.999~ = 1

[Dejhan_Tulip]

Originally Posted by deltacoder1020

constant of integration.

What do you mean with that ?? Explain please...

[Greg Ross]

Nothing's illegal, eh? To get from

a(b-a) = (b-a)*(b+a)

to

a = b+a

you have to divide by (b-a) which is zero! Which is the DEFINITION of illegal!

[Greg Ross]

Originally Posted by Dejhan_Tulip

What do you mean with that ?? Explain please...

Constant of Integration

[deltacoder1020]

Originally Posted by gwross

Nothing's illegal, eh? To get from

to

you have to divide by (b-a) which is zero! Which is the DEFINITION of illegal!

yep. oldest trick in the false-proof book. second oldest is trying to argue that x^2 = y^2 implies that x = y.

[deltacoder1020]

Originally Posted by Dejhan_Tulip

What do you mean with that ?? Explain please...

essentially, an indefinite integral will always be equal to the same indefinite integral plus a constant. it would have to be, as the following example shows:

f(x) = 2x + 1
g(x) = 2x

f'(x) = 2
g'(x) = 2

if we didn't have a constant of integration,
integral(f'(x)) = 2x = integral(g'(x)) = 2x
and thus f(x) would be implied to be = to g(x)

however, the correct form is this:
integral(f'(x)) = 2x + C1
integral(g'(x)) = 2x + C2

thus, f(x) = g(x) + C2 - C1, where C2 and C1 are integers.

essentially what this all means is that you can't simply "take out" an indefinite integral from an equation, and thus the fact that 0 + integral(f(x)) = 1 + integral(f(x)) does not imply that 0 = 1.

[Denman]

well said...
whenever you integrate an indefinate you must always have a constant. in definate integrals, this doesn't matter as when you solve it you get
(.........+k)-(.........+k) and they cancel out. (where k is a constant)

[Grommit]

Originally Posted by dk5sm5luigi

This is exactly why the proof that I posted with e^(i*pi) doesn't work.

No. This is quite different. Quaternions are an extension of the real numbers and you don't need to even mention them to talk about e^(i*pi). With quaternions, ab != ba in some circumstances. There are even octonions with a(bc) != (ab)c.

The reason why your e^(i*pi) thing doesn't work is that you are taking the log of a negative/complex value incorrectly. Why does log(exp(3pi*i) = i*pi? Because exp(3pi*i) = exp(pi*i) = -1 and in the natural extension of the log to the complex plane log(-1) = i*pi. So one must be very careful when using natural logs on complex arguments. There are several good books on the subject, in fact MIT has a complex analysis class available entirely free. This is necessary before one "proves" anything involving complex arguments using functions defined on the reals.

[pryoplasm]

*wipes a single tear from his eye*

just when i think i was some hot ****, i have never been so proud to be proven that i am not by far as much of a math geek

<3 the topic

[hansTP2S]

alright i thought of a different way to prove that .999...=1

first we have to prove that if (X+Y)/2=X or (X+Y)/2=Y then X=Y
this is simple

(X+Y)/2=X
X+Y=2X
0=X-Y
Y=X

alright now we move on

1 (.9999...+1)/2 ?= .999...
2 (1.999...)/2=
3 .999...=.999...
now you might be thinking: "whoa...how do you get from 2 to three...that looks crazy"

well 1.9/2=.95
1.99/2=.995
1.999/2=.9995

so 1.999.../2=.999...5

however i have given all you non believers false hope with that last 5

1.999.../2=.999...

because the nines will go on FOREVER, they wont stop, so that five does not exist the five never happens because the nines never stopped...thus .999...=1