.

This is quite different with what happens with the degree of all other polynomials, which is always defined in the same way (*) But don't think that if for the slightiest of reasons we were to fnd it useful to change the definition to do something we wanted, we would.

(*) Actually, that is not exactly true: we sometimes put degrees on polynomials which are different from the usual ones, but usually only on polynomials with more than one variable.

Mariano Suárez-Álvarez
  • 135,076
  • 1
    Can you give an example of where each kind is used? – Soham May 23 '16 at 06:03
  • 13
    Well, for example: the degree of the product of two nonzero polynomials is the sum of the degrees of the factors. If you want to extend this to include the possibility that the factors be zero, one needs to do something, and a natural option is to define the degree of $0$ to be $-\infty$. – Mariano Suárez-Álvarez May 23 '16 at 06:05
  • 9
    For some purposes it i useful to know that the degree of a polynomial $f$ to be the least number $i$ such that the $i$th derivative $f^{(i+1)}$ is equal to zero, and if this is going to work for the zero polynomial too, then you need to define its degree to be $-1$. &c. – Mariano Suárez-Álvarez May 23 '16 at 06:09
  • 8
    So it has yet to be found where $\deg 0=0$ is any good. – Vim May 23 '16 at 06:25
  • There are also contexts for which it makes sense to define the degree of the zero polynomial to be $+\infty$. For example, for nonzero polynomials $P$ and $Q$, it's true that if $P$ divides $Q$ then $\deg P \le \deg Q$; setting $\deg 0=+\infty$ is the only way to extend this fact to the zero polynomial. (Another reason: $\deg P$ equals the number of roots of a polynomial over $\Bbb C$.) – Greg Martin May 23 '16 at 08:19
  • @GregMartin: But in every Euclidean domain the Euclidean function (the statistic used to express that a remainder must be strictly smaller than the divisor) suffers from a similar problem: on one hand its value for the zero element must be strictly smaller than for every nonzero element$~b$ (because one can always have a division by$~b$ be exact), but on the other hand the zero element is always multiple of$~b$ so divisibility goes in the opposite sense for (only) this case. By the way this shows that $\deg(0)$ should be definitely strictly less than $0$ to be of any use. – Marc van Leeuwen May 23 '16 at 10:11
  • Yep, I agree that it's problematic, which is why it's typically undefined. – Greg Martin May 23 '16 at 19:18
  • Leaving it undefined does not resolve any problem, for you need to be careful never to use the degree of the zero polynomial. – Mariano Suárez-Álvarez May 23 '16 at 19:26
9

I think $-\infty $ make sense. Indeed, let $P$ a polynomial of degree $\geq 1$. Then, you have that $$\deg(PQ)=\deg(P)+\deg(Q),$$ for every polynomial $Q$. Now, if you define $\deg(0)=0$, you'll get $$\deg(0\cdot P)=0+\deg(P)>0,$$ which is not compatible with the degree formula. The only way to give a sense to this formula is to define $\deg(0)=-\infty $.

Same if you defined $\deg(0)=-1$, the formula won't be compatible if $\deg(P)\geq 2$.

Dietrich Burde
  • 130,978
Surb
  • 55,662
  • It makes sense if what you want is your first displayed formula to hold. If you are not interested in that, then it does not make a lot of sense for you migh be interested in some other property to hold which forces you to define $\deg 0$ differently. This is a matter of convention, and conventions are chosen with some purpose in mind. – Mariano Suárez-Álvarez May 23 '16 at 06:20
  • 1
    Indeed, only $-\infty$ can act as a black hole. But anyway this's only one aspect of mathematical practices. – Vim May 23 '16 at 06:24
  • 3
    @Vim This isn't really enough to justify $-\infty$: why not $+\infty$ which also acts as a black hole? (Yes, I understand why, but this isn't explained in the answer). In fact, naively $\infty$ is a very natural suggestion since it generalizes the theorem "a polynomial of degree $d$ has at most $d$ roots". – Erick Wong May 23 '16 at 06:34
  • @ErickWong I'd say its because there are some theorems that say "if $\deg P>\deg Q$, then blah blah" but unfortunately I've forgotten most of the algebra I learned so I can't name any :( – Vim May 23 '16 at 06:48
  • @Vim You're probably thinking of "if $\deg P > \deg Q$ then $\deg(P+Q) = \deg P$". If $\deg 0 = +\infty$ then we could take $P=0$, and then we'd have $\deg Q = +\infty$ for any $Q$. – Erick Wong May 23 '16 at 07:46
8

Defining it as $-\infty$ makes the most sense.

As mentioned in Surb's answer and comments, some properties of degrees are kept intact this way, e.g.

  1. $\deg(PQ)=\deg P+\deg Q$
  2. If $\deg P>\deg Q$ then $\deg(P+Q)=\deg P$

It also starts making more sense if you consider expressions that can take on negative powers as well. That is, instead of $\sum_{k=0}^na_kx^k$, consider $\sum_{k=-\infty}^{n}a_kx^k$. So you could have $3x^2+2x$ and $x+1+3x^{-2}$ and $2x^{-3}-\frac45x^{-5}$. Then degree is just "the supremum of all $k$s for which $a_k\neq 0$. The degrees of these 3 expressions are 2, 1 and -3 respectively. Then it's easy to see that 0, which has no nonzero coefficients, has a degree of $-\infty$.

The same works if you consider expressions than can also have fractional degrees. Then the degree of, say, $3\sqrt{x}-x^{-3}$ is $1/2$.

Of course, this inspires the definition of a "dual" degree, which is the infimum instead of the supremum. Then the degree of $3x^4+2x^3+5x^2$ will be 2, and the degree of $0$ will be $\infty$.

Keeping the degree of $0$ undefined is understandable (not everyone wants to deal with infinities). Defining it as $-1$ has merits (if you don't consider negative powers, $0$ is one step down from nonzero constants). But there is absolutely no sense in defining the degree as $0$. The $0$ polynomial has as much similarity with constants, as constants have with linear polynomials.

Meni Rosenfeld
  • 4,013