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 June 7th, 2013, 01:19 AM #1 Member   Joined: May 2013 From: Phnom Penh, Cambodia Posts: 77 Thanks: 0 impossible (prove that 0 = 1) Can you solve that 0=1 by using at least two different methods?
 June 7th, 2013, 02:18 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: imposible 0*0 = 0*1 Divide by 0. Gives 0 = 1. Like this, many ways can be found.
 June 7th, 2013, 08:37 AM #3 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Re: imposible $\begin{array}{ccc} \text{We have: } && +1 &=& (-1)(-1) \\ \\ \\ \\ \text{Take square roots:} && \sqrt{+1} &=& \sqrt{(-1)(-1)} \\ \\ \\ \\ && 1 &=& \sqrt{-1}\,\cdot\,\sqrt{-1} \\ \\ \\ \\ && 1 &=& i\,\cdot\,i \\ \\ \\ \\ && 1 &=& -1 \\ \\ \\ \\ \text{Add 1:} && 2 &=& 0 \\ \\ \\ \\ \text{Divide by 2:} && 1 &=& 0 \end{array}$
 June 7th, 2013, 11:32 AM #4 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,205 Thanks: 901 Math Focus: Wibbly wobbly timey-wimey stuff. Re: impossible (prove that 0 = 1) Don'tcha just hate explaining why that last one is wrong? -Dan
 June 7th, 2013, 03:25 PM #5 Member   Joined: May 2013 From: Phnom Penh, Cambodia Posts: 77 Thanks: 0 Re: impossible (prove that 0 = 1) So interesting with the solution of two of you. I have just found another way to show that 0=1. lim(x->0) of (2013/x +1) =lim(x->0) of (2013/x) +1= indefinite +1=indefinite Then, simplify indefinite we obtain 1=0.
 June 7th, 2013, 09:11 PM #6 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: impossible (prove that 0 = 1) Suppose it is not clear what sqrt(1) is, consider... $x^2 \= \ 1$ $sqrt{x^2} \= \ \sqrt{1}$ $|x| \= \ sqrt{1}$ $x \= \ \pm \sqrt{1}$ So even if $sqrt{1}$ is defined as +1 , you still get $x \= \ \pm 1$
 June 9th, 2013, 02:24 AM #7 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: impossible (prove that 0 = 1) 0 = 1 In the trivial ring. http://mathworld.wolfram.com/TrivialRing.html http://en.m.wikipedia.org/wiki/Trivial_ring
June 9th, 2013, 04:28 AM   #8
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Joined: Mar 2012
From: India, West Bengal

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Math Focus: Number Theory
Re: impossible (prove that 0 = 1)

Quote:
 Originally Posted by agentredlum 0 = 1 In the trivial ring.
Best answer of all, you rule

 June 9th, 2013, 05:18 AM #9 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Re: impossible (prove that 0 = 1) $\text{Maclaurin series: }\:\ln(1\,+\,x) \;=\;x\,-\,\frac{x}{2}\,+\,\frac{x}{3}\,-\,\frac{x}{4}\,+\,\frac{x}{5}\,-\,\frac{x}{6}\,+\,\cdots$ $\text{Let }x \,=\,1:$ $\ln2 \;=\;1\,-\,\frac{1}{2}\,+\,\frac{1}{3}\,-\,\frac{1}{4}\,+\,\frac{1}{5}\,-\,\frac{1}{6}\,+\,\cdots$ $\ln2 \;=\;\left(1\,+\,\frac{1}{3}\,+\,\frac{1}{5}\,+\,\ cdots\right)\:-\:\left(\frac{1}{2}\,+\,\frac{1}{4}\,+\,\frac{1}{6 }\,+\,\cdots\right)$ $\text{Add and subtract }\frac{1}{2}\,+\,\frac{1}{4}\,+\,\frac{1}{6}\,+\,\ cdots$ $\ln2 \;=\;\underbrace{\left(1\,+\,\frac{1}{3}\,+\,\frac {1}{5}\,+\,\cdots\right)\;+\;\left(\frac{1}{2}\,+\ ,\frac{1}{4}\,+\,\frac{1}{6}\,+\,\cdots\right)}\;-\;\underbrace{\left(\frac{1}{2}\,+\,\frac{1}{4}\,+ \,\frac{1}{6}\,+\,\cdots\right) \;-\;\left(\frac{1}{2}\,+\,\frac{1}{4}\,+\,\frac{1}{6 }\,+\,\cdots\right)}$ $\ln2 \;=\;\;\;\;\;\;\;\;\;\;\; \left(1\,+\,\frac{1}{2}\,+\,\frac{1}{3}\,+\,\frac{ 1}{4}\,+\,\cdots\right) \;\;\;\;\;\;\;\;\;\;\;\;\; - \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; 2\left(\frac{1}{2}\,+\,\frac{1}{4}\,+\,\frac{1}{6} \,+\,\cdots \right)$ $\ln2 \;=\;\left(1\,+\,\frac{1}{2}\,+\,\frac{1}{3}\,+\,\ frac{1}{4}\,+\,\cdots\right)\;-\;\left(1\,+\,\frac{1}{2}\,+\,\frac{1}{3}\,+\,\fra c{1}{4}\,+\,\cdots\right)$ $\ln2 \;=\;0$ $\text{Therefore: }\:e^0 \,=\,2 \;\;\;\Rightarrow\;\;\;1 \,=\,2 \;\;\;\Rightarrow\;\;\; 0 \,=\,1$
 June 9th, 2013, 07:22 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: impossible (prove that 0 = 1) If a hockey game ends 0-0, then overtime is played until 1-0; the loser gets 1 point in the standings, so 0 = 1

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