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3 answers

f certain does define a function.

Note that
f(jan) = 31,
f(feb) = 28 (or 29, depending if it's a leap year)
f(mar) = 31
f(apr) = 30
f(may) = 31

And so forth.

Sure, it's a function. However, it's not an _invertible_ function.

Since there are several months which have 31 days,
f^(-1) is not defined.

2007-01-28 05:57:04 · answer #1 · answered by Puggy 7 · 0 0

By most defintions of "function", yes it does.
It maps the months of the year into the set {28, 30, 31}.

This depends on the expression "number of days in feb" being well-defined. If you worry about leap years, then the expression is not well-defined, and f is not a function.

2007-01-28 05:57:14 · answer #2 · answered by morningfoxnorth 6 · 1 0

yes and so does f-1(x)

2007-01-28 05:59:50 · answer #3 · answered by james bond 2 · 0 0

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