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
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%
if(this.signature != "")
{
MessageBox.Show("This is my signature: " + Environment.NewLine + signature);
}
else
{
MessageBox.Show("404-Signature not found");
}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%
if(this.signature != "")
{
MessageBox.Show("This is my signature: " + Environment.NewLine + signature);
}
else
{
MessageBox.Show("404-Signature not found");
} -
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%
if(this.signature != "")
{
MessageBox.Show("This is my signature: " + Environment.NewLine + signature);
}
else
{
MessageBox.Show("404-Signature not found");
} -
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%
if(this.signature != "")
{
MessageBox.Show("This is my signature: " + Environment.NewLine + signature);
}
else
{
MessageBox.Show("404-Signature not found");
}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.
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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.
Why do so many people here fail to grasp the obvious, that "x2" is meant to represent x squared? It's almost as if the CP community isn't very high quality.
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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
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Well... 0.999... is 0+(9/10) + (9/100) + (9/1000) +... which is 0+9*((1/10)+(1/100)+(1/1000)+...) which is 9* (sum(1/10^n)) where the sum runs over n for n=1 to infinity. Now for the same sum, if n ran from 0 to infinity, there is a convenient formula, that says it is (I'll mark this with a capital S to distinguish it from the one we're trying to calculate): Sum(1/10^n)=1+1/10+1/100+...=1/(1-(1/10))=1/(9/10)=10/9 therefore the sum we want is 1/10+1/100+...=(10/9)-1=1/9 therefore 0.999...=9*(1/9)=1 So it's math. On the other hand, computer calculated numbers are approximations, and in the binary system at that, and this is why you get all those rounding errors. But I get the feeling you already know all this, and that you are just toying with us obsessive compulsive types.
Φευ! Εδόμεθα υπό ρηννοσχήμων λύκων! (Alas! We're devoured by lamb-guised wolves!)
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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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0.9999 is indeed approximately 1 0.9999... (note the dots there) is EXACTLY 1.
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yiangos wrote:
0.9999... (note the dots there) is EXACTLY 1.
Unfortunately, no. Even as 9s extend on infinitely, it only approaches 1, but never actually gets there. It is not EXACTLY 1, but a very close approximation.
Money makes the world go round ... but documentation moves the money.
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yiangos wrote:
0.9999... (note the dots there) is EXACTLY 1.
Unfortunately, no. Even as 9s extend on infinitely, it only approaches 1, but never actually gets there. It is not EXACTLY 1, but a very close approximation.
Money makes the world go round ... but documentation moves the money.
Please take a look at the thread. There is a number of mathematical proofs that an infinite series of 9s in the decimal part is actually EXACTLY equal to 1. Heck, even I offered a proof. There's no approximation there. 0.999... is EXACTLY equal to 1. Again, the dots play an important role. No dots, no equality. EDIT No I didn't offer a proof. Here's one. 0.999...= 9*(0.111...) 0.999...=9*[(1-1)+0.111...) 0.999...=9*[(1+0.111...)-1] 0.999...=9*[(1+0.1+0.01+0.001+...)-1] Now 1+0.1+0.001+0.0001+... = 1/(1-0.1)=1/0.9 This is the closed form for the sum of an infinite number of terms of aa geometric series. This is NOT an approximation. This is the actual final value of summing infinite terms of a geometric series. You can see e.g. at Geometric series - Wikipedia, the free encyclopedia[^] or you can look it up in wolfram (too tired to look it up myself right now). Therefore 0.999...=9*[(1/0.9)-1] 0.999...=(9/0.9) -9 0.999...=(90/9) -9 0.999...=10-9 0.999...=1 QED. There's no pesky math there (such as dividing by zero, as the OP did).
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Since I remember it from math about that time it has to be old. :-D How can I remember this but cannot find my car keys?
Mongo: Mongo only pawn... in game of life.
Corporal Agarn wrote:
How can I remember this but cannot find my car keys?
It's called "selective memory". Apparently your car keys are not essential to your survival, but finding the flaw in such a trick is. :-\ :-O :laugh: ;P
Φευ! Εδόμεθα υπό ρηννοσχήμων λύκων! (Alas! We're devoured by lamb-guised wolves!)