FL. Algebra on the Radio

Is it true that if you give me any number you please (say 3, for example), I can come up with a new number (in this case 3/2) so that the sum and products of our numbers are the same? ( 3+ 3/2 = 3 x 3/2 = 9/2 )

A couple more examples: if you give me 0, I’ll give you 0, and 0+0 = 0 x 0.

If you give me 2, I’ll give you 2 and 2 + 2 = 2 x 2.

If you give me -1, I’ll give you 1/2 since -1 + 1/2 = -1 x 1/2.

Can I always give a response to your number, or is there a way to stump me?


  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.
    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.

    1x = x + 1
    x = x + 1
    0 = 1 (no solution)

    You can stump me with the number 1.

  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!

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