July 3rd, 2017, 09:24 PM  #1 
Banned Camp Joined: Jul 2010 Posts: 118 Thanks: 0  Physmath
Square circumference.......c Square diagonal............... d Circle circumference .......... c Circle diameter .................. d Subject  Squares: c 1 = 3 mm , c 2 = 137 mm Prove that: c 1 \ c 2 = d 1 \ d 2 Subject  Circles: c 1 = 3 mm , c 2 = 137 mm Prove that: c 1 \ c 2 = d 1 \ d 2 Successfully 
July 3rd, 2017, 11:24 PM  #2 
Member Joined: Feb 2017 From: Canada Posts: 31 Thanks: 1 
Look cool in my head maybe will put it on paper

July 4th, 2017, 01:29 AM  #3 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,068 Thanks: 692 Math Focus: Physics, mathematical modelling, numerical and computational solutions 
So your problem statement is basically this: If the circumferences of a square and circle are defined as $\displaystyle c_1$ and $\displaystyle c_2$, the diagonal of a square is defined as $\displaystyle d_1$ and the diameter of a circle is defined as $\displaystyle d_2$, then $\displaystyle \frac{c_1}{c_2} = \frac{d_1}{d_2}$ This can be disproved by counterexample. Select $\displaystyle d_1$ and $\displaystyle d_2$ to be a value $\displaystyle a$. We should therefore expect the ratio $\displaystyle \frac{c_1}{c_2}$ to be 1. However: $\displaystyle c_1 = 4\frac{d_1}{\sqrt{2}} = 2 \sqrt{2} a$ $\displaystyle c_2 = \pi d_2 = \pi a$ Therefore the ratio $\displaystyle \frac{c_1}{c_2} = \frac{2 \sqrt{2} a}{ \pi a} = \frac{2 \sqrt{2}}{\pi} \approx 0.900316$, which is not equal to 1. 
July 4th, 2017, 03:02 AM  #4 
Banned Camp Joined: Jul 2010 Posts: 118 Thanks: 0  These are separate proofs
This is separate proof. Square circumference.......c Square diagonal............... d Subject  Squares: c 1 = 3 mm , c 2 = 137 mm Prove that: c 1 \ c 2 = d 1 \ d 2 and this is separate proof Circle circumference .......... c Circle diameter .................. d Subject  Circles: c 1 = 3 mm , c 2 = 137 mm Prove that: c 1 \ c 2 = d 1 \ d 2 Thanks 
July 4th, 2017, 07:13 AM  #5 
Banned Camp Joined: Jul 2010 Posts: 118 Thanks: 0  The proof of the squares
Square circumference.......c Square diagonal............... d Subject  Squares: c 1 = 3 mm , c 2 = 137 mm Prove that: c 1 : c 2 = d 1 : d 2 c 1 = 3 , d 1 = root of 2 *( 3 : 4 ) c 2 = 137 , d 2 = root of 2 *( 137 : 4 ) c 1 : c 2 = 3 : 137 d 1 : d 2 = root of 2 *( 3 : 4 ) : root of 2 *(137 : 4) = 3 : 137 Now remains to present the proof of circles Circle circumference.......c Circle diameter............... d Subject  Circles: c 1 = 3 mm , c 2 = 137 mm Prove that: c 1 : c 2 = d 1 : d 2 
July 4th, 2017, 07:28 AM  #6 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,972 Thanks: 2295 Math Focus: Mainly analysis and algebra 
Euclid and Archimedes each proved both of those statements.

July 4th, 2017, 07:40 AM  #7 
Banned Camp Joined: Jul 2010 Posts: 118 Thanks: 0  Euclid proved about a squares
Archimedes did not prove about circles.

July 4th, 2017, 07:57 AM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,972 Thanks: 2295 Math Focus: Mainly analysis and algebra 
I understand that he did, via similar triangles and limits. Euclid certainly proved it anyway.

July 4th, 2017, 08:14 AM  #9 
Banned Camp Joined: Jul 2010 Posts: 118 Thanks: 0  Right, Euclid proved about squares
but Archimedes did not prove, about circles The proof of circles c 1 : c 2 = d 1 : d 2 , simply does not exist. 
July 4th, 2017, 08:46 AM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,972 Thanks: 2295 Math Focus: Mainly analysis and algebra 
Yes it does. I linked an example in one of your threads the other day. I'm not surprised that you don't accept it because you refuse to accept $\pi$ as a constant, but that's your problem not anyone else's.
