1. I must say I didn't see either what you did wrong, the only possible thing I can see is that you started with the thing you where going to prove and developed it into something obvious, instead of doing the opposite, that was maybe what ThePrefectHacker ment by the preservation of the not equality.

2. Originally Posted by The Pondermatic
Really?

*looks back at his work, fails to see problems*

Heh, I guess i'm a bit of a newb.

What exactly did I do wrongly again?

I have to admitt, though, if my proof really is wrong, that's pretty cool that it's right anyway.
Originally Posted by TriKri
I must say I didn't see either what you did wrong, the only possible thing I can see is that you started with the thing you where going to prove and developed it into something obvious, instead of doing the opposite, that was maybe what ThePrefectHacker ment by the preservation of the not equality.
ThePerfectHacker has a habit of spotting proof errors that are often nearly invisible. (I can't spot any errors in the "proof" either.) How he manages this I have yet to discover. I suspect a complete dissection of his brain would provide the answer...

-Dan

3. Originally Posted by topsquark
ThePerfectHacker has a habit of spotting proof errors that are often nearly invisible.
Thank you I am honored.

How he manages this I have yet to discover. I suspect a complete dissection of his brain would provide the answer...
My secret is simple. Pure math. What do I mean by that? Occasionally everyone (I mean mathemations) are lazy (like usual) and you do not state out all the details. For example, the proper procedure to use the substitution rule for integrals is to show the inner function is differenciable and then show that the outer function poses an anti-derivative.... the traditional stuff. Most people ignore that (I do that sometimes) and make a mistake. When doing a math problem we need to go all the way back to the definitions. This is why Mathematical proves (like Perelman's) take an awufully long time to check. Because everyone returns back to the defintions of a topology and procede from there. This is why many mathemations did make mistakes (but easily correctable) in their long proofs because they were far too lazy to do that from the beginning.

4. A classic from Maclaurin Series . . .

We are given: .$\displaystyle \ln(1 + x) \:=\:x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots$

Let $\displaystyle x = 1:\;\;\ln(2) \;=\;1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots$

The right side is: .$\displaystyle \left(1 + \frac{1}{3} + \frac{1}{5} + \cdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)$

Add and subtract $\displaystyle \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)$

. . $\displaystyle \underbrace{\left(1 + \frac{1}{3} + \frac{1}{5} + \cdots\right) + \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)}$$\displaystyle - \underbrace{\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)} . . . . . \displaystyle = \quad\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right)\qquad\qquad -\qquad\qquad 2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right) . . . . . . . . . . .\displaystyle = \quad\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right) \;- \;\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right) . . . . . . . . . . . . . . . . . . . . . . . . . . . \displaystyle =\quad 0 We have just shown that: .\displaystyle \ln(2) \,= \,0 . . . . . . . . . . .Therefore: .\displaystyle e^0 \,=\,2\quad\Rightarrow\quad\boxed{ 1 \,= \,2} 5. Originally Posted by Soroban A classic from Maclaurin Series . . . We are given: .\displaystyle \ln(1 + x) \:=\:x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots Let \displaystyle x = 1:\;\;\ln(2) \;=\;1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots The right side is: .\displaystyle \left(1 + \frac{1}{3} + \frac{1}{5} + \cdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right) Add and subtract \displaystyle \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right) . . \displaystyle \underbrace{\left(1 + \frac{1}{3} + \frac{1}{5} + \cdots\right) + \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)}$$\displaystyle - \underbrace{\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)}$

. . . . . $\displaystyle = \quad\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right)\qquad\qquad -\qquad\qquad 2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)$

. . . . . . . . . . .$\displaystyle = \quad\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right) \;- \;\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right)$

. . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle =\quad 0$

We have just shown that: .$\displaystyle \ln(2) \,= \,0$

. . . . . . . . . . .Therefore: .$\displaystyle e^0 \,=\,2\quad\Rightarrow\quad\boxed{ 1 \,= \,2}$

The series is not absolutely convergent and the rearrangement theorem cannot be used.

6. Originally Posted by Soroban
A classic from Maclaurin Series . . .

We are given: .$\displaystyle \ln(1 + x) \:=\:x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots$

Let $\displaystyle x = 1:\;\;\ln(2) \;=\;1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots$

The right side is: .$\displaystyle \left(1 + \frac{1}{3} + \frac{1}{5} + \cdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)$

Add and subtract $\displaystyle \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)$

