Outpost Universe Forums
Off Topic => Test/Spam Forum => Topic started by: CK9 on September 02, 2004, 08:28:48 AM
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I can prove 2 + 2 = 4 false
here's how using smileies:
:huh: :o + :o ;) = 3, not four
verbal explanation:
if both sets of two use the same exact object (not two similar ones) it isn't four, now is it?
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;) ;) + ;) ;) = 4 ;)
Yours has 4 total, 3 different
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:huh: :o + :o ;) = :huh: :o2 ;)
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lol :D
Anyway:
We can all agree that girls take both time and money, so:
Girls = Time * Money
We all know time is money, so:
Girls = Money^2
Money is the root of all evil, thus:
Money = (evil)^1/2
Thus:
Girls = ((evil)^1/2)^2
==
Girls = evil
:P
I would do the 1 = 0 one, but I forgot it :(
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xfir, "money is not the root of all evil, it is merely the tool"
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Money isn't the root of all evil, it's the love of money that is the root of all evil.
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your missing the point
girls=evil lol
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So with what I said in mind...it isn't girls that are the root of all evil....it's the love of girls that is the root of all evil....so I guess I have some deep roots then :D
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how do you walk if you have roots?
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beta, you have roots too
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Girls = Low Bank Balance
Money seemingly flys out of your pockets, and goes in theirs.
:angry:
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what do I =?
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only a few girls i know = evil, like my sis.
The ones that odn't cost THAT much are good.
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no girls I know = evil
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Again:
Girls doesn't = No Money
Love of Girls = No Money
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Money = evil.
so when:
Girl's = No Money
then
Girls = No Evil!
lol
I don't mind spending money to by my friends/gf things. Got nothing better to buy. Then again... I don't have a g/f anymore. lol.
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lol, now i'll spend someone else's moola. lol, j/k. well, if girls = evil, then what do boys =?
men = life forms from Mars
women = life forms from Venus
mankind = we all have to live together and get over it
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well, I cannot say as this is still a family-frindly forum, lol
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If you've ever gotten your nipple bitten off by a beaver, you might be a redneck
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if your truck is your pool, you might be a redneck
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if you have a whole set of salid bowls and they all say kool whip on the side, you might be a redneck
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you might be a betaray if...
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u mite b blond if u set the computer screen ontop of a copy machine to print out a document.
(thanx luweeg64)
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lol
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lol, now i'll spend someone else's moola. lol, j/k. well, if girls = evil, then what do boys =?
if girls=evil then:
girls <=> boys
so then boys=good?
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no, some guys of a certain age = lustful
lmao
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k.
rofl.
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what's rofl? hey, i no plenty of boys who r lustful
lustful=sin
sin=evil
lustful=evil
so there fore, if u look upon a person as being hot,
u=evil
(u, in this case, may b boy or girl)
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but evil cannot exist without a good to compare it to and vise versa so a town full of sinners = evil compared to a town full of non-sinners
but compared to nothing at all
town full of sinners = normal for them
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"Good" is defined as that which the majority of people see as "good" whilst "evil" is defined as that which the majority of people see as "evil."
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what's rofl? hey, i no plenty of boys who r lustful
lustful=sin
sin=evil
lustful=evil
so there fore, if u look upon a person as being hot,
u=evil
(u, in this case, may b boy or girl)
hmmm... yup i am evil...
**PlayingOutpost0-24 snaffles a cookie off garrettsgirl
:D
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*thinks for a few moments* nope not evil (some pretty ugly people here, lol, but I'm looking at mental rather asthetic value myself.)
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ummmm...
CK9 == EVIL != NOT EVIL.
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Thats an odd way of doing math..
Calculus and Derivations:
f(x) = x^3 + 4x^2 - 5
f'(x) = 3x^2 + 8x
Thus, the average velocity where x = 5 is 115.
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*sigh*...
x^3=12^9
"Answer me that, and i show u a green dog..."
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x^3=12^9
x = 12 * 12 * 12 * 12 * 12 * 12 * 12 * 12 * 12 = 144 * 144 * 144 * 12 = 5159780352 ^ (1/3) = 1728
simple mental math
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Here's one, a simple proof every single person on this board should be able to do:
picture: (i'll try to draw it the best using ascii art, lol)
[doHTML]<font face="courier new"> ^ A
/|\
/ | \
B<----->C
D
</font>[/doHTML]
Given: triangle ABC is scalene, and segment AD bisects BC.
Prove: AD is not perpendicular to BC.
