Posted on November 16, 2008 by Jason Dyer

After I made my last post I browsed Jeff Miller’s page on Ambiguously Defined Mathematical Terms. A couple parts made my eyebrows go up (and may be fodder for future blog posts) but I wanted to focus on this question:

Is 0 in the domain of ?

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Fred Hollingshead, on November 16, 2008 at 2:44 pm said:If so, would it necessarily mean that 0 is in the domain of x*1/x and vice versa?

Jason Dyer, on November 16, 2008 at 2:53 pm said:It isn’t exactly the same situation because one of them involves a complex number (0 is in the domain of even just sqrt(x-1) if we are mapping to the complex numbers) and the other one involves a disallowed operation.

jd2718, on November 16, 2008 at 6:13 pm said:Maybe 0 no, 0+0i yes? I am weak here.

Jonathan

Mr. K, on November 16, 2008 at 7:49 pm said:It seems to fit to me:

sqrt (0-1) = i (in the complex domain)

0 * i = 0, which is real again.

The fact that you can’t be sure without the complex step is a distraction, I think.

This seems to me to be the same as a 2nd grade teacher telling a student that you can’t larger numbers from smaller ones.

Jason Dyer, on November 16, 2008 at 8:13 pm said:@Mr. K: I am tending to agree (see the solution to a cubic equation, for example, where even when every term is real it can have a complex part of the calculation).

Note that the original page does contain a counterargument:

I take this position because the domain of the product function fg should only be the intersection of the domain of f with the domain of g (e.g., see Larson & Hostetler’s College Algebra, 2nd edition, 1989).Part of the issue here is simply the definition of the word domain.

Jason Dyer, on November 17, 2008 at 9:02 am said:Something else to note is that most graphing calculators cannot handle a function that enters and then leaves the complex numbers, like . Consider when x=-1. Then the equation is , that is -1. In other words the graph of the equation should be the line y=x. But see what it looks like in Fooplot.

Are there any calculus proofs that only work when the domain of each part of the product function is closed in the reals?

Ross Isenegger, on November 18, 2008 at 4:51 pm said:I vote no. I think in Actionscript that 0*NaN (not a Number) is NaN. If it is thought of as f(x)*g(x), g(0) is undefined, so the product is undefined. I don’t think that multiplying by zero takes precedence, if you like.

Jason Dyer, on December 26, 2008 at 7:11 pm said:Just to note, Jeff has (as of December 22nd) changed the entry on domain:

DOMAIN OF A FUNCTION. This term seems not to be precisely defined in algebra textbooks. Mathematics Dictionary by James and James (1949) says that it is “the set of values which the independent variable may take on, or the range of the independent variable.” Merriam-Webster’s 11th Collegiate Dictonary has: “the set of elements to which a mathematical or logical variable is limited; specifically, the set on which a function is defined.” Heath Algebra 2 by Larson, Kanold, and Stiff (1998) defines the domain, or input, of a function as “the x-values of a function y = f(x).”