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|
| Author |
Message |
| 25 new of 62 responses total. |
brighn
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response 25 of 62:
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May 1 16:11 UTC 2002 |
BLASPHEMER! John, you are bound to HELL for such a post!
Not for saying it, but because you used syntax VB doesn't recognize. ;}
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flem
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response 26 of 62:
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May 1 16:38 UTC 2002 |
uh...
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jazz
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response 27 of 62:
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May 1 19:19 UTC 2002 |
Paganism != Visual Basic, too.
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mcnally
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response 28 of 62:
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May 1 23:16 UTC 2002 |
If it's not equal, does that mean that it's greater than or less than?
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gelinas
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response 29 of 62:
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May 2 00:23 UTC 2002 |
Yes, the statements
a <> b
and
a != b
and
a!b
are pretty much synonymous. Different systems use different operators.
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jp2
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response 30 of 62:
|
May 2 00:43 UTC 2002 |
This response has been erased.
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mcnally
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response 31 of 62:
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May 2 01:05 UTC 2002 |
re #29: they're not quite the same, actually..
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morwen
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response 32 of 62:
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May 2 01:32 UTC 2002 |
resp:24 Jazz, I have to know. What made you post that?
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jaklumen
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response 33 of 62:
|
May 2 10:02 UTC 2002 |
Megazone 23.. I guess I need to look that up.
I enjoy anime, but I am cash-poor. If I had mad money, I'd be
catching up with more stuff like that.. oh yeah, and all the stuff
that's been recommended to me in the music cf.
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remmers
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response 34 of 62:
|
May 2 10:53 UTC 2002 |
Re #28: "Not equal" does not always imply "greater than or
less than". The third possibility is "incomparable". e.g.
red != heavy
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brighn
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response 35 of 62:
|
May 2 15:00 UTC 2002 |
#32> "Muggle" is the word in Harry Potter used to refer to non-magickal sorts
(like his foster family). Many pagans are using "muggle" to refer to
non-pagans, becuase many pagans get confused easily and think that Harry
Potter is a religious document.
#34> That depends on what level you're approaching it from. What happens if
you try:
i% <> cheese$
in BASIC? I'm not really sure. While:
"red" != "dog"
is true while there's no clear sense that "red" is "greater than or less than"
"dog" as far as content goes, the assessment is based on the character values,
not the content (i.e., "dog" < "red" < "zoo"). But then, I imagine you knew
all that, I'm just saying...
|
aruba
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response 36 of 62:
|
May 2 16:24 UTC 2002 |
You can't use < or > to compare elements of a set unless you have defined an
ordering on the set. Arbitrary sets don't have a priori orderings.
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gelinas
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response 37 of 62:
|
May 3 03:13 UTC 2002 |
Mathematically, you are correct; nonetheless, I've used languages that use
the "<>" operator to mean "does not equal".
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jp2
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response 38 of 62:
|
May 3 03:21 UTC 2002 |
This response has been erased.
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brighn
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response 39 of 62:
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May 3 03:49 UTC 2002 |
#37> Where there's no ordering? For instance, while "cat" <> "dog", so too
it is that "Dog" <> "dog" in all the languages I'm familiar with.
I prefer perl's method, even though I thought it was sill the first time I
saw it (where != is for numerics and ne is for strings).
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gelinas
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response 40 of 62:
|
May 3 04:37 UTC 2002 |
I'd have to track down my code to figure out which language it was. If not
C, then probably Pascal. I wasn't doing that kind of stuff in Fortran.
|
oval
|
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response 41 of 62:
|
May 3 05:55 UTC 2002 |
<> meant not equal when i was working on an .asp site. i didn't write the
code, just had to work with it/manipulate it. i found it rather obnoxious ..
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remmers
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response 42 of 62:
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May 3 10:50 UTC 2002 |
Pascal's "not equal to" operator is <>. If I recall correctly,
it's only defined for scalar types (integers, reals, characters).
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jazz
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response 43 of 62:
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May 3 15:28 UTC 2002 |
What made me post it?
Apparently some folks really are confused about paganism, and think
Harry Potter is advocating paganism, or represents paganism in some way. I've
known quite a few pagans - though myself I'm an athiest - and it's not at all
accurate. Except for the bit about their being big hairy jolly folks.
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aruba
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response 44 of 62:
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May 3 16:03 UTC 2002 |
Re #38: You can ceratinly define an ordering on a set; that's what you're
doing when you define the magnitude of complex numbers. In fact, using th
axiom of choice you can prove that any set can be _well-ordered_, meaning
that every subset has a minimum value. But the proof is completely
non-constructive, and I believe no one knows how to well-order, say, the
real numbers.
|
orinoco
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response 45 of 62:
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May 3 19:28 UTC 2002 |
(Er... I think I must have totally misunderstood #44. I would have thought
the real numbers _were_ well-ordered -- any subset of the reals _does_ have
a definite smallest member, no?)
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flem
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response 46 of 62:
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May 3 20:28 UTC 2002 |
Nope. Consider {x | x > 0}. For any x in that set, x/2 is also in the set,
and is smaller.
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jp2
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response 47 of 62:
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May 3 22:04 UTC 2002 |
This response has been erased.
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aruba
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response 48 of 62:
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May 3 22:30 UTC 2002 |
No, flem's example is fine. The standard ordering on the real numbers is
not a well-ordering, because the set {x | x > 0} (for instance) has no
minimum value. (It's also an open set, as you say, though Greg didn't prove
it.)
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jp2
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response 49 of 62:
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May 3 22:32 UTC 2002 |
This response has been erased.
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