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September 15th, 2019, 06:46 AM  #1 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  Easy proof of 22/7>pi
Let $\displaystyle x=22/7\pi\;$ and suppose $\displaystyle 22/7<\pi$ or $\displaystyle x<0$. Since x is negative then $\displaystyle sgn(x)=1+2H(x)<0$. $\displaystyle sgn(x)=1+2\lim_{s\rightarrow \infty} \frac{1}{1+e^{sx}}=1+2\lim_{t_s \rightarrow \infty} \frac{1}{1+e^{t_s x}}$. Since the limit of $\displaystyle 1+e^{sx}$ converges then : $\displaystyle sgn(x)=1+2=1$ which is a contradiction of statement :$\displaystyle x<0 \: \equiv \: sgn(x)<0$ . $\displaystyle sgn(x)>0 $ or $\displaystyle x>0$ proves $\displaystyle \frac{22}{7}>\pi$. Last edited by idontknow; September 15th, 2019 at 06:56 AM. 
September 15th, 2019, 07:20 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,035 Thanks: 2271 
You incorrectly assumed that $\displaystyle \lim_{s\rightarrow \infty} \frac{1}{1+e^{sx}} = \lim_{t_s \rightarrow \infty} \frac{1}{1+e^{t_s x}}$.

September 16th, 2019, 05:19 PM  #3 
Senior Member Joined: Jan 2014 From: The backwoods of Northern Ontario Posts: 393 Thanks: 71 
I tend to use simple arithmetic whenever possible for my "proofs." I happen to have memorized the first eight digits of pi: 3.1415926 Then I took my trusty calculator to find the approximate value of 22/7 I got 3.142857143 
September 17th, 2019, 12:29 AM  #4  
Senior Member Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 132 Thanks: 49 Math Focus: Area of Circle  Quote:
 
September 30th, 2019, 02:33 AM  #5 
Senior Member Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98  We are supposing that $\displaystyle x<0$ , then $\displaystyle e^{sx}$ becomes $\displaystyle e^{tx}\;$, $\displaystyle t>0$.


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22 or 7>pi, 22 or 7>pi, easy, proof 
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