From - Mon Jun 23 13:10:43 1997
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Newsgroups: sci.math
Subject: Re: 0^0 and binomial folly
Date: 22 Jun 1997 11:03:52 -0500
Organization: Purdue University
Message-ID: <5ojid8$r26@seaman.cc.purdue.edu>
References: <5oe1ll$8qu@nntp.interaccess.com> <5oed53$pbd@seaman.cc.purdue.edu> <5oie4d$a03@nntp.interaccess.com>
In article <5oie4d$a03@nntp.interaccess.com>,
Joseph B. Dunphy wrote:
>Subject: Re: 0^0 and binomial folly
>Newsgroups: sci.math
>References: <5oe1ll$8qu@nntp.interaccess.com> <5oed53$pbd@seaman.cc.purdue.edu>
>Reply-To: stats@interaccess.com
>Distribution:
> Hey, stupid. The argument was that defining 0^0 was tantamount to
> defining 0^1 / 0^1. You got the sign on the 1 wrong. Not that I
> believe for a second that you did so accidentally.
Your claim was that 0^0 = 0/0, which is the same as 0 * 0^(-1). Either
way, it's nonsense. As David Kastrup correctly pointed out, the same
logic would imply that 0^1 = 0^(2-1) = 0^2/0^1 = undefined. The
problem is that you did not pay attention when you were learning the
laws of exponenents, and you failed to notice that there are some
situations where they don't apply.
>: waiting for you to point out the flaw in the references I gave you
>: (Suppes, Bernays and Jech) in a previous posting. Suppes explicitly
>: states that n^0 = 1 for every cardinal number n, and then adds as a
>
> As this is the first followup in this thread, that's no mean feat.
> Your integrity comes shining through. Unless you were waiting for
> me to respond in a thread in which I had already announced my
> departure. In which case, you're even more of an idiot than I
> thought.
Ok, so you didn't see my previous post. I would send it by email, but you
don't accept email. Therefore, here it is again. In the first quoted
paragraph, you were addressing David Kastrup.
------------------------ begin repost ----------------------------
In article <5obhur$hjm@nntp.interaccess.com>,
Joseph B. Dunphy wrote:
> David, if you haven't gotten it yet, the point to the arguments that
> you offered is that the adoption of such a definition, setting 0^0
> equal to one, would be incompatible with the standard laws of
> exponents, from high school math. As you haven't shown any benefit
> to adopting this construction, aside from the fact that it appeals
> to the anal retentive side of your character, the rest of us see no
> reason to adopt this definition.
Since you won't listen to David Kastrup, perhaps you will listen to a
more authoritative treatment of the standard laws of exponents than
that generally offered in high school math. Try a textbook on
axiomatic set theory. Some possibilities are:
Suppes: Axiomatic Set Theory
Bernays: Axiomatic Set Theory
Jech: Lectures in Set Theory
In case you didn't know it, the rules of arithmetic (including the laws
of exponents) are derived from set theory. They probably didn't
explain this to you in high school math, but it's true, nonetheless.
The laws of exponents derive from the basic definition that if a and b
are cardinal numbers, then a^b is the cardinality of the set of
mappings from b to a. What you call "the laws of exponents from high
school math" can be proved from this definition and the corresponding
definitions for addition and multiplication.
The law of exponents says that if n^a and n^b are defined, then n^(a+b)
is defined and n^(a+b) = n^a * n^b. Notice that the rule does not
guarantee that n^a and n^b are defined whenever n^(a+b) is defined.
This is the source of your error in claiming that 0^0 = 0^1 * 0^(-1).
Perhaps you learned a sloppy version of the law of exponents in high
school.
By the way, another consequence of the definition is that 0^0 = 1,
since there is exactly one way to map the empty set to itself.
Therefore, it is you who advocates deviating from the law of exponents
in claiming that 0^0 should not be 1. It is incumbent on you to "show
any benefit" to denying the definition.
--------------------------------- end repost --------------------------
> Want to use the cardinality of mapping sets to define exponents ?
> Fine. Define 2 ^ -1. Let's see a set with -1 members.
> Better still, how about e ^ i*pi. I can't wait to see that
> construction.
There are different definitions because there are different kinds of
numbers. Before you can define real exponentiation, you need to define
it on the rationals. Before that, you need to define it for the
integers. Before you can do that, you need to define it on the natural
numbers. The natural numbers are just the finite cardinals.
