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Unread 5 Nov 2002, 14:06   #1
ELeeming
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For the mathematically challenged - ELeeming's guide to things you didn't know part 1

Let a = b

hence a*b = b^2

hence a*b - a^2 = b^2 - a^2

hence a(b - a) = (b + a)(b - a)

hence a = b + a

hence a = 2a

hence 1 = 2

QED.
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Unread 5 Nov 2002, 14:10   #2
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bah... trying to divide by zero
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Unread 5 Nov 2002, 14:29   #3
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Re: For the mathematically challenged - ELeeming's guide to things you didn't know part 1

If a = 1 and b = a,

a(b - a) = (b + a)(b - a)

1*(1-1) = (1 + 1)*(1 - 1)
0 = 2 * 0
0 == 0
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Unread 5 Nov 2002, 14:31   #4
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1^2=(-1)^2
thus
1=-1
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Unread 5 Nov 2002, 14:33   #5
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Re: Re: Re: For the mathematically challenged - ELeeming's guide to things you didn't know part 1

Quote:
Originally posted by Iniluki



nononon

1*(1-1) = 1*(0)

which is zero

But he has just proved 0 = 0 so why carry with the equation from there? Its bound to go wrong.
Oh yeah.
Put the 0 a line too late :\
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Unread 5 Nov 2002, 14:49   #6
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ITS SO AMAZING


Let's continue on the journey:


God is the creator of everything.

God wrote the book of nature in the language of mathematics.

God created ELeeming

ELeeming is a genious and proves that 1=2.

ELeeming doth blaspheme and is Satan.


Hence: Whenever someone asks you how many fingers they show and you see one finger, say 'I can clearly see two fingers.' And when the person says that he's only having one finger on display, you beat him senseless defending God against this follower of ELeeming.



Say, where is the flaw in my logic?





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Unread 5 Nov 2002, 16:18   #7
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Unread 5 Nov 2002, 16:30   #8
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X = 0.999...

thus

10X = 9.999...

10X-X = 9.999... - 0.999...

9X = 9

X = 1

so

0.999... = 1
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Unread 5 Nov 2002, 16:37   #9
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Quote:
Originally posted by menth0l
X = 0.999...

thus

10X = 9.999...

10X-X = 9.999... - 0.999...

9X = 9

X = 1

so

0.999... = 1

0.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 99999999999999999999999999999999999999... = 1

You're still a tad out :/
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Unread 5 Nov 2002, 16:38   #10
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mbushell: lies.

0.999... = 1.
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Unread 5 Nov 2002, 16:38   #11
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Quote:
Originally posted by menth0l

0.999... = 1
You didn't happen to design the Pentium chip, by any chance?
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Unread 5 Nov 2002, 16:40   #12
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Quote:
Originally posted by menth0l
mbushell: lies.

0.999... = 1.
0.9 = 1, nope, 0.1 out.
0.99 = 1, nope, 0.01 out.
0.999 = 1, nope, 0.001 out.
0.9999 = 1, nope, 0.0001 out.
0.99999 = 1, nope, 0.00001 out.
0.999999 = 1, nope, 0.000001 out.

etc.
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Unread 5 Nov 2002, 16:40   #13
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This is why "Rounding up" was invented.

Them clever scientists eh?
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Unread 5 Nov 2002, 16:41   #14
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Re: For the mathematically challenged - ELeeming's guide to things you didn't know part 1

Quote:
Originally posted by ELeeming
Let a = b

hence a*b = b^2

hence a*b - a^2 = b^2 - a^2

hence a(b - a) = (b + a)(b - a)

hence a = b + a

hence a = 2a

hence 1 = 2

QED.
b^2 - a^2 != (b + a)(b - a)
(b + a)(b - a) = b^2 + a^2 - 2ab
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Unread 5 Nov 2002, 16:43   #15
menth0l
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Quote:
Originally posted by mbushell


0.9 = 1, nope, 0.1 out.
0.99 = 1, nope, 0.01 out.
0.999 = 1, nope, 0.001 out.
0.9999 = 1, nope, 0.0001 out.
0.99999 = 1, nope, 0.00001 out.
0.999999 = 1, nope, 0.000001 out.

etc.
we are talking about 0.999... tho.

nothing you have given applies.
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Unread 5 Nov 2002, 16:44   #16
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Re: Re: For the mathematically challenged - ELeeming's guide to things you didn't know part 1

Quote:
Originally posted by PsyDe

b^2 - a^2 != (b + a)(b - a)
(b + a)(b - a) = b^2 + a^2 - 2ab

actually, (a+b)(a-b) = a^2-ab+ba-b^2

=a^2-b^2
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Unread 5 Nov 2002, 16:46   #17
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Quote:
Originally posted by menth0l


we are talking about 0.999... tho.

nothing you have given applies.
I'm saying you'll always be 0.000....1 out
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Unread 5 Nov 2002, 16:49   #18
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Quote:
Originally posted by mbushell


I'm saying you'll always be 0.000....1 out
Unless the ... implies an infinite series of 9s, which is how I took it.

