Numerical analysis, floating-point arithmetic

[Chasing_Time]

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.

Homework Statement

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.

Homework Equations

The general form of a floating point number:

x = d_0.d_1 ... d_{t-1} * 10^e

The Attempt at a Solution

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

x = (1 + \frac {d_1}{2} + ... + \frac {d_{t-1}} {2^{t-1}}) * 2^e
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
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

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.

 

[Mark44]

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.

Homework Statement

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.

Homework Equations

The general form of a floating point number:

x = d_0.d_1 ... d_{t-1} * 10^e

The Attempt at a Solution

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

x = (1 + \frac {d_1}{2} + ... + \frac {d_{t-1}} {2^{t-1}}) * 2^e
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
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

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.

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.

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.