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Can pi be reduced to a rational number?


Dennisg
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What's the basis for claiming that pi is 3 at plank length?

 

What????

 

He thinks that the world is made up of planck-length-sized balls, and furthermore that mathematical entities are somehow dependent on our ability to physically make something that shape. From there, I get the impression he's thinking of arranging seven of those balls in tightest packing, i.e. a hexagon, which would have a diameter of two planck lengths, and a circumference (measured from the center of each ball) of six, making a ratio of three to one. Since, in that universe, this would be the most circle-like thing you could physically construct at that scale, he's decided it is a circle, therefore making "pi" equal to three.

 

He seems pretty adamant about the whole thing, too.

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What's the basis for claiming that pi is 3 at plank length?

 

A circle with a diameter = 2plank lengths can have only 6 points that define its circumance. A straight line drawn through any of these points (tangent) will only intersect one point – therefore the line between any two points is curved. One can then can calculate the area of the “circle” by calculating the area of the 6 “triangles” and then calculate pi = A/r squared.

 

He seems pretty adamant about the whole thing, too.

 

not really - just pushing things.

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RE:

 

therefore the line between any two points is curved.

 

and

 

calculating the area of the 6 “triangles”

 

How can these two statements be compatible? If the "line" between any two points is curved, how can this be a triangle? A triangle has straight sides. By definition. So, either the sides aren't curved or they aren't triangles.

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Look, [math]\pi=3.1415927...[/math]. However, given the constraint that the smallest length possible is the Planck Length, then a significantly small circle may have a circumference-to-diameter of nearly 3.0. Pi is defined as that ratio in constraint-less space; thus, even though you can theoretically construct a "circle" whose ratio is nearly 3.0, this does not change the value of the constant, pi.

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You cannot construct a circle whose ratio is nearly 3.0, not even theoretically. For example, I can construct a circle whose diameter is the plank length, and the ratio will be pi.

 

Or rather, since a circle can be ANY arbitrary size, I can theoretically construct a circle whose diameter is 10^-1000 m, and the ratio will still be pi.

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Look, [math]\pi=3.1415927...[/math]. However, given the constraint that the smallest length possible is the Planck Length, then a significantly small circle may have a circumference-to-diameter of nearly 3.0. Pi is defined as that ratio in constraint-less space; thus, even though you can theoretically construct a "circle" whose ratio is nearly 3.0, this does not change the value of the constant, pi.

 

Given those constraints, it might have a ratio of 3.0, but it wouldn't be a circle. It would be a hexagon. And even a hexagon has incommensurable ratios in it, so the whole exercise is pretty much a reductio ad absurdum.

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Given those constraints, it might have a ratio of 3.0, but it wouldn't be a circle. It would be a hexagon. And even a hexagon has incommensurable ratios in it, so the whole exercise is pretty much a reductio ad absurdum
.

 

 

Because of uncertainty at the plank scale it definitely couldn't be defined as a hexagon - a proto circle would be more apt. What this shows is that pi in not one mumber but depends an scale. At our scale the differences are so small the we cannot measure them.

 

Incidentally, does Dennisg believe that also becomes a rational number for really tiny triangles?

 

Hey don't rock the boat - you might fall off.:eyebrow:

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Given those constraints, it might have a ratio of 3.0, but it wouldn't be a circle.

 

Exactly. It wouldn't be a circle. I think we can all put this to rest now.

 

.

What this shows is that pi in not one mumber but depends an scale.

 

No, it does not!

Edited by I_Pwn_Crackpots
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Please back this assertion with some evidence.

 

Ok, pi is defined as the ratio of a circle's circumference to its diameter in Euclidian geometry. Euclidian geometry is not dependent on scale. Of course, four or five people have already said that, so I don't know what else you could possibly be looking for.

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Ok, pi is defined as the ratio of a circle's circumference to its diameter in Euclidian geometry. Euclidian geometry is not dependent on scale. Of course, four or five people have already said that, so I don't know what else you could possibly be looking for.

 

This answer is based on the assumption that there is no limit to "smallness" Since the plank length is that limit then Euclidian geometry needs to be updated.

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The circumference will decrease at relativistic speeds.

 

Say this circle is a merry-go-round. You are on the edge, spinning at relativistic speed, and you are trying to measure the circumference with a ruler (the circle is large enough that this isn't completely foolish). The ruler would contract, and therefore more ruler lengths would fit around the outside. You would measure the ratio to be larger than pi.

 

Maybe this is a question of reference frames, though. In a rest frame, the circumference would contract. No such specification was made in the original post, however.

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Guys, I would like to remind you that this is a mathematics, not relativity, forum. Bearing this in mind, the answer to your question is: no, [imath]\pi[/imath] is irrational because it is a mathematical definition. You can debate 'relativistic circles' or whatever until the cows come home, but that is not mathematics.

 

Unless someone can convince me to keep this thread open, I will close it later on this evening.

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dave, you should know that relativity is a highly mathematical theory and it does bear some relationship to the question at hand.

 

Yeah, but so are all the other physical theories. But ultimately they are all imperfect models of the universe. So I'm going to disagree with you here.

 

The circumference will decrease at relativistic speeds.

 

And so will the radius. As such, pi is always irrational.

 

Bearing this in mind, the answer to your question is: no, [imath]\pi[/imath] is irrational [b]because it is a mathematical definition[/b].

 

Quoted for emphasis.

Edited by I_Pwn_Crackpots
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