Then he is right.
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Unread 5 Nov 2002, 17:03   #19
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Quote:
Originally posted by mbushell


I'm saying you'll always be 0.000....1 out
prove my calculation wrong then.
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Unread 5 Nov 2002, 17:04   #20
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Quote:
Originally posted by acropolis
Unless the ... implies an infinite series of 9s, which is how I took it.

Then he is right.

And bingo was his name.
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Unread 5 Nov 2002, 17:15   #21
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Quote:
Originally posted by mbushell


I'm saying you'll always be 0.000....1 out

0.000...1 is not a possible figure, because the ... implies infinity.

what you have given is not infinite.
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Unread 5 Nov 2002, 18:33   #22
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menth0l is right

anyone got one that's NOT been posted a million times before?
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Unread 5 Nov 2002, 19:39   #23
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Quote:
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You didn't happen to design the Pentium chip, by any chance?
Thank God there was SOMETHING in this thread worth reading
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Unread 5 Nov 2002, 19:56   #24
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1 + oo = oo

2 + oo = oo

hence 1 = 2.


This is the logic menth0ls one uses.
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Unread 5 Nov 2002, 20:04   #25
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Unread 5 Nov 2002, 20:12   #26
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cos^2 x = 1-sin^2 x
1+cos x = 1 +(1-sin^2 x)^0.5
(1+cos x)^2 = [1 +(1-sin^2 x)^0.5]^2

when x=(2 * pi)/3:

(1-(1/2))^2 = [1+(1-(3/4))^0.5]^2
1/4 = [1+(1/4)^0.5]^2
1 = 9
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Unread 5 Nov 2002, 20:13   #27
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Exclamation

I don't like maths.
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Unread 5 Nov 2002, 20:17   #28
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The difference between 0.999... and 1 is infintesimal. Assuming you're not dealing with infintesimal variables, the difference is 0
(btw, the proof isn't too good, you have to be careful when substracting one infinite series from another)
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Unread 5 Nov 2002, 20:18   #29
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y = x^0.5 does not follow from y^2 = x
If +- means plus or minus:
y^2 = x => y = +-x^0.5
Quote:
Originally posted by Vince McMahon
cos^2 x = 1-sin^2 x
1+cos x = 1 +- (1-sin^2 x)^0.5
(1+cos x)^2 = [1 +- (1-sin^2 x)^0.5]^2

when x=(2 * pi)/3:

(1-(1/2))^2 = [1+-(1-(3/4))^0.5]^2
1/4 = [1+-(1/4)^0.5]^2
1 = 1 or 1 = 9
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Unread 5 Nov 2002, 20:36   #30
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Re: For the mathematically challenged - ELeeming's guide to things you didn't know part 1

Quote:
Originally posted by ELeeming
Let a = b

hence a*b = b^2

hence a*b - a^2 = b^2 - a^2

hence a(b - a) = (b + a)(b - a)

hence a = b + a

hence a = 2a

hence 1 = 2

QED.

you are allowed to multiply both sides with a variable, but then you have to check the results by filling them into the original equation. 1 of the results is wrong and the other remains.

x = 4
x^2 = 16
x = -4 or x = 4
fill the results in the original equation:
x = -4 or x = 4
-4 = 4 or 4 = 4
-4 != 4 so the only good result is x = 4


you are never allowed to divide a variable out.
x^2 = x
x^2-x = 0
x(x-1) = 0
x = 0 or x = 1
0 = 0 or 1 = 1

if you do it your way:
x^2 = x
x = 1

and you miss one result
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Unread 5 Nov 2002, 22:32   #31
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Quote:
Originally posted by ComradeRob


You didn't happen to design the Pentium chip, by any chance?

0.9 recuring does however equal 1, although I can't remember the exact proof off the top of my head :-)

I'm also glad someone remembers Intels wonderful inability to be able to do simple floating point arithmetic, although I always found the fact that they denied it existed, when you could do 2 * 2 = 3.99817 yourself amusing.

http://www.matthill.demon.co.uk/comedy/pentium.html

A small point of semi-related humour :-)
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Unread 6 Nov 2002, 10:23   #32
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Quote:
Originally posted by Nodrog
1 + oo = oo

2 + oo = oo

hence 1 = 2.


