Foo - "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?

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.

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.

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.
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?

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).

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)

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.:D

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

Or it could be a line to another dimension. One where the narrator is Rod Serling.:D

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.

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 :p ) but isn't dividing by zero impossible regardless of what number you are dividing?

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.

Maybe I'm remembering math all wrong (highly likely; if I could do math or science, I wouldn't be a lawyer :p ) 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. :)

Maybe I'm remembering math all wrong (highly likely; if I could do math or science, I wouldn't be a lawyer :p ) but isn't dividing by zero impossible regardless of what number you are dividing?

It's conceptualized as "Undefined", actually.

It's conceptualized as "Undefined", actually.

Ahhhhh, got it. Sort of like a lawyer's soul.