The Math Factor Podcast

2 Comments »

1. mathphan said,

   March 27, 2009 at 11:59 am

   This can be solved algebraically:
   [spoiler]Let the numbers be a and b
   ab = a + b

   Subtract b from both sides:
   ab – b = a
   Factor out b:
   b(a – 1) = a
   Divide both sides by a – 1:
   b = a / (a – 1)
   Basically for whatever number you give me, I subtract 1 from it and divide it into your original number.
   Examples:
   a = 3
   b = 3/2
   a = 7
   b = 7/6
   a = ?
   b = ? / (? – 1)
   Now looking at the domain of b / (b – 1), all real numbers are possible except b = 1.

   Double-check:
   1x = x + 1
   x = x + 1
   0 = 1 (no solution)

   Answer:
   You can stump me with the number 1.
   [/spoiler]

2. Stephen Morris said,

   March 30, 2009 at 7:03 pm

   This made me ask another question.

   Given a value for the product/sum can you always come up with a pair of numbers that gives you that product/sum?
   Is this pair unique?
   To put it another way: given a number, k, can you always come up with a pair of numbers, a and b, such that a+b = ab = k?
   Consider the expression (x-a)(x-b) = x^2 – (a+b)x + ab = x^2 – kx + k
   This is zero only when x=a or x=b. 
   So a and b are the solutions to the equation x^2 – kx + k = 0.
   We can find a and b by the standard forumala for a quadratic which gives
     a, b = {k +- sqrt(k(k-4)) }/2
   Now consider that k(k-4) term.  
   If it is negative we don’t have a solution, not in real numbers anyway.
   If it is zero we have a solution with a = b.  This happens if k= 0 or k=4 which gives a=b=0 or a=b=2 respectively.
   If it is positive we have a unique solution in the real numbers. k(k-4) is positive when k<0 or k> 4.
   So we can always find a and b when k<= 0 or k>= 4.
   Blimey! Enough with the quadratics already!

RSS feed for comments on this post · TrackBack URL

Leave a Comment

You must be logged in to post a comment.