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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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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
Rules for the FOSW ![ ^]
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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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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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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.
Φευ! Εδόμεθα υπό ρηννοσχήμων λύκων!
(Alas! We're devoured by lamb-guised wolves!)
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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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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).
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 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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Actually 1.0 = 0.9999... I did the derivation of this proof as an exercise in a algebra class back in the early 80s.
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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
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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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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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Quote: Dividing by (x-y), obtain...
...the silent reproach of a million tear-stained eyes. 
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Dominic Burford wrote: since we started with y nonzero.
oh, the irony.
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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
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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.
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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."
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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 lamb-guised wolves!)
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I suppose it's arguable but dual state does not imply equality. It sounds like they should have named them confusedbits.
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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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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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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. 
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This appears to work, but only because you are dividing by zero since (x-y) is, by definition of the first line, equal to zero.
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