Good luck!
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i will solve it AS SOON as i get what those words mean in Hungarian...
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we're doin proofs in geometry rit now... i've never done so much of one thing in my whole life...
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i prove another thing false... well i say i am not bored... BUT I AM!
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to zfir's "equation"
if the quote says "the love of money is the root of all evil"
then "money is the root of all evil"
is also part of the quote and
can thus be another quote
to hackers "proof"
^ A
/ | \
/ | \
B<----->C
D
line AD must be perpendicular to line BC in order to bisect line BC in a scalene triangle so there is NO SOLUTION to prove that line AD is not perpandicular to line BC
unless your in one of beta's imaginary universes
edit: hacker how did you get the triangle to match up, i put spaces in but it just deleted them
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use [code|] [/code] (without the | line.)
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By its definition, a scalene triangle has no equal angles!
Rough redraw:
(http://www.xfir.net/offsite/scalene.png)
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nice picture, x
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AutoCad
Anyway, By the Definition of Scalene triangles, no two sides nor angles are equal.
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Hint: use indirect proofs.
Hmm, autocad. reminds me, i should get a copy from the school for personal use sometime...
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Easy way to do the X^3 = 12^9:
X^3 = (12^3)^3
x = 12^3
Which is where I usually stop since I'm more of a pure math type of person and rarely bother with arithmetic involving more than 1 digit numbers. :) (Calculations are for computers, the beauty is in the proof).
But alright, x = 12 * 12 * 12 = 144*12 = 1728.
As for the initial question, that's not really addition. What you're doing is taking the cardinality of set union, where repeated items only count once. It's a simple matter to show:
| A union B | <= |A| + |B|
In fact, this follows from the more general result:
|A| + |B| = | A union B| + | A intersect B|
f(x) = x^3 + 4x^2 - 5
f'(x) = 3x^2 + 8x
Thus, the average velocity where x = 5 is 115.
Just to be picky :P , that's the instantaneous velocity. But yeah, wonderful stuff.
As for 0 = 1, all these proofs contain some subtle flaw that people aren't meant to notice. If you look hard enough, you can always find a false step, or a false assumption.
Consider:
(Step1) 1 = 1^(2*1/2)
(Step2) = (1^2)^1/2
(Step3) = ((1)^2)^1/2
(Step4) = ((-1)^2)^1/2, since (-1)^2 = 1 = (1)^2
(Step5) = (-1)^(2*1/2)
(Step6) = (-1)^1
(Step7) = -1
Which of course you can use to your hearts desire to "prove" other results, such as:
1 = -1
1 + 1 = -1 + 1
2 = 0
1 = 0 (divide both sides by 2)
So, anyone care to guess which of the above steps was false?
Anyways, these "proofs" often like to make use of square roots but there are other examples that misuse the Well-Ordering princiople, or a proof by smallest counter example.
As an example of the Well-Ordering Principle, (every nonempty set of natural numbers contains a least element) we can "prove" that every natural number is "interesting". (And no, this isn't exactly a false proof, just really silly).
Proof: Suppose for the sake of contradiction that not every natural number is interesting. Let X be the set of all the non-interesting natural numbers. By assumption, X is non-empty. By the Well-Ordering principle, let x be the smallest element of X.
Now, 0 is Certainly interesting, it's the identity element for addition, and anything *0 = 0. Similarily, 1 is the multiplicative identity, it's the only unit in the set of natural numbers, 2 is the only even prime. Certainly all interesting numbers.
So what is x? It's the first natural number that is not interesting, and that makes it very interesting. => <= :)
(Granted, the last line is pretty ridiculous, but hopefully you can appreciate the humor in it.)
As for Hacker's question, if we are allowed the assumption that two sides of a triangle and the inclosing angle completely define the triangle (which is certainly true).
Then suppose AD bisects BC and AD is perpendicular to BC, then we have the |BD| = |DC| (the lengths are equal). But then since AD is perpendicular to BC, we have angle BDA = 90 degrees = angle CDA. But then we have for triangle, BDA, sides BD and AD enclosnig a 90 degree and, while for triangle CDA, sides DC and AD enclose a 90 degree angle, but |DC|=|BD| and certainly |AD|=|AD|, so these two triangles are quivalent. Hence the third sides BA and CA have equal length. But that contradicts triangle ABC is scalene. Thus if AD bisects BC, it is not perpendicular.
Here's some funny puzzles for the calculus types:
Note: For all functions below, the domain must be defined on all real numbers.
