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Old 04-12-2006, 02:33 PM   #1
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How is this possible? (Math Problem)

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Old 04-12-2006, 02:34 PM   #2
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is the guy on the right holding a used condom?
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Old 04-12-2006, 02:34 PM   #3
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is the guy on the right holding a used condom?
you really should find something else to do with your day...
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Old 04-12-2006, 02:38 PM   #4
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I believe something starts getting fucked around 4th row.
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Old 04-12-2006, 02:38 PM   #5
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a=0

simple misdirection when you tried to cancel out the variable a
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Old 04-12-2006, 02:49 PM   #6
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On line 4 when you divide by (a-b), you're dividing by zero since a=b. Division by zero is undefined.
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Old 04-12-2006, 02:59 PM   #7
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You cannot prove this while using the assumption to prove the proof.
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Old 04-12-2006, 03:03 PM   #8
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This is correct 68% of the time.

28% of women know that.
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Old 04-12-2006, 03:04 PM   #9
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Old 04-12-2006, 03:07 PM   #10
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fuck i've gt a sore head just lookin at that!
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Old 04-12-2006, 03:07 PM   #11
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Originally Posted by High Plains Drifter
On line 4 when you divide by (a-b), you're dividing by zero since a=b. Division by zero is undefined.
Exactly what I was thinking!
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Old 04-12-2006, 03:08 PM   #12
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Originally Posted by High Plains Drifter
On line 4 when you divide by (a-b), you're dividing by zero since a=b. Division by zero is undefined.
i think he's right...

row 4 is jacked up
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Old 04-12-2006, 03:17 PM   #13
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I believe something starts getting fucked around 4th row.
You see the ripple in the Matrix as well?
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Old 04-12-2006, 03:52 PM   #14
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meh fucking algebra , that's shit pissed me off big time in high school
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Old 04-12-2006, 05:03 PM   #15
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yeah those algebraic rules they are following are not correct

on a more defined note if i have 5 apples and take away 4 apples i still have 5 apples
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Old 04-12-2006, 05:41 PM   #16
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wow at least some of you are intelligent
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Old 04-12-2006, 06:57 PM   #17
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math is fucking lame....
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Old 04-12-2006, 07:05 PM   #18
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(a+b) doesn't = b, a+a doesn't = a and 2a doesn't = a you twit. Substitute a REAL number and you'll see what crap that is.
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Old 04-12-2006, 07:09 PM   #19
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It's not, now get back to work
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Old 04-12-2006, 07:15 PM   #20
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yeh divide by zero. lol at used condom comment
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Old 04-12-2006, 07:18 PM   #21
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yeh divide by zero. lol at used condom comment
There is no divsion in that. Please show me where the division is.
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Old 04-12-2006, 08:03 PM   #22
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What soulbleed means is that to get to the final answer shown, you would have to divide both sides by a which is supposed to be assigned the value 0.
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Old 04-12-2006, 08:15 PM   #23
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What soulbleed means is that to get to the final answer shown, you would have to divide both sides by a which is supposed to be assigned the value 0.
(a+b)(a-b) = b(a-b)

How in hell did they come there? This is not equal at all!

(a+b) != b

So, even before making the division by 0, the equation is fucked up.
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Old 04-12-2006, 08:19 PM   #24
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If you start off with the assumption that a=0 and b=0, the logic works until the end.
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Old 04-12-2006, 08:47 PM   #25
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you can factor a-b out if you state it is not equal to zero

the issue for me is where they divide a out of the question

they should have brought it over and then factored it out to show that a=0
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Old 04-12-2006, 08:57 PM   #26
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a+a=a only when a=0

Therefore 0=0
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Old 04-12-2006, 08:58 PM   #27
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Originally Posted by High Plains Drifter
On line 4 when you divide by (a-b), you're dividing by zero since a=b. Division by zero is undefined.
yup, line 3 is essentially saying 0 = 0. anything less itself is zero.
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Old 04-12-2006, 09:10 PM   #28
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The more formal definition of this invalid proof (ie. the one that isn't a sloppy photoshopped graphic) has the values 'a' and 'b' defined as non-zero qualities. Besides the division by zero, the proof is sound.

Code:
Proof that 2 equals 1
Let a and b be equal non-zero quantities 
a = b 
Multiply through by a 
aČ = ab 
Subtract bČ  
aČ minus bČ = ab minus bČ  
Factor both sides 
(a - b)(a + b) = b(a minus b) 
Divide out (a minus b) 
a + b = b 
Observing that a = b 
b + b = b 
Combine like terms on the left 
2b = b 
Divide by the non-zero b 
2 = 1 


The fallacy is in line 5: the progression from line 4 to line 5 involves division by (a minus b), which is zero since a equals b. Since division by zero is undefined, the argument is invalid.
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Old 04-12-2006, 09:24 PM   #29
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The more formal definition of this invalid proof (ie. the one that isn't a sloppy photoshopped graphic) has the values 'a' and 'b' defined as non-zero qualities. Besides the division by zero, the proof is sound.
No it's not. You can't avoid the divison by ZERO. If a=b and from 4 to line 5you are dividing out (a-b) you are ALWAYS going to get ZERO no matter what non-zero numbner you use for a and b in the beginning.
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Old 04-12-2006, 09:27 PM   #30
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you can factor a-b out if you state it is not equal to zero
It has too.

Use ANY number let's try 3

a=3 and a=b so b=3 so (a-b)=(3-3) which is ZERO.
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Old 04-12-2006, 10:18 PM   #31
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Didn't someone prove that 1+1=3???
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Old 04-12-2006, 10:36 PM   #32
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Didn't someone prove that 1+1=3???
Yea... it was an affiliate trying to prove a sponsor was shaving.
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Old 04-12-2006, 10:43 PM   #33
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To take a different route and disprove the logic, if on line 5 you were to subtract b from both sides, you would get a = 0.
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Old 04-12-2006, 11:37 PM   #34
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Damn...just when I discovered the infinite digit in Pi - along comes THIS problem.
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Old 04-13-2006, 01:29 AM   #35
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Non-linearity is a bitch.


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If not evolutionary, what then?






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Old 04-13-2006, 02:30 AM   #36
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On line 4 when you divide by (a-b), you're dividing by zero since a=b. Division by zero is undefined.

yea, i agree
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Old 04-13-2006, 05:56 AM   #37
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now that you know it, how does it help you in real life besides wasting time?
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