Airships,warp gates and you

Wrin · **Posted:** Mon Jan 20, 2003 10:52 pm

*hands over 150 gp* I'd like three...room for my snacks will be required. ^_^ This should be good!

Tarnok · **Joined:** Tue Apr 30, 2002 5:00 pm **Posts:** 7

Psycojes:

Actually the associative property of addition allows us to add numbers in any order so 1 + 1 - 1 + 1 - 1.... = (1+1)-(1+1)-.....

There are, however, difficulties in summing infinite series, so the conclusion that 0 = 1 is wrong.

Marcos · **Joined:** Tue Dec 31, 2002 5:00 pm **Posts:** 22

Quote:

On 2003-01-20 18:16, Muad'Dib wrote:
Hooray for sco08y! He is correct.
All of these "proofs" have the common error of sneaking in a division or multiplication by zero, thus making everything after them meaningless.

Not all of them. Geometric proof that 1 = 2:
Take a circle and a semicircle of the same radius. Map each point on the semicircle to two diametrically opposite points on the circle. This sets up a 1:2 correspondence between points on the semicircle and points on the circle.

Draw a line tangent to one endpoint of the semicircle and offset it so it no longer touches the semicircle. For each point on the semicircle, construct a tangent line that intersects the offset line. Each line will intersect the offset line at exactly one point, giving a 1:1 correspondence between points on the semicircle and points on the offset line.

Draw a line tangent to the circle and parallel to the offset line from the semicircle. For each point on the circle, draw a line tangent to the circle that intersects the original tangent line. Each line will intersect the tangent line at exactly one point, setting up a 1:1 correspondence between points on the circle and points on the line.

For each point on the offset line, construct a line perpendicular to that line intersecting the circle tangent line. This sets up a 1:1 correspondence between points on the offset line and points on the circle tangent line.

These three 1:1 correspondences can be used to construct a 1:1 correspondence between points on the circle and points on the semicircle. Since there is also a 1:2 correspondence between points on the semicircle and points on the circle, 2 must equal 1.

Q.E.D.

OmnipotentEntity · **Posted:** Tue Jan 21, 2003 4:30 am

I'll try this with out the obvious screw up this time. (You know I meant "b" Deuce)

Everyone here with euler's formula?

e^((pi)i)+1=0

A more general from for it is

e^(xi) = cos(x)+ i*sin(x)

You can derive this by using Taylor series. A quick Google search should yield the proof with ease.

Let's try substituting different values in for x, say 2(pi)

e^(2(pi)i) = cos(2(pi)) + i*sin(2(pi))
e^(2(pi)i) = 1 + 0i = 1
e^(2(pi)i) = e^0
2(pi)i = 0
sqrt(-4(pi)^2) = 0

So, the graph of y = sqrt(x) now has a negative solution, at f(-4(pi)^2) = 0, but the inverse of y = sqrt(x) is y = x^2 which now apparently has two solutions at f(0), 0 and -4(pi)^2, so it no longer passes the vertical line test and is therefore now no longer a function.
BTW, anyone who can figure out the logical fault in this one deserves a golden star-shaped cookie wrapped in powder blue panties (by the BiShou compromise), it stumped my math teacher...

Pyromancer · **Posted:** Tue Jan 21, 2003 5:07 am

Quote:

On 2003-01-21 03:09, Marcos wrote:

Quote:

On 2003-01-20 18:16, Muad'Dib wrote:
Hooray for sco08y! He is correct.
All of these "proofs" have the common error of sneaking in a division or multiplication by zero, thus making everything after them meaningless.

Not all of them. Geometric proof that 1 = 2:
Take a circle and a semicircle of the same radius. Map each point on the semicircle to two diametrically opposite points on the circle. This sets up a 1:2 correspondence between points on the semicircle and points on the circle.

Draw a line tangent to one endpoint of the semicircle and offset it so it no longer touches the semicircle. For each point on the semicircle, construct a tangent line that intersects the offset line. Each line will intersect the offset line at exactly one point, giving a 1:1 correspondence between points on the semicircle and points on the offset line.

Draw a line tangent to the circle and parallel to the offset line from the semicircle. For each point on the circle, draw a line tangent to the circle that intersects the original tangent line. Each line will intersect the tangent line at exactly one point, setting up a 1:1 correspondence between points on the circle and points on the line.

For each point on the offset line, construct a line perpendicular to that line intersecting the circle tangent line. This sets up a 1:1 correspondence between points on the offset line and points on the circle tangent line.

These three 1:1 correspondences can be used to construct a 1:1 correspondence between points on the circle and points on the semicircle. Since there is also a 1:2 correspondence between points on the semicircle and points on the circle, 2 must equal 1.

Q.E.D.

Well, I'm not very good at geometric proofs, but I'll take a shot at it.

As a circle can be considered as a polygon of infinitely many sides, taking "every point on a semicircle" would likewise result in an infinite number of points. Since infinity does not follow the rules for real numbers, 2x infinity = infinity. The argument is invalid when applied to a regular polygon with any finite number of sides.

P-M

-><-

[ This Message was edited by: Pyromancer on 2003-01-21 04:08 ]

Marcos · **Joined:** Tue Dec 31, 2002 5:00 pm **Posts:** 22

Quote:

On 2003-01-21 03:30, OmnipotentEntity wrote:
I'll try this with out the obvious screw up this time. (You know I meant "b" Deuce)

Everyone here with euler's formula?

e^((pi)i)+1=0

A more general from for it is

e^(xi) = cos(x)+ i*sin(x)

You can derive this by using Taylor series. A quick Google search should yield the proof with ease.