1) Find a function which is discontinuous at infinitely many points
2) Find a function which is discontinuous at an uncountably infinite number of points.
3) Find a function which is continuous at only a countably infinte number of points.
3b) but whose integral is continuous everywhere.
4) Find a function which is continuous at exactly one point.
5) Fnid a function which is discontinuous everywhere.
6) Find a continuous function which if not differentiable anywhere.
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i can still prove things false :D
1) need... dictionary...
2) need... dictionary...
3) need... dictionary...
3b) need... dictionary...
4) need... dictionary...
5) need... dictionary...
6) need... dictionary...
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1) Find a function which is discontinuous at infinitely many points
lol, that would be the stair-step equation
2) Find a function which is discontinuous at an uncountably infinite number of points.
same as #1, just stated in different terms that make it redundant
3) Find a function which is continuous at only a countably infinte number of points.
define countably infinate...
3b) but whose integral is continuous everywhere.
4) Find a function which is continuous at exactly one point.
impossible, as for it to be continuous, it must have other poins of reference that would be continuous as well
5) Fnid a function which is discontinuous everywhere.
saw it just the other day...then we had to d some classwork and I forgot it
6) Find a continuous function which if not differentiable anywhere.
lol, if I had time, I would answer this, but I would need at least half an hour to write and check it
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proving things true
head=butt
just look at Kerry
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Bah, my computer ate my post! Well, here it is again I guess.
1) Find a function which is discontinuous at infinitely many points
lol, that would be the stair-step equation
Yep, that's exactly what I was thinking of.
f(x) = FloorOf(x)
Damn, the lack of math symbols that have plagued computers for so long!
2) Find a function which is discontinuous at an uncountably infinite number of points.
same as #1, just stated in different terms that make it redundant
Nope, the step function is only discontinuous at countably infintly many points. (At all the integers).
3) Find a function which is continuous at only a countably infinte number of points.
define countably infinate...
Ahh, now we get to the good question.
Formally, a countable set can have a bijection formed between it and a subset of the natural numbers. (A bijection is a one-to-one and onto function, basically it creates a pairing between elements of one set with elements of another set. Basically, a countable set can be "listed" in such a way that no element is missed. Also, sets are considered to be subsets of themselves. So a subset of the natural numbers would include the natural numbers themselves.)
For instance, you can list the integers so that none are missed:
0, 1, -1, 2, -2, 3, -3, 4, -4, ...
(Hence they have the pairing (1, 0), (2, 1), (3, -1), (4, 2), (5, -2), (6, 3), (7, -3), ...)
You can also list the positive rational numbers in a special way, where the rational numbers are laid out in a 2D array with numerator and denomintors increasing along corresponding axis. You can then traverse diagonallay through the array. Note: You can't traverse along each row or column since they are infinte and you would never get to the next row or column, hence missing all those numbers. Also, as a technicality, you should skip over repeated numbers such as 2/2 = 1/1. Anyways, once you have a list of positive rational nubmers, you can form a list of all rational numbers in the same was as the integers were listed above.
Non countable sets would include the irrational numbers and the real numbers. There is no way to pair them with the natural numbers and not miss any of them.
Also, just because a set is strictly larger or smaller than another set, doesn't mean they have different cardinality. That holds only in the finite case. For instance, the natural numbers are a subset of the integers which are a subset of the rational numbers, but they all have the same cardinality (countably infinite). Thus, even though there are "more" rational numbers than natural numbers, these sets have the same "size". (At least if you're defining size in terms of cardinality, which it usually is).
impossible, as for it to be continuous, it must have other poins of reference that would be continuous as well
Entirely possible. Continuity is defined at a point, the properties at that point need not hold for surrounding points. Granted, the definition requires considering "nearby" points, but there is no specification as to how near those points are. Essentially what you need to do is find a way to break up "nearby" points everywhere except at the one point you want to be continuous. At this point, "nearby" points bear similarity to the continuous point, but not so much to each other.
6) Find a continuous function which if not differentiable anywhere.
lol, if I had time, I would answer this, but I would need at least half an hour to write and check it
This one is kinda tough. I don't think I could write a proof of it in half an hour without my textbook handy. I'm tempted to give a hint to this one. Essentially what you want is sharp corners, so the function is not differentiable at the point, but the function is still continuous. But to be non differentiable everywhere, you need these sharp corners essentially everywhere (well, "almost everywhere"). There are a few hints I'd like to give about how to go about doing this but I figure I should let the question stand for a little while longer.