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Homework Help: Numerical analysis, floating-point arithmetic

  1. Feb 10, 2010 #1
    Hi all, this (probably easy) problem from numerical analysis is giving me trouble. I can't seem to get started and need some poking in the right direction.

    1. The problem statement, all variables and given/known data

    Consider the following claim: if two floating point numbers x and y with the same sign differ by a factor of at most the base B (1/B <= x/y <= B), then their difference x-y is exactly representable in the floating point system. Show that this claim is true for B = 2 but give a counter example for B > 2.

    2. Relevant equations

    The general form of a floating point number:

    [tex] x = d_0.d_1 ... d_{t-1} * 10^e [/tex]

    3. The attempt at a solution

    I have tried exploring the binary case, noting that d_0 must be = 1 in base B=2:

    [tex] x = (1 + \frac {d_1}{2} + ... + \frac {d_{t-1}} {2^{t-1}}) * 2^e [/tex]
    [tex] y = (1 + \frac {d_1}{2} + ... + \frac {d_{t-1}} {2^{t-1}}) * 2^{e-1} = (\frac {1}{2} + \frac {d_1} {4} + ... + \frac {d_{t-1}} {2^t}) * 2^e[/tex]
    [tex] x - y = (1 + \frac {d_1 - 1} {2} + ... + \frac {d_{t-1} - d_{t-2}} {2^{t-1}} - \frac {d_{t-1}} {2^t})*2^e[/tex]

    Is this "exactly representable" in the floating-point system? I don't know what else to do or what to use as a counter example. Am I even on the right track? Thanks for any help.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 11, 2010 #2


    Staff: Mentor

    Your arithmetic is off here. Since you have set this up with x being two times y, the difference x - y better be equal to y.
    Certainly x - y is exactly representable in a base-2 floating-point system, as long as x and y are.

    I don't have any examples in mind that would serve as counterexamples, but if you work with some specific numbers in base 3 or higher bases, you might be able to come up with one. By "specific numbers" I mean that you should work with numbers like 2.0121 X 32 (base-3), rather than symbolically representing the digits with d1, d2, etc. That's where I would start.
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