| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Here's a proof that 2=1 that was presented to me during one of the
math classes at High School. Although it contains a very basic mistake,
I've managed to stump many engineers, and yes, some math majors!!
To make things worse, the only person who figured it out was an
Architect!! I hope DEC people do better!!
Here it is (I'll say why it's wrong sometime later)
Let: x=y
Then:
x=y
x^2=x*y
x^2-y^2=x*y-y^2
(x-y)*(x+y)=y*(x-y)
x+y=y
y+y=y
2y=y
therefore 2=1
Enjoy.
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 675.1 | Solution | JON::MORONEY | Light the fuse and RUN! | Fri Mar 13 1987 12:52 | 33 |
Let: x=y
Then:
x=y
x^2=x*y
x^2-y^2=x*y-y^2
> (x-y)*(x+y)=y*(x-y) If x=y, this line is 0 * something.
Both sides are 0, since (x-y) = 0.
> x+y=y To get this, you have divided both sides
by 0. (naughty naughty!)
y+y=y
2y=y
therefore 2=1
Here's a similar one, proving 2<1
Let: 0<x<y
Then:
x<y
x^2<x*y
x^2-y^2<x*y-y^2
(x-y)*(x+y)<y*(x-y)
x+y<y
y+y<y
2y<y
therefore 2<1
I used to do these all the time to a math teacher when I was a kid.
-Mike
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| 675.2 | yes, devide by zero will let you show 1=2 | THEBUS::KOSTAS | Wisdom is the child of experience. | Fri Mar 13 1987 13:08 | 8 |
re. -1
well you may find a few of the problems in this notes file.
In your example, division by zero is the problem.
/kgg
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| 675.3 | take a look at note 466.* for similar ideas | THEBUS::KOSTAS | Wisdom is the child of experience. | Fri Mar 13 1987 13:14 | 4 |
re. .0
take a look at some of the ideas in the note 466.*
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| 675.4 | very old! | IOSG::DEMORGAN | Mon Jun 15 1987 07:12 | 1 | |
This was a hoary old chestnut thirty years ago. | |||||