In case you want to just skip that and use the log function directly,
let's see how you would go abaout defining the log function without
using any of the basic operations of arithmetic, such as addition and
multiplication. Were you planning to define it as an integral? What's
an integral? What's a Riemann sum?
> You've pulled the standard freshman screw up of taking the
> development in a single book you found to be canon. WRONG. It is
> one way of constructing one's definitions, and not necessarily the
> best.
I gave you three references, and I could easily give you many more.
During a previous incarnation of this argument I visited the math
library and checked every textbook I could find on axiomatic set
theory. They all agree on the definitions of a+b, a*b, and a^b, where
a and b are cardinal numbers. For example, a*b is the cardinality of
the Cartesian product.
You won't find these definitions in your high school textbooks, because
they don't attempt to give rigorous definitions of the basic operations
of arithmetic. You also won't find these definitions in places like
abstract algebra textbooks, where it is assumed that everyone at least
is familiar with Z and Q and the fact that they obey certain rules, but
the derivation of those operations is left to courses on mathematical
foundations.
>: Theorems are written to be provable statements. No matter how tidy or
>: untidy a statement may be, it is not useful (i.e., is not a theorem and
>: therefore cannot be applied to anything) if it can't be proved. The
>: usual statement of the binomial theorem cannot be proved if you
>: artifically exclude 0^0 from the domain of definition. Let's see your
>: statement of the binomial theorem that is still provable in spite of
>
> Oh, gosh. Requiring both x and y to be nonzero. The nervous
> breakdowns of Purdue students as they're required to go through
> the nervewracking process of determining if this holds in a
> given case. The humanity !
I haven't seen your restatement yet. Have you considered that you also
need to exclude the case (a-a)^n? Have you tried to explain yet why
punching holes in definitions and corresponding ones in theorems is
easier than leaving the definitions and the resulting theorems intact?
>: this anomaly. I think I will prefer my version, which will be much
>: simpler to state, prove, understand, and apply.
>:
> Really ? You think that using set theory to prove the binomial
> theorem is easier than using mathematical induction ? Let me
> guess, you're using differential geometry to prove the Pythagorean
> theorem, right ?
Did I say you couldn't use mathematical induction? There are a few
things you apparently haven't thought about, though:
1. Before you can apply mathematical induction, you have to state
what it is you are trying to prove. If you don't know the
definitions of the basic operations of arithmetic, you can't
even do that much. The definitions come from set theory.
2. The principle of mathematical induction itself comes from set
theory. It has a form know as transfinite induction that
applies to the ordinals. Since the finite ordinals are
identical to the natural numbers, there is a special case of
induction that applies to the naturals. Therefore, even in
mentioning mathematical induction, you were implicitly using
set theory.
>: > As for calculating the derivative of f(x) = x ^ 1 at x = 0 using the
>: > formula (d/dx)(x ^ n) = n * (x ^ (n-1)), again, why would anyone want
>: > to do that ?
>:
>: The same comment applies. If the formula is not correct for all cases,
>: then it needs to be restated in order to make it correct. Your version
>: will be needlessly complicated.
>
> You are a freshman, aren't you ? It isn't right for all cases. Try
> n = 0.
I was a freshman 35 years ago, but even before that I knew the
difference between necesary and unnecessary complications.
>: If you think mathematical arguments are settled only by appeals to
>: authority, then you really are a newcomer at this. Let's see your
>
> It just gets better, and better, doesn't it ? You respond to my
> refusal to take the authority of that childish little FAQ file
> seriously
> as a statement that mathematical arguments are settled by appeals
> to authority, as you put it, turning the statement on its head.
> How could Purdue's standards ever slip so low as to let such a
> sorry excuse for a student like you in, anyway ?
>
>: refutation to the fact that 0^0 = 1 is a theorem (actually a corollary)
>: of ZFC.
Just because you can't answer the question, there's no need to throw a
hissy fit. Your disproof of ZF (I should have said ZF rather than ZFC)
again, is...?
Merely because an argument appears in the FAQ, you can't automatically
conclude that it is wrong. First, you must disprove it. That requires
logic.