. . $\displaystyle \underbrace{\left(1 + \frac{1}{3} + \frac{1}{5} + \cdots\right) + \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)}$$\displaystyle - \underbrace{\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)} . . . . . \displaystyle = \quad\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right)\qquad\qquad -\qquad\qquad 2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right) . . . . . . . . . . .\displaystyle = \quad\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right) \;- \;\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right) . . . . . . . . . . . . . . . . . . . . . . . . . . . \displaystyle =\quad 0 We have just shown that: .\displaystyle \ln(2) \,= \,0 . . . . . . . . . . .Therefore: .\displaystyle e^0 \,=\,2\quad\Rightarrow\quad\boxed{ 1 \,= \,2} A lot of working just to be incorrect so early on A lot of working just to be incorrect so early on 7. Originally Posted by The Pondermatic Proof that it isn't: Line 1 √(-1 * -1) ≠ √(-1) ^ 2 Line 2 √(1) ≠ i ^ 2 Line 3 1 ≠ -1 Heh. That was probably obvious, but I love doing short proofs like that. ...Well, I guess I just said the opposite of what the origonal problem said. It makes my brain hurt. Square roots have two possible answers therefore proofs like this can never work \displaystyle \sqrt{1} = 1 is true because 1 x 1 = 1 \displaystyle \sqrt{1} = -1 is true because -1 x -1 = 1 It does not follow that 1 must equal -1 as square root is not a 1 to 1 function Plus assuming what you want to be true implies something that is true does not make original assumption true or else eg All multiplications give the answer zero 6 x 0 = 0 Second line is true and follows from first line. However first line certainly isn't true 8. Originally Posted by Glaysher Square roots have two possible answers Then why do we always use the pluss/minus sign in solutions to second order equations? Edit: Okey, I looked up the definition of square root and saw it could be both the values. According to wikipedia, there is the principal square root, apperantly, and the other square root. Another thing I didn't quite understand was if the \displaystyle \sqrt{\ } symbol denoted the principal square root or not. First it says it does, then it says that "The square root symbol (\displaystyle \sqrt{\ }) was first used during the 16th century". 9. Originally Posted by TriKri Sorry, this post can be removed. You can remove your own posts (though I don't know about an actual thread) by editing your message. There is an option at the bottom of the preview box that asks if you wish to delete the message. -Dan 10. Originally Posted by TriKri Then why do we always use the pluss/minus sign in solutions to second order equations? Edit: Okey, I looked up the definition of square root and saw it could be both the values. According to wikipedia, there is the principal square root, apperantly, and the other square root. Another thing I didn't quite understand was if the \displaystyle \sqrt{\ } symbol denoted the principal square root or not. First it says it does, then it says that "The square root symbol (\displaystyle \sqrt{\ }) was first used during the 16th century". As always context is key. If you are given a calculation to do usually you just take the principal square root. If you are solving an equation and at some point you have to square root you need to take into account both possible solutions eg \displaystyle x^2 = 4 \displaystyle x = \pm \sqrt{4} \displaystyle x = \pm 2 You use the \displaystyle \pm sign to show you are considering both possible solutions 11. Originally Posted by Glaysher As always context is key. If you are given a calculation to do usually you just take the principal square root. If you are solving an equation and at some point you have to square root you need to take into account both possible solutions eg \displaystyle x^2 = 4 \displaystyle x = +/- \sqrt{4} \displaystyle x = +/- 2 You use the +/- sign to show you are considering both possible solutions the +/- is more commonly written as: \displaystyle \pm 12. Then I guess \displaystyle \sqrt{1} \neq -1 since \displaystyle \sqrt{\ } is the principal square root. 13. Originally Posted by Soroban A classic from Maclaurin Series . . . We are given: .\displaystyle \ln(1 + x) \:=\:x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots Let \displaystyle x = 1:\;\;\ln(2) \;=\;1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \cdots The right side is: .\displaystyle \left(1 + \frac{1}{3} + \frac{1}{5} + \cdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right) Add and subtract \displaystyle \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right) . . \displaystyle \underbrace{\left(1 + \frac{1}{3} + \frac{1}{5} + \cdots\right) + \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)}$$\displaystyle - \underbrace{\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)}$

. . . . . $\displaystyle = \quad\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right)\qquad\qquad -\qquad\qquad 2\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots\right)$

. . . . . . . . . . .$\displaystyle = \quad\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right) \;- \;\left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots\right)$

. . . . . . . . . . . . . . . . . . . . . . . . . . . $\displaystyle =\quad 0$

We have just shown that: .$\displaystyle \ln(2) \,= \,0$

. . . . . . . . . . .Therefore: .$\displaystyle e^0 \,=\,2\quad\Rightarrow\quad\boxed{ 1 \,= \,2}$

Isn't the Maclaurin series for ln(x+1) only valid for -1 < x < 1?

14. Originally Posted by Quick
the +/- is more commonly written as: $\displaystyle \pm$
Didn't know the LaTex command and didn't have time to look it up.

15. May that person, whoever it was, rest in peace. $\displaystyle \pm$

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