
Actually 1.0 = 0.9999... I did the derivation of this proof as an exercise in a algebra class back in the early 80s.





1 === 3/3 === X/X
The thing is base 10 decimal (0.333) does not have a correct way to represent 3/3, except putting bar above the value that repeats infinitely
_
0.333





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.





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





Quote: Dividing by (xy), obtain... ...the silent reproach of a million tearstained eyes.





Dominic Burford wrote: since we started with y nonzero.
oh, the irony.





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





Since I remember it from math about that time it has to be old.
How can I remember this but cannot find my car keys?
Mongo: Mongo only pawn... in game of life.






If you divide by zero you get sucked into the mathematical black hole where all logic is lost. But if it were true that 1=0, would that simplify chip design?
"A little knowledge is a dangerous thing, drink deeply or taste not."





MKJCP wrote: But if it were true that 1=0, would that simplify chip design?
I wonder, I've heard that qubits can be in the "0" and "1" state at the same time, so does that qualify for "1=0" at any level?
Φευ! Εδόμεθα υπό ρηννοσχήμων λύκων!
(Alas! We're devoured by lambguised wolves!)





I suppose it's arguable but dual state does not imply equality. It sounds like they should have named them confusedbits.





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





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(xy)
 =  = 2 < left side of the equation
(xy) (xy)
and obviously you cant separate (xy) from the right side (xy  y2)/(xy) != y
sorry guys example is flawed and 1 != 0





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.





This appears to work, but only because you are dividing by zero since (xy) is, by definition of the first line, equal to zero.





huh? I don't think x = y implies x2 = xy. How did you get that?





"x2" is supposed to mean x squared.





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.





If 1 = 0, then 1  1 = 0  1, which means 1 = 0. Contradiction, so 1 = 0 can't be true.





This is a code site so I wanted to prove this with C# and it fell apart for me.
[TestMethod]
public void TestMethod1()
{
var y = 5;
var x = y;
Assert.AreEqual(x * 2, x * y);
}





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.





Well, the first problem with this is
x = y
then x2 = xy IFF x = y = 2
This was stated just above, but I didn't get that far when I was reading the comments...





Seeing you're playing with polynomials there, aren't you?
following the logics..
x = y
x^2 = xy
x^2y^2 = xyy^2
(xy)(x+y) = (xy)y
x+y = y
2y = y
however you draw the wrong conclusion here, go on like this:
2yy = yy
y = 0