Let's try substituting different values in for x, say 2(pi)

e^(2(pi)i) = cos(2(pi)) + i*sin(2(pi))
e^(2(pi)i) = 1 + 0i = 1
e^(2(pi)i) = e^0
2(pi)i = 0

I think I've got it. ex as a real-valued function is invertable, with inverse ln(x). ex as a complex-valued function is not invertable. To get from line 3 to line 4, you have a hidden step of "take the inverse of both sides of the equation", which is not allowed, since you are using ex as a complex-valued function.

Marcos · **Joined:** Tue Dec 31, 2002 5:00 pm **Posts:** 22

Quote:

On 2003-01-21 04:07, Pyromancer wrote:
Well, I'm not very good at geometric proofs, but I'll take a shot at it.

As a circle can be considered as a polygon of infinitely many sides, taking "every point on a semicircle" would likewise result in an infinite number of points. Since infinity does not follow the rules for real numbers, 2x infinity = infinity. The argument is invalid when applied to a regular polygon with any finite number of sides.

P-M

-><-

That's basically it. I'm multiplying and then dividing by infinity.

Crashman · **Posted:** Tue Jan 21, 2003 8:05 am

>.<

Augh. Too much math. I could never figure out half the things you people are talking about. Jeez, and here I thought everything was just kinda dopey and happy around here...then I get slapped in the face by Algebra (faulty mathmatical system!) and Advanced Geometry. Gimme something basic, something my puny mind can grasp, and I'll be happy. All these formulae, quite frankly, mean nothing to me, just letters and numbers and symbols.

ANYHOO, I think this is the same timeline. It'd just be a waste to have seperate timelines. And I frankly think the Errant hunter would be the winner in any combat scenario, because frankly her job would be damn dangerous otherwise. Can you imagine how many half elven lunatic mages she would probably encounter?
>.>
<.<
Ooh. There's the possibility that Meiji would be the first half-elven psychotic mage she encounters, and it's likely both from her demeanor and the fact she's half elf that she'll be seen as an Errant. Onbe with a familiar, no less. Speaking of which, refresh my memory: what was a familiar for?

OmnipotentEntity · **Posted:** Tue Jan 21, 2003 1:11 pm

Quote:

On 2003-01-21 04:13, Marcos wrote:

Quote:

On 2003-01-21 03:30, OmnipotentEntity wrote:
I'll try this with out the obvious screw up this time. (You know I meant "b" Deuce)

Everyone here with euler's formula?

e^((pi)i)+1=0

A more general from for it is

e^(xi) = cos(x)+ i*sin(x)

You can derive this by using Taylor series. A quick Google search should yield the proof with ease.

Let's try substituting different values in for x, say 2(pi)

e^(2(pi)i) = cos(2(pi)) + i*sin(2(pi))
e^(2(pi)i) = 1 + 0i = 1
e^(2(pi)i) = e^0
2(pi)i = 0

I think I've got it. ex as a real-valued function is invertable, with inverse ln(x). ex as a complex-valued function is not invertable. To get from line 3 to line 4, you have a hidden step of "take the inverse of both sides of the equation", which is not allowed, since you are using ex as a complex-valued function.

Wow! That's actually exactly it... *tosses a cookie to Marcos* I'm impressed.

And for my next trick, I will pull a rabbit out of my pants.

Wandering Idiot · **Posted:** Tue Jan 21, 2003 1:15 pm

Yeah, zero and infinites do tend to throw monkey wrenches into standard algebra :)

Quote:

On 2003-01-21 07:05, Crashman wrote:
Gimme something basic, something my puny mind can grasp, and I'll be happy.

This from the man who posted about *negative dimensions*. You see what you people and your infinite sets have done? Infinite sets are evil! Eeevil, I tell you! (But kind of fun, nonetheless :)

And yeah, I'm really do think that Meji are Sarine might be in the same time period. Which suits me just fine, 'cause then I get to sell tickets to the inevitable fight-upon-first-meeting-of-two-good-guys that Sarine and Meji are sure to have (although classifying Meji as a "good guy" might be questionable at this point...)

Crashman · **Posted:** Tue Jan 21, 2003 6:21 pm

Er, uh, well......my theories relied on logic and simple reasoning. This stuff is all learned mathmatics that require computation, something folks with ADD aren't too keen on. Don't gimme any of that "They grow out of it" stuff either, SOME do, not all, darn ya'!

*whew* Another possible disaster avoided by using my immense linguistic capabilities to avert events that may have been detrimental to my mental and or physical well being.
>.>

Clay_Allison · **Posted:** Tue Jan 21, 2003 7:05 pm

Quote:

On 2003-01-21 12:11, OmnipotentEntity wrote:
And for my next trick, I will pull a rabbit out of my pants.

You ever try to get one into your pants? THAT would be a trick!

Immanio · **Posted:** Tue Jan 21, 2003 7:52 pm

Quote:

On 2003-01-21 18:05, Clay_Allison wrote:

Quote:

On 2003-01-21 12:11, OmnipotentEntity wrote:
And for my next trick, I will pull a rabbit out of my pants.

You ever try to get one into your pants? THAT would be a trick!

He's probably too busy getting into their pants :smile:

Bealz · **Posted:** Wed Jan 22, 2003 12:37 am

nm

[ This Message was edited by: Bealz on 2003-01-23 01:11 ]

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Airships,warp gates and you

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