This is the logic menth0ls one uses.
No his logic is more like:

a*0 = b*0
=> a = b
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Unread 6 Nov 2002, 10:59   #33
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Quote:
Originally posted by menth0l
X = 0.999...

thus

10X = 9.999...

10X-X = 9.999... - 0.999...

9X = 9

X = 1

so

0.999... = 1


10 x 0.999 = 9.990

9.990 - 0.999 = 8.991, not 9

No matter how many 9's you stick on the end, it still doesn't work :O)
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Unread 6 Nov 2002, 14:05   #34
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Unread 6 Nov 2002, 14:18   #35
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Unread 6 Nov 2002, 14:19   #36
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Marilyn Manson has ascended to a higher existance and no longer needs rep points to prove the size of his e-penis.Marilyn Manson has ascended to a higher existance and no longer needs rep points to prove the size of his e-penis.Marilyn Manson has ascended to a higher existance and no longer needs rep points to prove the size of his e-penis.Marilyn Manson has ascended to a higher existance and no longer needs rep points to prove the size of his e-penis.Marilyn Manson has ascended to a higher existance and no longer needs rep points to prove the size of his e-penis.Marilyn Manson has ascended to a higher existance and no longer needs rep points to prove the size of his e-penis.Marilyn Manson has ascended to a higher existance and no longer needs rep points to prove the size of his e-penis.Marilyn Manson has ascended to a higher existance and no longer needs rep points to prove the size of his e-penis.Marilyn Manson has ascended to a higher existance and no longer needs rep points to prove the size of his e-penis.Marilyn Manson has ascended to a higher existance and no longer needs rep points to prove the size of his e-penis.Marilyn Manson has ascended to a higher existance and no longer needs rep points to prove the size of his e-penis.
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'Ello, ello, ello, what's goin on 'ere then?
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Unread 6 Nov 2002, 14:19   #37
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i was about to complain about that!
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Unread 6 Nov 2002, 14:22   #38
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The struggle between the mod and the admin!
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Unread 6 Nov 2002, 19:48   #39
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&nbsp;&nbsp;&infin;
&nbsp;&nbsp;&Sigma;&nbsp;&nbsp;&nbsp;log(cos(2^(-n)))
n=1

And a little infinite sum that might actually be fun for the mathematically challenged to quickly and easily work out... (Old homework.)

And I didn't even use "oo" to represent &infin;...
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Unread 6 Nov 2002, 19:52   #40
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Quote:
Originally posted by Miserableman




10 x 0.999 = 9.990

9.990 - 0.999 = 8.991, not 9

No matter how many 9's you stick on the end, it still doesn't work :O)

Stop being a silly moo.
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Unread 6 Nov 2002, 19:52   #41
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Quote:
Originally posted by Cyp
&nbsp;&nbsp;&infin;
&nbsp;&nbsp;&Sigma;&nbsp;&nbsp;&nbsp;log(cos(2^(-n)))
n=1



natural log i assume? (to base e)


<2nd edit>oops, misread it </2nd edit>

Last edited by Ragnarak; 6 Nov 2002 at 19:58.
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Unread 6 Nov 2002, 19:55   #42
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Quote:
Originally posted by Ragnarak


natural log i assume? (to base e)
Yes. (The math teacher says that real mathematicians don't write ln. Or said something similar. C++ compiler seems to agree.)
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Unread 6 Nov 2002, 20:18   #43
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This is making my head hurt.
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Unread 6 Nov 2002, 20:23   #44
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some people need to be reminded of the 'rules'

(a+b)^2 = a^2 + 2ab + b^2

a^2 - b^2 = (a + b) . (a - b)

dont confuse them, kids
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Unread 6 Nov 2002, 20:37   #45
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Quote:
Originally posted by XxBrollYxX
1 + 1 = 11
HAH
try n beat THAT
add the val( )


damn i sound like my computer class teacher
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Unread 6 Nov 2002, 20:40   #46
Sarina_Joy
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Quote:
Originally posted by XxBrollYxX
1 + 1 = 11
HAH
try n beat THAT

I know a guy who uses that tactic to solve the Countdown number puzzle. He's a tad thick.
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Unread 6 Nov 2002, 23:20   #47
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skiddy contributes so much and asks for so littleskiddy contributes so much and asks for so littleskiddy contributes so much and asks for so littleskiddy contributes so much and asks for so littleskiddy contributes so much and asks for so littleskiddy contributes so much and asks for so littleskiddy contributes so much and asks for so littleskiddy contributes so much and asks for so littleskiddy contributes so much and asks for so littleskiddy contributes so much and asks for so littleskiddy contributes so much and asks for so little
My head hurts now.
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