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I think it may be time to upgrade your 386. Maybe to even a Pentium. 
There are only 10 types of people in the world, those who understand binary and those who don't.
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Or at least press the "Turbo" button. 
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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x = y.
Then x2 = xy.
Subtract the same thing from both sides:
x2 - y2 = xy - y2.
Dividing by (x-y), obtain
x + y = y.
Since x = y, we see that
2 y = y.
Thus 2 = 1, since we started with y nonzero.
Subtracting 1 from both sides,
1 = 0.
"There are two ways of constructing a software design: One way is to make it so simple that there are obviously no deficiencies, and the other way is to make it so complicated that there are no obvious deficiencies. The first method is far more difficult." - C.A.R. Hoare
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Stop dividing by zero (x-y=0)...
(yes|no|maybe)*
"Fortunately, we don't need details - because we can't solve it for you." - OriginalGriff
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Why are you dividing by (x-y) when you started with x=y? 
You have just been Sharapova'd.
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Dominic Burford wrote: x2 - y2 = xy - y2.
Dividing by (x-y), obtain
x + y = y.
Wrong... aint it?
x2 -y2 / (x-y) => x+y
xy -y2 / (x-y) => y+y
xy / x-y => 1*1/1-1 => (actually how could you divide this when there is a minus...)
y2 / x-y => y/x-1 => (actually how could you devide this when there is a minus...)
But i guess thats bullshit 2 because it "should"? be x2/x-y - y2/x-y which ruins everything 
Okay screw everything... you are wrong, thats all i can say
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HobbyProggy wrote: xy -y2 / (x-y) => y+y
Nope.
xy-y2 = y(x-y)
You have just been Sharapova'd.
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Ah yeah i started to correct it because i know its not correct you have to divide every part of that equation that leads to funny result
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if(!string.IsNullOrEmpty(this.signature))
ftfy
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Thank god signature is not an object you first have to convert .ToString() right?
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A high school teacher showed this to me some 10+ years back (feeling old now).
All I can say is no.
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The only thing i can agree with is that 1 != 3/3 (at least not exactly) because 1/3 is 0,333... and multiplied with 3 it is just 0,99999.... which is technically 1 but not 100%
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Quote: 0,99999.... which is technically 1 but not 100%
Uh?
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Whats the matter ?
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By definition, not by maths
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By maths. There's no difference.
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Allright, tell that the next flightcomputer that breakes down because he multiplied 0,333 and didn't get to 1 
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You know, float are not real numbers.
(likewise Computer Science is not Mathematics) 
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CPallini wrote: (likewise Computer Science is not Mathematics)
But its both logic And 0,999 is not 1 it's even written different 
But i guess we could debate ages about that
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Right: we could follow all the nines until we find a difference.
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Take your pick: How Can 0.999... = 1? | Purplemath[^]
For example:
x = 0.999...
10x = 9.999...
10x - x = 9.000...
9x = 9
x = 1
EDIT: As Nagy said[^], 2+ hours ago.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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1/6 = 0,166666
->*6
0,9996
x = 0,9996
10x = 9,996
10x -x = 8,9964
9x = 8,9964
x = 0,9996
?
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HobbyProggy wrote: 1/6 = 0,166666
->*6
0,9996
Nope - 6 × 0.1666... = 0.9999...
It's simple multiplication:
1 x 6 = 6 (0.6)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.96)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.996)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.9996)
6 x 6 = 36 = add 3 to the column on the left (6 + 3 = 9), and put 6 in this column (0.99996)
etc.
Because you're repeating the operation an infinite number of times, there's no point where you stop and leave the last digit as 6. There's always another digit to the right which needs to be multiplied by 6, carrying the 3 into the current column.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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