--
Dave Seaman dseaman@purdue.edu
++++ stop the execution of Mumia Abu-Jamal ++++
++++ if you agree copy these lines to your sig ++++
++++ see http://www.xs4all.nl/~tank/spg-l/sigaction.htm ++++
---------------------------------------------------------------------------
From - Mon Jun 23 13:38:05 1997
From: ags@seaman.cc.purdue.edu (Dave Seaman)
Newsgroups: sci.math
Subject: Re: 0^0 and binomial folly
Date: 22 Jun 1997 13:11:38 -0500
Organization: Purdue University
Lines: 109
Message-ID: <5ojpsq$rd1@seaman.cc.purdue.edu>
References: <5oe1ll$8qu@nntp.interaccess.com> <5oed53$pbd@seaman.cc.purdue.edu> <5oie4d$a03@nntp.interaccess.com> <5ojcmm$fkr@nntp.interaccess.com>
In article <5ojcmm$fkr@nntp.interaccess.com>,
Joseph B. Dunphy wrote:
> Of course, if, like Seaman, you're satisfied with a definition that
> only makes sense for integral exponents, then one could use the
> standard definition that we all saw in high school.
You have to start somewhere. The definition for the natural numbers
extends easily to the integers, the rationals, and the reals. You
didn't explain the starting point for your definition; how do you
define the log function without using the basic operations of
arithmetic? Do you understand what a circular argument is?
> If b > 0, then a ^ b is equal to the product of a string of b
> factors, all equal to a, that is,
>
> a ^ b = a * a * a * .... ( b copies of a appearing)
Serge Lang, in his algebra text, uses exactly this argument to define
a^n for the case where a is an element of a monoid and n is a natural
number. He concludes that a^0 = e, the identity element of the monoid,
since the resulting product is empty.
The real numbers are a multiplicative monoid with identity element 1.
Therefore, a^0 = 1 for every real a. Corollary: 0^0 = 1.
> Or, to state this inductively,
>
> a ^ 1 = a
It works perfectly well if you change that to a^0 = 1 as a starting
point for the recursion.
> a ^ (n+1) = a * (a ^ n) for all positive integers n
>
> If b < 0, or b = 0, and b = p - q, where p and q ARE positive
> integers, then
>
> a ^ b = ( a ^ p ) / ( a ^ q )
In the sense that if the right side is defined, then so is the left,
and the expressions are equal. I notice you had to exclude the case b
> 0 from your rule. You don't need to do that if you accept that 0^0 =
1. Then you can say that 0^1 = 0^2 / 0^1 only in the sense that if the
right side is defined, then so is the left. You can't argue the
converse, because it isn't true.
> Note that any definition to apply this, which IS a definition
> in standard use, will leave 0 ^ 0 expressed as the ratio
> 0 / 0, with absurd results if we let this equal 1. Namely that
> this would mean that
>
> 1 = 0 / 0 = (0 * 0)/ 0 = 0 * (0/0) = 0 * 1 = 0.
It doesn't follow. See the explanation above. Since 0 / 0 is
undefined, you can't draw any conclusion about the status of 0^(0-0);
it may be defined, or it may not.
At the risk of being repetitive, your argument would imply that 0^1 =
0^(2-1) is undefined. I know you called David Kastrup an idiot for
pointing this out, but I have yet to see your refutation.
> As for his literature citation, which he miraculously expected
> me to see in a thread that I had already openly taken my leave
> of, I am not going to take time out of my way to dig up this
> source, or worse, possibly go through interlibrary loan to get
> ANOTHER one of the dozens of undergrad level set theory
> textbooks out there, just to confirm what seems likely.
> Namely, that someone has confused the concept of power set,
>
> A ^ B, where A and B are sets
>
> with that of exponent, a ^ b, where a and b are real or
> complex numbers. Either that, or he's been encouraged in this,
> by someone making the unnatural choice of defining a ^ b to
> be the cardinality of A ^ B, if card (A) = a, and
> card (B) = b. Where this is true, this is a theorem to be
> proved, not a definition to be made.
Some use the notation
B
A
(i.e. A with a superscript B, written on the left), to mean the set of
mappings from B to A. If a and b are cardinal numbers, then a^b is the
cardinality of that set of mappings, which makes sense because cardinal
numbers are also sets. There are variations in the way it is stated,
but all the textbooks I have checked have used the same fundamental
concepts. I have yet to see any textbook define exponentiation of
cardinals in a fundamentally different way.
> Parting note : I do not want to hear from someone who thinks that I've
> been too harsh with Seaman. The moment he resorted to
> distorting points in an attempt to refute them, he
> forfeited any right to be dealt with civilly. To be
> polite with a snotty little twerp like that is to hand
> him the floor, as anyone who's been on the net for a
> while knows.
