Question:
A well defined function problem?
Steve
2010-04-03 22:38:10 UTC
What makes a function well defined??

Student C tries to define a function h: Q---> Q by rule

h(m/n) = m^2/n, for all integers m and n with n =! 0.

Student D claims that h is not well defined. Justifiy student D's claim.


Is it because of the m^2 that makes the arguement invalid??
Three answers:
alwbsok
2010-04-03 22:53:44 UTC
The problem with this function is that you can have the function value of the same rational number map to two different numbers under this "function"! So, for example:



h(0.5) = h(1/2) = 1^2 / 2 = 1/2

h(0.5) = h(2/4) = 2^2 / 4 = 1



This means that h(0.5) alone is not well-defined. Depending on how you represent the number 0.5, we get different answers. It's not the number that's different, it's just the way we write it down. *That's* what makes it not well defined.



I claim that it's not well defined for every rational number other than 0. See if you can convince yourself of this fact.



A better way to define such a function, for example, would be to define:



h(r) = m^2 / n



where r = m / n and m and n are relatively prime (i.e. share no factors greater than 1). This is well-defined, because the representation of any rational number in lowest terms is unique.



So, for example, 1/2 will still map to 1/2, but 2/4 won't map to 1. Why? Because 2/4 is not in lowest terms. We must first simplify back to 1/2, and map it to 1/2.



Thus, the mapping is well-defined.



============================



Actually, as a point of interest, my Real Analysis lecturer, in one lesson, defined a function:



f(x) = {1 / n, if x is rational, and x = m / n in lowest terms

......... {0, if x is irrational



It was an example of a function that was continuous on the irrational points, but discontinuous on the rational points. It's a well defined function, and it has that interesting property.



EDIT: Expanded the discussion of the well-defined h.
ladwig
2016-09-30 17:04:15 UTC
Well Defined Function
gintable
2010-04-03 22:40:40 UTC
No, it is the fact that he is trying to define a function of two variables as a function of exclusively the ratio of the two variables.



It is better to state it as:

h(m, n) = m^2/n


This content was originally posted on Y! Answers, a Q&A website that shut down in 2021.
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