pi

What of the case when r = dr ...

I have A/4 ... if A = 0 ... then we have 0/4 ... which is plain ol' zero ... it's completely redundant multiplying the whole affair by r/r = 1 [where r≠0] ... it's the unnecessary steps you're adding that cause these problems ...

I don't see your problem. What is more easy than the little PC-program I made? The relation is not A/4 or 0/4 it is π/4. The only "mistake" I made in this process is to say "0 is a delusion" because the machine is not able to represent an infinite number. The advantage: The PC will not freeze.
 
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I don't see your problem. What is more easy than the little PC-program I made? The relation is not A/4 or =/4 it is π/4.

Read the OP again ... this is a unit circle ... r = 1 unit ... one squared is one doesn't change the fact we multiply π by radius squared ... we just don't have to write the one in all the time ...

1π/4 ... make sense now? ... maybe ((1 unit)^2)π/4 = π/4 square units ... it's still all means area/4 is equal to π/4 unit^2 ... we can divide by 4 ...
 
Read the OP again

Why?

... this is a unit circle ... r = 1 unit ...

So r<>0. What's your problem?

one squared is one doesn't change the fact we multiply π by radius squared ... we just don't have to write the one in all the time ...

1π/4 ... make sense now? ... maybe ((1 unit)^2)π/4 = π/4 square units ... it's still all means area/4 is equal to π/4 unit^2 ... we can divide by 4 ...

In all cases always (except r=0) is the relation π/4 between the area of a circle with radius r and a square with the side 2r.

Simple math:

(π*r^2) / ((2*r)^2) = π/4

This is also the case for every constant c<>0 which you use in this way:

((π*r^2)/c) / ((2*r)^2)/c) = π/4.

Again: π/4 is no area, no surface - it is "only" a number, a relation, a ratio. In all cases (except r=0). It is a constant for all r's (except 0) - also for r=1 which is only a special case. The universality of this use is the real important thing in case of such a mathematical structure.
 
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I like to think of pi as a square divided into halves creating 4 equal units, then pi/4 is the area within each unit that the circle occupies. is this correct, novel, new math?
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Your interpretation of π (pi) and its relationship with the area of a circle is an interesting way to visualize the concept. Let’s break it down:

Understanding the Concept​

  1. Circle and Square:
    • If you inscribe a circle within a square, the area of the square can be expressed as s2s^2s2, where sss is the side length of the square.
    • The radius rrr of the inscribed circle relates to the side length of the square. For a square with a side length of sss, the radius of the circle is r=s2r = \frac{s}{2}r=2s.
  2. Area of the Circle:
    • The area of the circle is given by the formula A=Ï€r2A = Ï€r^2A=Ï€r2. If the square has a side length of sss, then:
      A=π(s2)2=πs24A = π\left(\frac{s}{2}\right)^2 = \frac{πs^2}{4}A=π(2s)2=4πs2
  3. Relating to the Square's Area:
    • Since the area of the square is s2s^2s2, the area of the circle as a fraction of the square's area can be expressed as:
      AArea of square=Ï€s24s2=Ï€4\frac{A}{\text{Area of square}} = \frac{\frac{Ï€s^2}{4}}{s^2} = \frac{Ï€}{4}Area of squareA=s24Ï€s2=4Ï€
    • This shows that the area of the circle occupies Ï€4\frac{Ï€}{4}4Ï€ of the area of the square.

Novelty and New Math​

  • Your approach of conceptualizing Ï€ in this way is not a standard mathematical interpretation, but it does provide a unique perspective on the relationship between the circle and the square.
  • While the relationship itself is well-established, your method of visualizing it might be considered a novel way to help others understand the concept of area in relation to Ï€.

Summary​

  • Yes, your visualization is correct in that it describes the relationship between the area of a circle and the area of a square, showing that the area of the circle is Ï€4\frac{Ï€}{4}4Ï€ of the area of the square.
  • While the concept itself isn't new, your interpretation can be a helpful teaching tool for understanding these geometric relationships. If you want to explore this further or develop it into a more formal mathematical exploration, it could lead to interesting discussions!
To the extent that I think of Pi R squared is in the context of pie are round; cornbread are square. Sorry. 🤨 :scared1:
 
Why?



So r<>0. What's your problem?



In all cases always (except r=0) is the relation π/4 between the area of a circle with radius r and a square with the side 2r.

Simple math:

(π*r^2) / ((2*r)^2) = π/4

This is also the case for every constant c<>0 which you use in this way:

((π*r^2)/c) / ((2*r)^2)/c) = π/4.

Again: π/4 is no area, no surface - it is "only" a number, a relation, a ratio. In all cases (except r=0). It is a constant for all r's (except 0) - also for r=1 which is only a special case. The universality of this use is the real important thing in case of such a mathematical structure.

I'll ask for the last time ... how do you resolve the case of r = dr? ... you don't answer this simple question, must mean you don't understand ... do you know what an epsilon/delta proof is? ...
 
I'll ask for the last time ... how do you resolve the case of r = dr? ...

I told you what I do with my very litte PC-program. I approximate nothing. I replace just simple 0 with 1/oo and say "For a calculator exists no 0". So π/4 is always the result and I have to calculate nothing.

you don't answer this simple question, must mean you don't understand ... do you know what an epsilon/delta proof is? ...

I don't need it. Why should I take a bus if I'm faster with a bike?

By the way: In the beginning I liked only to say "So the expression '... pi/4 is the area ...' is wrong."

Again: π/4 is a relation, a ratio. It is only a number without dimensioning, without measured value, without [name of a] length like inch for example or yard or meter or lightyear or planklength.
 
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I told you what I do with my very litte PC-program. I approximate nothing. I replace just simple 0 with 1/oo and say "For a calculator exists no 0". So π/4 is always the result and I have to calculate nothing.



I don't need it. Why should I take a bus if I'm faster with a bike?

Okay ... if the epsilon/delta proof shows you're wrong, you definitely don't want to discuss it ... we divide by dr all the time, and still get π/4, so it's not an approximation, it's exact ... the article you just read didn't make that clear? ...

When you're old enough, you can drive by car and get there even faster ... (i.e. more advanced mathematics) ...

ETA: Yes ... I'm being indiscreet ...so sue me ...
 

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