I find this fascinating. Just what points have I distorted? For that
matter, what point of yours have I not refuted? When have I resorted
to ad hominem attacks as a substitute for logic? I have endeavored
to be polite throughout theis discussion.
--
Dave Seaman dseaman@purdue.edu
++++ stop the execution of Mumia Abu-Jamal ++++
++++ if you agree copy these lines to your sig ++++
++++ see http://www.xs4all.nl/~tank/spg-l/sigaction.htm ++++
From - Tue Jun 24 13:20:19 1997
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From: stats@interaccess.com ("Joseph B. Dunphy")
Newsgroups: sci.math
Subject: Re: 0^0 and binomial folly
Date: 23 Jun 1997 19:03:15 GMT
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Subject: Re: 0^0 and binomial folly
Newsgroups: sci.math
References: <5oe1ll$8qu@nntp.interaccess.com> <5oed53$pbd@seaman.cc.purdue.edu> <5oie4d$a03@nntp.interaccess.com> <5okk2f$dvk@gap.cco.caltech.edu>
Reply-To: stats@interaccess.com
Distribution: inet
Ilias Kastanas (ikastan@alumni.caltech.edu) wrote, as the world waited in
breathless awe:
: >
: > Hey, stupid. The argument was that defining 0^0 was tantamount to
: > defining 0^1 / 0^1. You got the sign on the 1 wrong. Not that I
: > believe for a second that you did so accidentally.
:
:
: Is defining 0^2 = 0 tantamount to defining 0^3 / 0^1 ? So it
: is not much of an argument, after all.
Try reading arguments before replying to them, Ilias.
In the process of constructing 0^0, that's exactly
what would happen. I've already shown why, and
I refuse to repeat myself.
:
:
: >: waiting for you to point out the flaw in the references I gave you
: >: (Suppes, Bernays and Jech) in a previous posting. Suppes explicitly
: >: states that n^0 = 1 for every cardinal number n, and then adds as a
: >
: > As this is the first followup in this thread, that's no mean feat.
: > Your integrity comes shining through. Unless you were waiting for
: > me to respond in a thread in which I had already announced my
: > departure. In which case, you're even more of an idiot than I
: > thought.
:
:
:
: Call it a stab in the dark, but I get the impression you are
: harboring a secret dislike for Dave.
Call it a stab in the dark, but I think that you're an idiot.
The man announces that he's been awaiting my response
to a nonexistant post. I call this unethical, and you conclude
that it must be something personal.
:
: > Want to use the cardinality of mapping sets to define exponents ?
: > Fine. Define 2 ^ -1. Let's see a set with -1 members.
: > Better still, how about e ^ i*pi. I can't wait to see that
: > construction.
:
:
:
: Eh, for cardinals, finite and infinite. What do -1 or pi have to
: do with this? Shall we make fun of prime factorization, because it does
: not apply to irrationals?
Eh, prime factorization isn't supposed to be applicable to
nonintegers. Exponentiation is.
Ever hear of De Moivre's theorem ? Ever PASS a complex analysis
course ?
Clueless. You're just clueless.
:
:
:
: > For your information, dimwit, a more popular definition of a ^ b
: > is given by
: >
: > a ^ b = exp(b*ln(a)), where exp is defined to be the inverse
: > of the natural log, and the natural log is defined to
: > be the antiderivative of 1/x , equalling 0 at x =1.
: > See Apostol. See Conway for expansion to Complex
: > plane.
:
:
:
: I did look at Apostol, "Math. Analysis", 2nd ed. , p. 21. He defines
: z^0 = 1, for all complex z. No, z=0 is not excluded. (Next line, negative
: powers, it is).
Try looking up Calculus. He did write more than one book, you
know. Here's an idea. You might try looking up the elementary
concepts in the more elementary books. Remember when you
couldn't find the multiplication tables in that book on Galois
theory ? Well, that's why things didn't work out.
Am I to take it that the definition that is given for nonzero
exponents, should be discarded to make way for as seperate
definition for the case of the 0th power ?
Or maybe, just maybe, you've confused a derivation with a
definition. So, let's see. He explicitly states something in
Calculus. He neglects to mention something that every high school
freshman already knows later, and thus must be repudiating what
he wrote earlier by his silence. Oh no, it couldn't possibly
be an oversight.
God, you're an idiot.
