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Old 07-24-07, 03:53 PM   #1
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"proofs" that 1=0 (fallacious of course)

I used to have three fallacious proofs that 1=0.

The first one was regarding infinite series:

0 = (1-1) + (1-1) + ...
= 1 + (-1 + 1) + (-1 + 1) + ...
ergo, 0=1

The second one was based on dividing by zero:

0 x 0 = 0
0 x 1 = 0

0 x 0 = 0 x 1

divide both sides by zero
0/0 x 0 = 0/0 x 1
ergo 0=1


The last one, I forgot, it was integral based. Anyone have any ideas which it was?
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Old 07-24-07, 05:19 PM   #2
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Let's be a bit more subtle.

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

since a = b...

2b = b
2 = 1

2 - 1 = 1 - 1
1 = 0

Yes... it's still a division by zero fallacy, but at least we don't outright show a big fat 0 in the denominator.
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Old 07-24-07, 05:37 PM   #3
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Originally Posted by jschen View Post
Let's be a bit more subtle.

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

since a = b...

2b = b
2 = 1

2 - 1 = 1 - 1
1 = 0

Yes... it's still a division by zero fallacy, but at least we don't outright show a big fat 0 in the denominator.
At least this version followed math rules and could be followed.


The ones in O.P. just took leaps with no math rules being followed....it made no sense.
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Old 07-24-07, 05:46 PM   #4
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I think he was trying to use the propery of one.

Anything/itself=1 (except 0)

5/5=1, 1/1=1, but the exception 0/0=0 still, unless I'm remembering math all wrong.
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Originally Posted by mlts22 View Post
I used to have three fallacious proofs that 1=0.

The first one was regarding infinite series:

0 = (1-1) + (1-1) + ...
= 1 + (-1 + 1) + (-1 + 1) + ...
ergo, 0=1

The second one was based on dividing by zero:

0 x 0 = 0
0 x 1 = 0

0 x 0 = 0 x 1

divide both sides by zero
0/0 x 0 = 0/0 x 1
ergo 0=1


The last one, I forgot, it was integral based. Anyone have any ideas which it was?
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Old 07-24-07, 05:54 PM   #5
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Originally Posted by Tom Stormcrowe View Post
I think he was trying to use the propery of one.

Anything/itself=1 (except 0)

5/5=1, 1/1=1, but the exception 0/0=0 still, unless I'm remembering math all wrong.
You are remembering math correctly. It is deliberately wrong. Designed to make students look at the intermediary steps in proofs (atleast thats where I first saw it).
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Old 07-25-07, 11:33 AM   #6
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Originally Posted by asherlighn View Post
You are remembering math correctly. It is deliberately wrong. Designed to make students look at the intermediary steps in proofs (atleast thats where I first saw it).
Come to think about it, the result would either be 0 or undefined, depending on the application!
If it's a slope, the result would be undefined.
y=mx+b, with y=o and x=0, it would be an undefined result(vertical line right on the y axis, through the 0,0 intersection or origin)
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Old 07-25-07, 12:05 PM   #7
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Originally Posted by Tom Stormcrowe View Post
Come to think about it, the result would either be 0 or undefined, depending on the application!
If it's a slope, the result would be undefined.
y=mx+b, with y=o and x=0, it would be an undefined result(vertical line right on the y axis, through the 0,0 intersection or origin)
Or it could be a line to another dimension. One where the narrator is Rod Serling.
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Old 07-25-07, 12:49 PM   #8
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Originally Posted by mlts22 View Post
I used to have three fallacious proofs that 1=0.

The first one was regarding infinite series:

0 = (1-1) + (1-1) + ...
= 1 + (-1 + 1) + (-1 + 1) + ...
ergo, 0=1
Actually, if you continued this out it would actually be:

0 = 1 + (-1 + 1) + (-1 + 1) + ... + -1 + ....

0 = 1 - 1 = 0
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Old 07-25-07, 07:24 PM   #9
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Or it could be a line to another dimension. One where the narrator is Rod Serling.
Nope, because for that to occur, you'd have to factor in the tau axis and teh axis if you want to include multiverse theory, as well as x,y, and z on multiple axis', thereby winding up with something more like 0/∞, or ∞/0.
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Old 07-26-07, 11:58 AM   #10
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Originally Posted by Tom Stormcrowe View Post
I think he was trying to use the propery of one.

Anything/itself=1 (except 0)

5/5=1, 1/1=1, but the exception 0/0=0 still, unless I'm remembering math all wrong.
Maybe I'm remembering math all wrong (highly likely; if I could do math or science, I wouldn't be a lawyer ) but isn't dividing by zero impossible regardless of what number you are dividing?
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Old 07-26-07, 12:19 PM   #11
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Originally Posted by jschen View Post
Let's be a bit more subtle.

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

since a = b...

2b = b
2 = 1

2 - 1 = 1 - 1
1 = 0

Yes... it's still a division by zero fallacy, but at least we don't outright show a big fat 0 in the denominator.
Yep. This proves it, computer's can't work.
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Old 07-26-07, 02:28 PM   #12
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Maybe I'm remembering math all wrong (highly likely; if I could do math or science, I wouldn't be a lawyer ) but isn't dividing by zero impossible regardless of what number you are dividing?


Yep.. looks like he forgot to use L'Hôpital's Rule.
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Old 07-26-07, 02:35 PM   #13
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Maybe I'm remembering math all wrong (highly likely; if I could do math or science, I wouldn't be a lawyer ) but isn't dividing by zero impossible regardless of what number you are dividing?
It's conceptualized as "Undefined", actually.
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Old 07-27-07, 01:25 PM   #14
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It's conceptualized as "Undefined", actually.
Ahhhhh, got it. Sort of like a lawyer's soul.
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