1 = 0
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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. :wtf:
"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 Home | LinkedIn | Google+ | Twitter
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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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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HobbyProggy wrote:
xy -y2 / (x-y) => y+y
Nope. xy-y2 = y(x-y)
You have just been Sharapova'd.
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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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. :wtf:
"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 Home | LinkedIn | Google+ | Twitter
http://www.wolframalpha.com/input/?i=x2%3Dxy[^] you can use this site to experiment with your theories?
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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. :wtf:
"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 Home | LinkedIn | Google+ | Twitter
A high school teacher showed this to me some 10+ years back (feeling old now). All I can say is no.
Visit my blog at Sander's bits - Writing the code you need. Or read my articles at my CodeProject profile.
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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.
Visit my blog at Sander's bits - Writing the code you need. Or read my articles at my CodeProject profile.
Simplicity is prerequisite for reliability. — Edsger W. Dijkstra
Regards, Sander
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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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. :wtf:
"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 Home | LinkedIn | Google+ | Twitter
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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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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. :wtf:
"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 Home | LinkedIn | Google+ | Twitter
Dominic Burford wrote:
since we started with y nonzero.
oh, the irony.
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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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}if the numerator and denominator are equal then the value must be 1. Then the second part is almost right, but you missed a very important point, a recurring value is only an *approximation* rather than the absolute. 1/3 is absolute, but the decimal 0.3. is only an approximation. Proof, without bad maths of the 9's reccuring - 0.9. == 1 Multiply by 10: 9.9. = 10 Subtract the original: 9.9. - 0.9. = 9 9 = 9 QED. This is accepted as proper maths, the 1=0 using division by zero is, however, not.
veni bibi saltavi
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if the numerator and denominator are equal then the value must be 1. Then the second part is almost right, but you missed a very important point, a recurring value is only an *approximation* rather than the absolute. 1/3 is absolute, but the decimal 0.3. is only an approximation. Proof, without bad maths of the 9's reccuring - 0.9. == 1 Multiply by 10: 9.9. = 10 Subtract the original: 9.9. - 0.9. = 9 9 = 9 QED. This is accepted as proper maths, the 1=0 using division by zero is, however, not.
veni bibi saltavi
Nice :) But i guess as long as we have an approximation i'll go with its not exactly 1 :P
Nagy Vilmos wrote:
This is accepted as proper maths, the 1=0 using division by zero is, however, not.
I already said that in an other statement that hes wrong
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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. :wtf:
"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 Home | LinkedIn | Google+ | Twitter
Ah, the ol' divide by zero trick. I think my math teacher did that when I was in 8th grade. Almost 40 years ago. And I'm sure it's older than that. ;) Marc
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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 :sigh: But i guess we could debate ages about that
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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 :sigh: But i guess we could debate ages about that
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