:
:
: Exps and logs are necessary for the general case, to be sure; but
: not for integer powers. They can complicate things too. Consider f(x) = x^x
: Then df/dx = x^x(1 + log(x)), which is not real if x < 0. But for x any
: negative rational -m/(2n+1), f(x) _is_ real. These rationals are dense; so
I didn't really just read that argument, did I ? Stanford couldn't
possibly have graduated someone capable of making it, could it ?
You think that finding a limit of the slopes of the secant lines
between point of A DENSE SUBSET is enough to establish a derivative,
do you ? Try the indicator function for the rationals.
No, no, no, no. Flamethrower mode off, serious mode on. Ilias, you
couldn't possibly have meant to print that, could you ?
Ah, no. I think you've stepped deeply into the elite company of the
cube doublers with that one. Put aside flamethrower, reach for
bazooka. You ARE a troll, aren't you ?
The definition off limit involved the value of ALL of the points in
a neighborhood about a point, not just those on a dense set. Were
you just assuming that x^x was continuous on R- ? It isn't even
well defined ! Which value of (-0.5)^(-0.5) did you have in mind ?
Are we looking at one branch of x^x ? Do you appreciate that there
is even a question to be asked here ?
: at any such curve point we can draw chords to neighbors and take a limit to
: get a "tangent", a "df/dx"... obviously real! Why doesn't the standard df/dx
: give us this information?
:
: Maybe there is a practical point then not to run to x^y if x^k will do.
:
:
: ...
: >: If a and b are cardinal numbers, not both zero, then a^b is
: >: hereby defined to be the cardinality of the set of mappings
: >: from b to a. Although the definition would make perfectly good
: >: sense in the case that a = b = 0 and would imply that 0^0 = 1,
: >
: > and none whatsoever if any of the above are not integers,
: > but bring that up, and you can expect a chilly reception
: > from some demented undergrad in West Lafayette.
:
:
:
: Again -- it is a natural definition, it includes the integers,
: and 0 is an integer.
:
: This doesn't make it automatically right for all purposes; but it
: shows it deserving discussion.
The only thing deserving discussion is where you two go to
have head examine. No, I not make fun of your enghlishes.
Why you ask.
:
: ....
: >: > We don't write theorems to satisfy someone's obsession with making
: >: > everything perfectly tidy, because as a brief examination of the
: >: > subject will show, very often one will have to choose one form of
: >: > untidyness or another. Theorems are written to be used. Who in his
: >: > right mind would ever want to expand ( x + 0 ) ^ n using the binomial
: >: > theorem, when one can simply collapse the 0 into the x, and be done
: >: > with it ? Everyone did know that x + 0 = x, right ? guys ? huh ?
:
:
: Uh? x + 0 = x ?... Hmmm... I don't know... oh, all right...
:
: Seriously, though, it would be a pain when using (x + y)^n in a
: proof to have to pause every time and handle exceptions.
Words fail me. You pose a hypothetical worry that has never been
seen.
:
:
: >: Theorems are written to be provable statements. No matter how tidy or
: >: untidy a statement may be, it is not useful (i.e., is not a theorem and
: >: therefore cannot be applied to anything) if it can't be proved. The
: >: usual statement of the binomial theorem cannot be proved if you
: >: artifically exclude 0^0 from the domain of definition. Let's see your
: >: statement of the binomial theorem that is still provable in spite of
: >
: > Oh, gosh. Requiring both x and y to be nonzero. The nervous
: > breakdowns of Purdue students as they're required to go through
: > the nervewracking process of determining if this holds in a
: > given case. The humanity !
:
:
:
: Overcome by pity and fear, I sought catharsis in Graham, Knuth &
: Patashnik, "Concrete Mathematics", AW 1989. Page 162:
:
: "Some textbooks leave the quantity 0^0 undefined, because the
: functions x^0 and 0^x have different limiting values when x decreases to
: 0. But this is a mistake. We must define x^0 = 1, for all x, if the
: binomial theorem is to be valid when x=0, y=0, and/or x = -y. The theorem
: is too important to be arbitrarily restricted! By contrast, the function
Why ? Because they say so ?
Now, this IS an argument by authority, and a pretty feeble one at
that. You haven't merely cited a definition, but you've presented a
quote as if it was an argument, something that Dave has done and I
have not. Making use of a textbook that isn't exactly the last word
in its field.
: 0^x is quite unimportant."
