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Is $$\dfrac{x^2 + 2x}{x}$$ a polynomial?

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    $\begingroup$ One argument against it being a polynomial would be that the domain doesn't include zero, and a polynomial has always domain = $\mathbb{R}$ $\endgroup$ – Guillermo Mosse Jan 5 '17 at 14:42
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    $\begingroup$ @SeñorBilly A polynomial does not "always" have domain $\mathbb{R}$. Remember, domains are specified, not intrinsic. $\endgroup$ – AlohaSine Jan 5 '17 at 14:46
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    $\begingroup$ @TonyK Definitely. If the expression is considered as an element of $\mathbb{R}(x)$ (the field of fractions of the polynomial ring $\mathbb{R}[x]$), then it is certainly a polynomial, because we can “cancel out the $x$”. As a function of a real variable, it is not a polynomial function, because its domain is not $\mathbb{R}$. $\endgroup$ – egreg Jan 5 '17 at 14:51
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    $\begingroup$ @user1952009 You're comparing apples and oranges. $\endgroup$ – egreg Jan 5 '17 at 14:53
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    $\begingroup$ @user1952009 Not really: 9/3 = 3 is always true; $x^2/x=x$ is only correct if $x \ne 0$; since the left-hand side is undefined for $x=0$. $\endgroup$ – StackTD Jan 5 '17 at 15:04
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Some people would say that the rational number $7/1$ is not really equal to the integer $7$, but merely canonically identified with it. But (after reaching a certain level of sophistication) mathematicians say that $7/1$ and $7$ are indeed equal.


Some people would say that the rational function $$ \frac{x^2+2x}{x} \tag{*}$$ is not really equal to the polynomial $$ x+2, \tag{**}$$ but merely canonically identified with it. But (after reaching a certain level of sophistication) mathematicians say that (*) and (**) are indeed equal.

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  • $\begingroup$ It may depend on the context, but there is some subtlety to this analogy (as pointed out in the comments). If you, after canonical identification, indeed would say they are equal, then what about (what happens at) $x=0$...? $\endgroup$ – StackTD Jan 5 '17 at 15:14
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Well, it's a polynomial in the variable $t=\tfrac{x^2+2x}{x}$... But you probably mean a polynomial in the (real?) variable $x$. What precise definition of polynomial are you using?


I would say no, because it is not of the form $$a_0+a_1x+a_2x^2+\ldots+a_nx^n$$ for any $n \in \mathbb{N}$ and real numbers $a_i$ ($0 \le i \le n$).

Note that you cannot just simplify $$\frac{x^2+2x}{x} = x+2$$ as this equality is only valid for non-zero $x$, so not for all $x$.

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