[QUOTE=Lilyo;43876704]Why is everyone trying to use advance formulas to prove this wrong. There will always be enough space between the circle and the squares in order for the squares to add up to 4, regardless of how many squares you cut it down to. The length of each square will decrease as the # of squares increase, keeping the permitter at 4. For the circle, if D=1 then C=3.14159...[/QUOTE]
It's not a proof. He just (attempted to show) that pi = 4 when d = 1. Even if it was correct, can he do it for d = 2? d = n? d = any number, rational, irrational, imaginary? What about if we were determining pi without even considering circles?
That's why proofs are important - they're absolutely true for any case.
[QUOTE=JohnnyMo1;43884090]I am moving this to fast threads because it's pretty much just a question and it fits better there. You were already banned for putting it in debate. Next time, ask a mod where it ought to go before you remake it if you're unsure.
Okay, the problems here are basically:
1) A picture is not a proof. "It looks like the circle" doesn't mean the shape becomes a circle as you add more lengths.
2) You are using the wrong metric.
The first point is self-explanatory. For the second point, by restricting to horizontal and vertical lines like that, you are essentially [I]no longer in standard Euclidean space[/I]. You are in the plane with the "taxicab metric" or "Manhattan metric," where distances are measured by the sum of the differences in both coordinates of the point. Notice that in normal polygonal approximations of the circle, you've still got diagonal lines going on, because they're allowed, and you're simply taking a limit as the number of sides increases. That's why these approximations are valid, they're still in the Euclidean plane.[/QUOTE]
this is a type of response i was waiting for. I know nothing about taxicab metric, but it sounds like your saying 'because i'm restricting the shape of the pseudo circle to being made of only orthogonal lines, i'm not in standard euclidean space, because euclidean space allows for lines to be non orthogonal, they can be diagonal'. is this what you are saying?
The corners tend towards 0, but will never actually reach 0
The corners of the square will always have length, they'll each just be too small to measure
If you could infinitely zoom in you would always be able to see the corners looking like they do in the 4th image
[QUOTE=noh_mercy;43884494]this is a type of response i was waiting for. I know nothing about taxicab metric, but it sounds like your saying 'because i'm restricting the shape of the pseudo circle to being made of only orthogonal lines, i'm not in standard euclidean space, because euclidean space allows for lines to be non orthogonal, they can be diagonal'. is this what you are saying?[/QUOTE]
Yes, pretty much just that.
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