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# "proofs" that 1=0 (fallacious of course)

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# "proofs" that 1=0 (fallacious of course)

07-24-07, 03:53 PM
#1
mlts22
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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?
07-24-07, 05:19 PM
#2
jschen
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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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07-24-07, 05:37 PM
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Originally Posted by jschen
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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07-24-07, 05:46 PM
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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.
Originally Posted by mlts22
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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07-24-07, 05:54 PM
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asherlighn
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Originally Posted by Tom Stormcrowe
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).
07-25-07, 11:33 AM
#6
Tom Stormcrowe
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Originally Posted by asherlighn
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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07-25-07, 12:05 PM
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Originally Posted by Tom Stormcrowe
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.
07-25-07, 12:49 PM
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Originally Posted by mlts22
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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07-25-07, 07:24 PM
#9
Tom Stormcrowe
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Originally Posted by polara426sh
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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07-26-07, 11:58 AM
#10
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Originally Posted by Tom Stormcrowe
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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07-26-07, 12:19 PM
#11
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Originally Posted by jschen
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.
07-26-07, 02:28 PM
#12
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Originally Posted by bikingshearer
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.
07-26-07, 02:35 PM
#13
Tom Stormcrowe
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Originally Posted by bikingshearer
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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07-27-07, 01:25 PM
#14
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Originally Posted by Tom Stormcrowe
It's conceptualized as "Undefined", actually.
Ahhhhh, got it. Sort of like a lawyer's soul.
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