Let's look at a unit circle and a square, whose sides are tangent to the circle.

The perimeter of the circle is \(2\pi\) and the perimeter of the square is 8.

Now add 4 more squares inside the gaps between the square and the circle.now the perimeter of the new figure doesn't change.But we can add infinitely many squares, so the perimeter of the figure will approach the perimeter of the circle.

So \(2\pi=8\) or \(\pi=4\)

Actually, the above is a false statement.The limiting perimeter doesn't actually tend to the circumference.

No, it doesn't.Imagine any rectangle and put a smaller rectangle in it such that two of its sides touch the sides of the bigger rectangle.Now the perimeter of the figure, constructed by two sides of the bigger one, two sides of the smaller one and two others is the same as the one of the bigger rectangle.We can apply that philosophy infinitely many times in the pseudo-proof.

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TopNewestThe well-known 'proof':

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Let's look at a unit circle and a square, whose sides are tangent to the circle.

The perimeter of the circle is \(2\pi\) and the perimeter of the square is 8.

Now add 4 more squares inside the gaps between the square and the circle.now the perimeter of the new figure doesn't change.But we can add infinitely many squares, so the perimeter of the figure will approach the perimeter of the circle.

So \(2\pi=8\) or \(\pi=4\)

Actually, the above is a false statement.The limiting perimeter doesn't actually tend to the circumference.

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How does the perimeter of the new figure not change? I'm pretty sure it does.

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No, it doesn't.Imagine any rectangle and put a smaller rectangle in it such that two of its sides touch the sides of the bigger rectangle.Now the perimeter of the figure, constructed by two sides of the bigger one, two sides of the smaller one and two others is the same as the one of the bigger rectangle.We can apply that philosophy infinitely many times in the pseudo-proof.

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Can a circle have a right angle???

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I have a problem with this as the main idea :) A proof with a mistake ( or not ) Part 2

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