30 replies to this topic

[herenvardo]

Don't you realize that 2=1 is indeed true??
It's known that all (real) numbers fulfill a/a=1, and 0/a=0.
So, for a=0, we have 0/0=1 and 0/0=0. Since 0/0 must be equal to itself, then 1 must be equal to 0.
From here we can take that 1+0 = 1. But since 1=0, we can follow that 1+1=1 (we just replaced the 0).
But we also know that 1+1=2. If 1+1=2 and 1+1=1, then it's obvious that 2=1, because 1+1=1+1

Of course, that's not true at all. But if someone says that there is an error in the "proof" above, they will be wrong...
There are actually three gaps (a/a=1 and 0/a=0 are not true for all reals, but for all except zero; and 0/0 is an indetermination so it doesn't need to be equal to itself). I just hope you had some laugh with it

See you some other day, with my proof of PI=1 (that is among my favorites)

------
There are 10 kinds of people in the world: those who know binary, those who don't, and those who also know ternary.

[bookworm11]

For the second one, my friend explained it to me simply like this:

In order to find the end of 0.999999..., you have to put an infinite number of nines and then a zero at the end. But where is the end to infinity? There is none. Therefore, it is impossible to end 0.99999... and so it must be the same as 1.

[Tumohe]

Assume x = y which means 1 = 1

x = y then multply both sides by x so...
x² = xy then subtract y² so...
x² - y² = xy - y² then expand these quandratric equations to...
( x + y ) ( x - y ) = x ( x - y ) which is just another representation of the previous line.
Now, we currently have ( x - y ) on both sides, so lets divide both sides by ( x - y ) to get what is left...

(x+y)(x-y) = x(x-y) .
(x-y) . . . . (x-y)

That leaves us with the following
x + y = x

if we go back to the beginning and say that x equals y so that x = 1 and y = 1 then

1 + 1 = 1 OR
2 = 1


Logical except for the fact that you can not divide by x-y because if x=y then x-y=0 and you can not divide a number by nothing - ie, nothing goes into one an infinite number of times plus one.

[kh2expert]

where it says 'x=2' if you take that and plug it in to 'x=1'you get'2=1' which is inequal so there is your mistake

Sorry dude,



I do not understand your objection, and I think you're attempting to be contrarywise. However, here it is in proof form to overcome your objections.

Proof: 0.9999... = Sum 9/10^n
(n=1 -> Infinity)

= lim sum 9/10^n
(m -> Infinity) (n=1 -> m)

= lim .9(1-10^-(m+1))/(1-1/10)
(m -> Infinity)

= lim .9(1-10^-(m+1))/(9/10)
(m -> Infinity)

= .9/(9/10)

= 1


Perhapse you'd like to add two 0.99999 and get 2?


0.999... + 0.999... = Sum[n = 1 to infinity](9*10^-n)
+ Sum[n = 1 to infinity](9*10^-n)

= Sum[n = 1 to infinity](18*10^-n)

= Sum[n = 1 to infinity](10*10^-n + 8*10^-n)

= Sum[n = 1 to infinity](10^(-n+1))
+ Sum[n = 1 to infinity](8*10^-n)

= Sum[n = 0 to infinity](10^-n)
+ Sum[n = 1 to infinity](8*10^-n)

= 1 + Sum[n = 1 to infinity](10^-n)
+ Sum[n = 1 to infinity](8*10^-n)

= 1 + Sum[n = 1 to infinity]((8+1)*10^-n)

= 1 + Sum[n = 1 to infinity](9*10^-n)

= 1.999...
But you can't, you get 1.9999... interesting eh?


Perhapse if that does not satisfy, you could see it this way.

1.000000000...
-0.999999999...
=0.000000000...


You may ask about the "1" in the last decimal place, and I'll write that when I'm done writing an infanate string of "0"s. So the 4th decimal place has a "0", and the 4 billionth has a "0", and the decimal place corrisponding to any number you could or could not name or imagine has a "0." It can be seen that a number with a "0" in every decimal place must be equal to 0.


And you don't have to take my word for it, Blizzard Entertainment is weighing in on the discussion with their own proof...

http://www.blizzard....ss/040401.shtml

Or there is a quite technical and compleate work up of many difrent proofs and systems to arrive at such conclusions at Wikipedia

http://en.wikipedia.org/wiki/0.999...

[kh2expert]

where it says 'x=2' if you take that and plug it in to 'x=1'you get'2=1' which is inequal so there is your mistake

[JH J]

Assume x = y which means 1 = 1

x = y then multply both sides by x so...
x² = xy then subtract y² so...
x² - y² = xy - y² then expand these quandratric equations to...
( x + y ) ( x - y ) = x ( x - y ) which is just another representation of the previous line.
Now, we currently have ( x - y ) on both sides, so lets divide both sides by ( x - y ) to get what is left...

(x+y)(x-y) = x(x-y) .
(x-y) . . . . (x-y)

That leaves us with the following
x + y = x

if we go back to the beginning and say that x equals y so that x = 1 and y = 1 then

1 + 1 = 1 OR
2 = 1


Logical except for the fact that you can not divide by x-y because if x=y then x-y=0 and you can not divide a number by nothing - ie, nothing goes into one an infinite number of times plus one.

Ohh..!!
I saw this before!
It was on my test (Geometry) as an extra credit (I didn't get it... )

I also saw a similar problem in my geometry book (another extra credit)
which described how 2X2=5 (something similar to that...)

All of these are just breaking only one rule, which is that you can't divide by zero.

However...I'm kind of confused on what people are talking about when it comes to things like .9999...
...Oh well...

[95yj]

Actually, I've seen a similar mathematical fallacy stated as the following. I like it better because the error is hidden a little better and the proof looks a little cleaner. The proof is pretty much the same and is identical to the problem posted at the begging of the thread in that the algebra is correct but the proof violates a basic rule of mathematics.

Proof that any number 'a' is equal to a smaller number 'b':

a = b + c

a(a - b) = (b + c)(a - b)

a^2 - ab = ab - b^2 + ac - bc

a^2 - ab - ac = ab - b^2 - bc

a(a - b - c) = b(a - b - c)

a = b

Edited by 95yj, 16 November 2008 - 04:54 PM.

[young_asker22]

x(x-2)=x-2
x(2-2)=2-2
0=0

[mridhu]

Alright, try this one on for size:

X=.9999999...

10x=9.99999...
(10x)-x=9
(9x)/9=1
x=1

.999999...=1

X=.9999999...

10x=9.99999...
(10x)-x=9
10x = 9+x
10x-9 = x
10-9 = x/x
1 = 1

so, it always results in 1=1

[Royal]

Alright, try this one on for size:

X=.9999999...

10x=9.99999...
(10x)-x=9
(9x)/9=1
x=1

.999999...=1

There is no problem with this statement; it is true.

Think about it this way:
(1/3)=.333333.....
(2/3)=.666666.....
(1/3)+(2/3)=.999999....
or
(1/3)+(2/3)=(3/3)=1

The original question is wrong because he divides by (x-1) which would be division by 0.