1 = 0
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HobbyProggy wrote:
Because the 6 at the end is ...
But there isn't a
6at the "end", because there isn't an "end"! :laugh: Think of it like this:using System;
static class Program
{
static void Main()
{
while (true)
{
Console.WriteLine("9");
}Console.WriteLine("Squirrel!"); }}
If you run that program, how long will you have to wait before it prints "Squirrel"?
"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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Do you mean they are imaginary? :doh: :doh: :doh: This is getting more and more difficult by moments. I think I am grabbing a bier, maybe seeing it double makes it easier :rolleyes: ;P :laugh: :laugh: I'll get my coat
M.D.V. ;) If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about? Help me to understand what I'm saying, and I'll explain it better to you Rating helpful answers is nice, but saying thanks can be even nicer.
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So this works only for 0.999999 ? Because the 6 at the end is in this case important to show that it wont be a 1 in the end :) But nevermind, this is all just playing with numbers :)
if(this.signature != "")
{
MessageBox.Show("This is my signature: " + Environment.NewLine + signature);
}
else
{
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}mmm working by 9, not working by 6 what about 69? does it works? Oh, wait... I think I stop now since I am about to break the KSS rule :laugh: :laugh: I am still on the way getting my coat
M.D.V. ;) If something has a solution... Why do we have to worry about?. If it has no solution... For what reason do we have to worry about? Help me to understand what I'm saying, and I'll explain it better to you Rating helpful answers is nice, but saying thanks can be even nicer.
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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
Imperative to Functional Programming Succinctly Contributors Wanted for Higher Order Programming Project!
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.
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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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} -
Take your pick: How Can 0.999... = 1? | Purplemath[^] For example:
x = 0.999...10x = 9.999...
10x - x = 9.000...
9x = 9
x = 1EDIT: 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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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 != "")
{
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}
else
{
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}Actually, yes 0.9999..... is in fact 1: x = 0.99999.... 10 x = 9.999999.... multiply by 10 10 x = 9 + 0.99999.... split right side into arithmetic expression 10 x = 9 + x replace 0.9999.... with "x" 9 x = 9 subtract "x" from both sides x = 1 divide by 9. without dividing by "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
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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%
if(this.signature != "")
{
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else
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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%
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. 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.
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
Quote:
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>. }
