UW Colleges Math Department - "Lucky Larry" Contributions

From Dave Schilling (10/27/94):

AMATYC publishes the journal THE  AMATYC REVIEW (The Official 
Journal of the American Mathematical Association of Two-Year Colleges). 
In addition to its regular articles, it features a section called
"Lucky Larry" where examples of bad mathematics lead to good results.
I enjoy them. We've all seen these kind of examples, but they still offer
some "mathematical humor" in our mundane lives.  I remember Gary Britton
shared a few of his humorous responses a while back. 

	Here's a "Lucky Larry" solution that I recently had on and exam.   

Problem:    lim  ln(lnx)         @ means infinity... assume a quotient 
	    x-@   lnx    

Solution:  "Infinity over Infinity"...Use L'Hopital's Rule, simplify to 

 	    lim    1    = 0 
	    x-@  ln x     

Student's solution:   "cancel"  (lnx)/lnx and replace with 1: 

            You get:   lim  ln 1 = 0         Correct answer! 
                       x-@ 

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From Peter Stonitsch (10/27/94):

Another nasty example of the Lucky Larry type ...

Problem: "Simplify (27/8)^(2/3) ."

Lucky Larry: "(27/8)^(2/3) = (27/8)*(2/3) = 54/24 = 9/4 "

This problem occurred as a question on a departmentalized 105 final
 several years ago. I recall it vividly because I was assigned to grade
 that question.


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From  Bill Demmon (10/27/94):

     I'm grading exams & this one will be no surprise:
     
          Lim sin(4t)/t =4Lim sint/t=4 as t approaches 0.

     The humor dissipates quite rapidly when the student becomes
 adamant about wanting full credit!


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From Gene Robkin (10/27/94):

Unfortunately, this last one is not a Lucky Larry example but rather
an example of a missing step combined with a bit of notational slight
of hand.

The assertion that    Lim ((sin kt)/ct) = (k/c) Lim ((sin t)/t) = k/c   as
t goes to 0 is a theorem not an accident or a lucky guess. The fact 
that students write this sort of thing as a result of not being able to 
do theexpected manipulation does not change the primary fact that it
is a theorem.

It is bedeviling stuff like this that caused me to put a header in all of 
my exams telling the students that the work shown must support the 
results claimed or the scores will be low or zero.  I have forced students
to demonstrate that their Lucky Larry guess was not just a guess.  
Those who can show something get some credit even weeks after the 
exam is handed back.  I feel if I can motivate them enough to work 
something out then giving some credit is not an issue.  I do make them 
do the demonstration as I watch.  That way they can't just hand in 
something they copied somewhere.

A genuine Lucky Larry requires an illegal manipulation that does not work
in the general case but only for the specific situation in which it is applied.

Them's the facts folks.


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From J. Sriskandarajah (11/22/94):

The following example is from Mathematics and Technology in the 
Classroom, Sept. 1994:

      Solve: (x+3)(2-x) = 4

        Solution:       x+3 = 4 or 2-x = 4

        Thus x = 1 or x = -2


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From Gene Robkin (11/22/94):

Sris's latest example got me to thinking about the LL's a bit more.  
Instead of regretting their existence let's turn them around as a 
teaching resource.

The example of  (x + a)(b - x) = c is a LL with x = c - a or b - c if and 
only if 1 = a + b - c.

So the challenge to an algebra class is to analyse the LL and derive 
the conditions which make it work.

It seems worthwhile now to make a serious effort to collect them.  I 
can visualize the text now.  The Mathematical Tools of Lucky Larry.

At the least, a list would be of use in the courses for prospective 
teachers.

I think I will post an enquiry to some of the math newsgroups and 
see what develops.

Comments?


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From J. Sriskanadarajah (11/23/94):

Here are some mathematical tools you might like to use in your text? Pl. 
Check their validity and use with CARE!!! Solution guaranteed??

Ex 1.    Solve: 3x+4 / 6x+7 = x+1 / 2x+3

        This is of the form N1/D1 = N2/D2 with all N1,N2,D1,D2 linear and
the equation reduces to a linear equation and N1 + N2 = k(D1 + D2) then 
the only solution is given by setting N1 + N2 (or D1 + D2) = 0.

        Thus 4x+5 =0 and the unique solution is x = -5/4

Ex 2.   Solve: 16x-3 / 7x+7 = 2x-15 / 11x-25

        This equation satisfies all the above conditions except that the
reduced equation is not linear, but quadratic. If the additional condition
N1 - D1 = k(N2 - D2) is satisfied then the two solutions of the reduced
quadratic equation are given by:

(1) set N1 + N2 (or D1 + D2) = 0 and
(2) set N1 - D1 (or N2 - D2) = 0

Thus (1) implies 18x-18 = 0 or x = 1
and  (2) implies  9x-10 = 0 or x = 10/9

Ex 3.   Solve: 1/(x-7) + 1/(x+9) = 1/(x+11) + 1/(x-9)

        This is of the form 1/D1 + 1/D2 = 1/D3 + 1/D4
with all numerators being equal(numerical) and D1,D2,D3,D4 all linear
s.t  D1 + D2 = D3 + D4 and the reduced equation is also linear,then the
unique solution to this linear equation is given by setting

        D1 + D2 or D3 + D4 = 0

Thus 2x+2 = 0 or x = -1

More to come...Source: Vedic Mathematics

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From Kent Kromarek (12/6/94):

  For your file.

  [6^(2/3)]*[6^(4/3)] = 36^(6/6) = 36

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