MHB Value of Irrational Number π (Part 2)

AI Thread Summary
The value of the irrational number π is precisely 3.1415926535 when rounded to ten decimal places. The approximation 22/7 is commonly used for π, and historical context reveals that Archimedes established bounds for π, while Boethius introduced this approximation to the western world. The discussion explores how to determine the decimal agreement between 22/7 and π, with suggestions for using calculators or long division. There is also interest in finding the best approximation of π using fractions with numerators and denominators less than 100. The conversation emphasizes the challenge of achieving this without a calculator, prompting ideas for programming solutions.
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The value of irrational number π, correct to ten decimal places (without rounding), is 3.1415926535. By using your calculator, determine to how many decimal places the following quantity (22/7) agrees with π.

Extra notes from textbook:

Archimedes (287-212 B.C.) showed that
(223/71) < π < 22/7. The use of the approximation (22/7) for π was introduced to the western world through the writings of Boethius (ca 480-520), a Roman philosopher, mathematician, and statesman. Among all fractions with numerators and denominators less than 100, the fraction (22/7) is the best appriximation to π. Do you agree?

I was wondering if this question can be answered without a calculator. Can we show that (22/7) in terms of decimal places agrees with pi?
 
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Comparing $22/7$ with $$\pi$$ is simple if you have a calculator or use long division. Choosing the best approximation with denominators less than 100 is trickier. I would write a program for this.
 
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Evgeny.Makarov said:
Comparing $22/7$ with $$\pi$$ is simple if you have a calculator or use along division. Choosing the best approximation with denominators less than 100 is trickier. I would write a program for this.

How it is done without using a calculator?
 
RTCNTC said:
How it is done without using a calculator?

Evgeny.Makarov said:
use long division.
...
 
Evgeny.Makarov said:
...

The original question has been edited.
 

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