Bill Allombert on Tue, 17 Jul 2012 19:40:34 +0200 |
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Re: (Mod(1,2)*x*y)^0 |
On Tue, Jul 17, 2012 at 06:16:25PM +0100, Phil Carmody wrote: > --- On Tue, 7/17/12, Bill Allombert wrote: > > ? (Mod(1,2)*x*y)^0 > > %3 = 1 > > > > Is it intended ? > > Violating distributivity doesn't sound like a good idea: > > ? Mod(1,2)^0*x^0*y^0 > Mod(1, 2) > > Simpler expressions show the loss of type too: > ? (Mod(1,2)*x)^0*y > y I think this one bug is fixed in PARI 2.5 already. PARI 2.3 did not have the ^0 code so returned ? (Mod(1,2)*x)^0 %1 = 1 which was fixed. Cheers, Bill.