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Re: Round-off error?
On 11/4/13 at 11:17 PM, fateman at cs.berkeley.edu (Richard Fateman) wrote: >On 11/1/2013 11:21 PM, Bill Rowe wrote: >>shows the two machine precision numbers differ by 1 bit and >>consequently are considered equal by Equal as documented. >>This type of issue is inherent in floating point arithmetic done on >>any modern computer. >This is not at all true. Mathematica chooses to implement a fuzzy >version of equality that does not correspond to any hardware >floating-point operation on any modern computer that I am aware of. Just to clarify my comment. My intended meaning was (n+1)*dt will differ from n*dt+dt for some values of n and dt, a characteristic of any modern computer and a fundamental characteristic of floating point arithmetic. I had not intended to imply Mathematica's design for Equal is universal.