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
The second statement is false. if x = y, then x2 does not equal xy; x to the second power equals xy. For example: if 1=1 then 1(2) does not equal 1(1) Bullshit from the start, not to mention the division by zero
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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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}Am I the only one that doesn't see the issue that we are not dividing 1 by 3.... 1 = 3/3... If you're going to divide by 3 on the 1... wouldn't you need to also need to divide 3/3 by 3 also... 1/3 = (3/3)/3.... which would both be .03333... ??
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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
Sorry guys but second line is false its only true when x=1 or 2 x = y. Then x2 = xy is wrong if x=3 , xy=9 x2=6 6 !=9 you can't divide this way if its proper algebra left side is x2 - y2 2(x-y) ------- = ------ = 2 <- left side of the equation (x-y) (x-y) and obviously you cant separate (x-y) from the right side (xy - y2)/(x-y) != y sorry guys example is flawed and 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
Unfortunately, this "proof" falls apart at the 2nd line. x2 = xy only holds true for 0, 2, and -2. The correct equation is x^2 = xy. Still, a pretty interesting use of mathematics in an attempt to destroy all of our beliefs in what is thought to be true. :thumbsup:
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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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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
huh? I don't think x = y implies x2 = xy. How did you get that?
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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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}HobbyProggy wrote:
because 1/3 is 0,333
No. 1/3 does not equal 0.333. 1/3 is APPROXIMATELY 0.333. and 0.99999 is not technically 1. It is APPROXIMATELY 1 There is even a different symbol for it. 1/3≈0.333 (1/3 is approximately equal to 0.333) 0.99999≈1 (0.99999 is approximately equal to 1)
Money makes the world go round ... but documentation moves the money.
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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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A high school teacher showed this to me some 10+ years back (feeling old now).
10+ years is making you feel old? I saw that in high school 35+ years old..... (*) Even then, was able to figure out the flaw.
Truth, James
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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
As has been pointed out here, at some point you're dividing erroneously. But, you can prove that 0 = any number you want, using the derivation of differential calculus, as described here[^]: if you plug numbers into the equation at stage 3, you can "prove" pretty much whatever you want to prove. But of course, math is a language, not reality - and you can speak nonsense in any language.
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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
If 1 = 0, then 1 - 1 = 0 - 1, which means -1 = 0. Contradiction, so 1 = 0 can't be true.
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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
This is a code site so I wanted to prove this with C# and it fell apart for me.
[TestMethod]
public void TestMethod1()
{
//x = y
//Assuming y = 5.
var y = 5;var x = y; //Then x2 = xy Assert.AreEqual(x \* 2, x \* y); //Assert.AreEqual failed. Expected:<10>. Actual:<25>. } -
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
Sorry, that doesn't work. x2 is NOT equal to xy. That's like saying 2x = x*y which is ONLY true if x and y are both 2. Cute, but does not float.
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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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}HobbyProggy wrote:
because 1/3 is 0,333..
No it isn't. Mathematics defines and recognizes rational numbers. Mathematics defines and recognizes decimal numbers. They are two different things. Your equating them as the same does not remove certainly hundreds of years of mathematics and mathematicians that recognize and correctly differentiate the two. Certainly when I took mathematics courses that taught mathematics as it was implemented in computers they emphasized the assumptions and limitations of finite math. Actually understanding those assumptions and limitations makes the difference clear. Note clearly that in the above those were mathematics classes and not computer science classes. The difference is often where the former teaches the math and the later teaches how to use the computer. (Yes I took both.)
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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
Seeing you're playing with polynomials there, aren't you? following the logics.. x = y x^2 = xy x^2-y^2 = xy-y^2 (x-y)(x+y) = (x-y)y x+y = y 2y = y however you draw the wrong conclusion here, go on like this: 2y-y = y-y y = 0
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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
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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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} -
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:
Go to ParentThe 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%
This is nonsense ... .999... is exactly equal to 1, just as .333... is exactly equal to 1/3.
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huh? I don't think x = y implies x2 = xy. How did you get that?
"x2" is supposed to mean x squared.
