1. Using the notation and terminology of Project IV,  show that a function from  V  to the set of real numbers is cool (i.e., both even and odd) iff it is the zero function.
   
   Idea of SOLUTION. First, assume  f  is cool. We will show that   f  is zero. Consider  x  in  V. The fact that  f  is cool implies that
           f(-x) = f(x) = -f(x),
   which implies that  f(x) = 0.  Thus,  f  is the zero function.
   Conversely, suppose that  f  is the zero function. Then it is easy to show that  f  is the both even and odd, so  f  is cool.
   
   

2. Give an example of a vector space for which the product of odd functions is always odd.
   
   Idea of SOLUTION. Let  V  be a vector space containing only the zero vector. (Here, the  zero vector  could be either the number 0 or the zero vector in any other vector space.)
   
   Consider an arbitrary odd function   f  on  V.  Then (as can easily be shown for ANY odd function)  f(0) = 0.  Since 0 is the only element of the domain of  f,  f  is the zero function. Thus the product of any two odd functions is also the zero function, and therefore the product of odd functions is odd.
   

[ To prove the falsity of the statement “the product of odd functions is odd”, ] most students chose, often without pointing this out explicitly, to give a counterexample only for the case where the vector space is the set of real numbers (with the usual operations of addition and multiplication). A typical counterexample started as follows:

Choose functions  f  and  g by letting  f(x) = g(x) = x.

3. Explain what essential ingredient is missing from this definition of the functions. (This has nothing to do with vector spaces; it is about the polite way to define functions and introduce variables.)
   
   SOLUTION. Two related issues here:  The variable  x  is not introduced (who is  x?),  and the domain of the “functions” is not given (you don't have a function until you have a domain).
   
   
4. Given the kind of functions which are being talked about in Project IV,  explain why this is not an appropriate choice of functions to be used as counterexamples unless one assumes that the vector space  V  is the set of real numbers.
   
   SOLUTION. THIS IS A VERY IMPORTANT POINT. Pay attention to the definitions and rules for the problem you are solving. The point here is to give an example of odd functions whose product is not odd. OUR DEFINITION OF ODD FUNCTIONS SPECIFIED THAT AN ODD FUNCTION WAS A FUNCTION FROM A VECTOR SPACE TO THE REAL NUMBERS. Thus our counterexample should have outputs in the real numbers. The functions given have outputs in the real numbers if and only if the inputs are also real numbers, so the examples make sense in this context only if the domain is the set of real numbers.
   
   It is also true that if we are going to multiply these functions, the outputs must be things that can be multiplied, and, in general, multiplication is not defined in vector spaces. This becomes a problem is you try to multiply these functions, BUT IT DOES NOT BECOME A PROBLEM UNTIL THEN. The main point is, these functions are unsatisfactory to begin with as examples of odd functions unless the domain is the set of real numbers.

We have now chosen functions to be used as counterexamples. These functions are odd (provided the domain is chosen correctly); this should at least be noted and really (in the spirit of Part I of the project) it should be proved. But it is true that these are odd functions (provided the domain is chosen correctly). Now we want to determine the product of these two functions and, if this is to be a counterexample, we want the product not to be odd. So here goes:

Then  fg(x) = f(x)g(x) = xx = x²,
which is even.

Thus,  fg  is not odd, and so the product of two odd functions is not necessarily odd.

5. Explain why this is bad logic (i.e., bad reasoning), especially in the context of Problem 1(c) in Part II of Project IV. The issue here is NOT the failure to introduce  x. The egregious issue here is NOT directly related to Part I or the approach used. The issue here is bad logic (i.e., bad reasoning — some or all of the conclusion as stated is not a consequence of the preceding statements), especially in the context of the entirety of Problem 1 in Part 2 of Project IV.
   
   SOLUTION. There is a difference between reasoning which leads to conclusions which actually do not follow from the premises, on the one hand, and sloppy or incomplete reasoning, on the other hand.
   
   In the statement of the problem, some of the points which were NOT the issue were listed. Similarly, the fact that there is no formal proof of the fact that  fg  is even is not the issue. It is TRUE that  fg  is even; this follows from the premises, even though the proof is not given.
   
   The issue is going from the CORRECT observation that  fg  is even to the next statement, “Thus,  fg  is not odd”.
   “EVEN” does not imply “NOT ODD”.
   IT IS POSSIBLE FOR A FUNCTION TO BE BOTH EVEN AND ODD.
   That is one of the main points of the original Project IV.
   IT IS A DIRECT CONSEQUENCE OF PROBLEM 1 ABOVE, since the zero function is both even and odd.
   Thus, if you want to prove that a function is not odd, IT DOESN'T DO ANY GOOD TO PROVE IT IS EVEN.
   
   
6. Give a correct proof that the function  fg  above is not odd. (Smell the clover in Problem 2 of Part I of Project IV.)
   
   COMMENT on SOLUTION. As pointed out in the solution of Problem 2 of Part I of Project IV, to prove that a function is not odd, you should find a point in the domain of the function for which the condition of oddness is not satisfied. So to prove that  fg  is not odd, you should find a real number  x  such that
         fg(-x) ≠ -fg(x).
   For the functions considered above, any nonzero number will do.

Last modified May 3, 2008 7:45 AM

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