Another "Proof" that 0 = 1

jinydu · 2007-06-01, 06:49

I thought of this today. It relies on divergent series. Enjoy

$\Large{1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+...}$

$\Large{=\frac{1}{1-\frac{1}{z}}}$

$\Large{=\frac{z}{z-1}}$

$\Large{=-z\frac{1}{1-z}}$

$\Large{=-z(1+z+z^2+z^3+...)}$

$\Large{=-z-z^2-z^3-\ ...}$

Adding the negative of the right-hand side to both sides gives:

$\Large{...+\frac{1}{z^3}+\frac{1}{z^2}+\frac{1}{z}+1+z+z^2+z^3+...=0}$

Regarding the right-hand side as a Laurent series with all coefficients equal to zero, we can equate corresponding coefficients to get 1 = 0 (infinitely many times).

As a bonus, setting z = 1 gives $\infty=0$ .

Orgasmic Troll · 2007-06-01, 10:44

cute

ATH · 2007-06-01, 11:18

Place where it fails:

$1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+... = \frac{1}{1-\frac{1}{z}}$

only works for z>1. While:

$\frac{1}{1-z} = 1+z+z^2+z^3+...$

only works for 0<z<1.

Edit: Sorry /spoiler doesn't work with /tex.

JuanTutors · 2007-07-22, 15:40

I've never seen this proof. Very unique.

mfgoode · 2007-07-24, 10:07

Quote:

Originally Posted by ATH

Place where it fails:

$1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+... = \frac{1}{1-\frac{1}{z}}$

only works for z>1. While:

$\frac{1}{1-z} = 1+z+z^2+z^3+...$

only works for 0<z<1.

Edit: Sorry /spoiler doesn't work with /tex.

Hats off to you ATH. You put the last nail to the coffin.

A good paradox jinydu. Shows you are thinking, and better still , original !

Mally

2007-06-01, 06:49	#1
jinydu Dec 2003 Hopefully Near M48 2×3×293 Posts	Another "Proof" that 0 = 1 I thought of this today. It relies on divergent series. Enjoy $\Large{1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+...}$ $\Large{=\frac{1}{1-\frac{1}{z}}}$ $\Large{=\frac{z}{z-1}}$ $\Large{=-z\frac{1}{1-z}}$ $\Large{=-z(1+z+z^2+z^3+...)}$ $\Large{=-z-z^2-z^3-\ ...}$ Adding the negative of the right-hand side to both sides gives: $\Large{...+\frac{1}{z^3}+\frac{1}{z^2}+\frac{1}{z}+1+z+z^2+z^3+...=0}$ Regarding the right-hand side as a Laurent series with all coefficients equal to zero, we can equate corresponding coefficients to get 1 = 0 (infinitely many times). As a bonus, setting z = 1 gives $\infty=0$ . Last fiddled with by jinydu on 2007-06-01 at 06:51

2007-06-01, 11:18	#3
ATH Einyen Dec 2003 Denmark 3⁵×13 Posts	Place where it fails: $1+\frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^3}+... = \frac{1}{1-\frac{1}{z}}$ only works for z>1. While: $\frac{1}{1-z} = 1+z+z^2+z^3+...$ only works for 0<z<1. Edit: Sorry /spoiler doesn't work with /tex. Last fiddled with by ATH on 2007-06-01 at 11:20

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2007-06-01, 10:44	#2
Orgasmic Troll Cranksta Rap Ayatollah Jul 2003 641 Posts	cute

2007-07-22, 15:40	#4
JuanTutors Mar 2004 7²×11 Posts	I've never seen this proof. Very unique.