:
:
: Not bad. (x+y)^a = Sum(all k) (a pick k) x^k y^(a-k) if either
: 'a' is an integer >= 0, or if |x/y| < 1, 'a' arbitrary. Wouldn't
: exceptions be a shame!?
Try again. (x + y)^n = (usual expansion) whenever x and y
are nonzero, and n is a nonnegative integer.
So very, very complicated. Your head must be spinning by now.
:
: >: this anomaly. I think I will prefer my version, which will be much
: >: simpler to state, prove, understand, and apply.
: >:
: > Really ? You think that using set theory to prove the binomial
: > theorem is easier than using mathematical induction ? Let me
: > guess, you're using differential geometry to prove the Pythagorean
: > theorem, right ?
:
:
:
: Set theory was for 0^0 and m^n. The binomial theorem proof uses
: supercompact cardinals.
The binomial theorem is high school algebra, and uses simple
mathematical induction. The only set theory that makes an appearance
lies in the observation that the standard ordering well orders Z+.
Wow. Gee, hope something that deep doesn't get us into any
antinomies !
:
:
: >: > As for calculating the derivative of f(x) = x ^ 1 at x = 0 using the
: >: > formula (d/dx)(x ^ n) = n * (x ^ (n-1)), again, why would anyone want
: >: > to do that ?
: >:
: >: The same comment applies. If the formula is not correct for all cases,
: >: then it needs to be restated in order to make it correct. Your version
: >: will be needlessly complicated.
: >
: > You are a freshman, aren't you ? It isn't right for all cases. Try
: > n = 0.
:
:
: Well, n=1 is fine if 0^0 = 1; for n=0, 0^0 undefined would mean x^0
Ah ! Dave has found a soulmate !
The very passage that you excerpted refers to the case n = 0,
not n = 1. Hoping that noone would notice, huh ?
Try it out. A formal application yields
(d/dx) (x^0) = 0 * (x^ (0-1)) = 0 * x^ -1
At x = 0, this becomes 0 * 0 ^ -1, even though the right
hand derivative of x^0 is perfectly well defined (zero, in fact)
at x = 0 if we accept this definition. A bit of a conundrum,
isn't it ?
: undefined at 0... no, thanks. The formula isn't really wrong, f'(x) = 0/x
: easily implies the 0/0 at x=0 can be determined.
:
: Now, what about the derivative of 0, written 0^x ?!
What about it ? You've made the stunning discovery that if you set
up an easy problem in a stupid way, you can keep yourself from
solving it. Wow. Every student who has ever flunked out of
calculus has discovered this for himself.
And, the question can be posed with equal force, what about the
right hand derivative of x^0 at x=0 ? You do know what a right
hand derivative is, don't you ?
:
: >: If you think mathematical arguments are settled only by appeals to
: >: authority, then you really are a newcomer at this. Let's see your
: >
: > It just gets better, and better, doesn't it ? You respond to my
: > refusal to take the authority of that childish little FAQ file
: > seriously
: > as a statement that mathematical arguments are settled by appeals
: > to authority, as you put it, turning the statement on its head.
: > How could Purdue's standards ever slip so low as to let such a
: > sorry excuse for a student like you in, anyway ?
:
:
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: This sort of thing, personal attacks and insults, usually shows
: insecurity, can make people stop taking you seriously, and does nothing to
What goes around, comes around. Deal with it.
But let me see if you can comprehend this, Ilias.
He took a simple statement, and completely inverted its meaning.
In doing so, he acted like an idiot. So, I called him an idiot.
End of subject. If that hurt the pride of a poor excuse for a
student like Dave, here, good. He's an unethical piece of
human trash that has no business pursuing an academic career.
And don't you dare respond to that statement until you've
familiarised yourself with everything that has occured to date.
As for "convincing people of my point of view", or however you put
it - this subject isn't controversial. You're merely clueless.
: of the form 0^0 are 1, period. Or 0 log(0) is 0, etc. Just an idea;
: you can probably find better ones.
:
Yes. By going down to the local shelter, and talking to someome
in detox, I can find better.
By the way, you're confusing 0 log 0
with lim (x log x) as x -> 0
GOODBYE. Expect no further response, send no e-mail. It's a dead
subject. Got it ? Good. Welcome to my killfile, you brain dead
troll. Dave's been waiting there for you. You'll have a lot to
talk about. And take this scatterbrained topic with you. I'm
killfiling it too. I've got work to do. Now get lost.
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: Ilias
: