Pi in fraction form

[folmerveeman]

When messing around with my calculator today (ti-84) I found out that the seventh power of Pi is actually a perfect fraction, I don't have my calculator at my house right now so I can't give it, but you can do it on your own. It's something like 22 million divided by 7030

Is this something major? Or just due to the Ti-84's limited significancy?

[heyitsguay]

That's just an artifact of the calculator's finite number of decimal places, it's not really a fraction. Pi is a transcendental number, so no equation involving powers and sums of pi will ever yield a rational number.

[folmerveeman]

Oh ok, too bad! I thought it was pretty cool regardless

So is that fraction the way the calculator defines Pi then?

[heyitsguay]

Well, I don't know for sure, but it seems unlikely. Why store pi as the seventh root of some nasty fraction when you can store it as the finite decimal sequence 3.1415926535898 or whatever? Then, it can just be looked up when needed - computing seventh roots is costly, computationally.

[Meteoric]

More likely, your calculator stores pi out to however many digits, and just calculated pi^7, got 3020.29323 or some such, and then when you asked it to convert that to a fraction, it turned that into 302029323/100000 and reduced. Depending on your calculator and how many decimal places it uses, it might get a slightly different decimal which would thus reduce to a different fraction, but the idea is the same.

[gmalivuk]

Except that fraction is clearly already in lowest terms, since the numerator isn't divisible by 2 or 5.

More likely what it does is find the fraction with the smallest denominator within some narrow range around the number it's given. For example, 21362534/7073 is what I get within 10^-8 of pi^7. This is close enough to "something like 22 million divided by 7030" for me to suspect it is the actual result the first poster's calculator gave. (Well, on second thought, I realize that I'm doing this with a program that uses pi to arbitrary precision, which most handheld calculators definitely don't do. So it's possible the calculator in question did give something else, but this fraction is a much much better approximation given the size of its denominator.)

Edit: Incidentally, if you were going to store pi as some function of a rational number, and weren't going to use the ones from the continued fraction expansion of pi itself, using 2143/22 for pi^4 is quite nice. It is within 1.25x10^-7 of the exact value, despite having such a low denominator.

[Proginoskes]

Pi can be expressed in fractional form, though: {\pi \over 1}

This reminds me of the story about Ramanujan, who once dreamt that a goddess visited him and told him that \pi = \root 4 \of {2143 \over 22}. (I guess goddesses don't know everything.)

[tomandlu]

I realise that there's no way to ask this question without it seeming really stupid, but I'll give it a go anyway... is there some intrinsic quality of a circle that dictates that Pi must be an irrational number or is it just 'bad luck'? I can't think of a way to phrase that without it looking bat-shit insane, so the answer may be 'unask the question'... perhaps a better way of saying it would be to say "what is the proof that pi is irrational'... (for laymen).

[vilidice]

The best way for a layman to approach it is probably with triangles (really most of the circle stuff "comes from" trig anyway):

but using trig Lambert's proof is probably the easiest to work through:
http://en.wikipedia.org/wiki/Proof_that ... irrational

a more extensive discussion is found here:
http://paramanands.wordpress.com/2011/0 ... rts-proof/

[Talith]

I wonder if there's some nice contradiction that you can come to if you assume that {e^(nx)}_n>0 is not a dense subset of the circle (actually, if pi were rational, this set would be finite, not just non-dense). The contradiction would then imply that pi is irrational.

[tomandlu]

The best way for a layman to approach it is probably with triangles (really most of the circle stuff "comes from" trig anyway):

but using trig Lambert's proof is probably the easiest to work through:
http://en.wikipedia.org/wiki/Proof_that ... irrational

a more extensive discussion is found here:
http://paramanands.wordpress.com/2011/0 ... rts-proof/

Many thanks - actually, I came back to post, after a search, that the answer would appear to be "yes, there is, but no, you won't understand it."

Nevertheless I'll take another look using the links you posted. Brains exploding due to sub-standard construction at ten.

[SunAvatar]

I realise that there's no way to ask this question without it seeming really stupid, but I'll give it a go anyway... is there some intrinsic quality of a circle that dictates that Pi must be an irrational number or is it just 'bad luck'?

It's not a stupid question at all. There is proof that π is irrational (in fact several proofs are on Wikipedia), but this is incidental to what you are really asking. Ultimately what it comes to is this: almost all real numbers are irrational, in the sense that if you randomly plop down an infinite string of decimal digits, one after another, you will almost certainly not get a rational number. Since there's no particular reason that π must be rational, it would be a coincidence bordering on the miraculous if it were. It turns out that no such miracle occurs, and so π is irrational.

[Proginoskes]

I realise that there's no way to ask this question without it seeming really stupid, but I'll give it a go anyway... is there some intrinsic quality of a circle that dictates that Pi must be an irrational number or is it just 'bad luck'?

It's not a stupid question at all. There is proof that π is irrational (in fact several proofs are on Wikipedia), but this is incidental to what you are really asking. Ultimately what it comes to is this: almost all real numbers are irrational, in the sense that if you randomly plop down an infinite string of decimal digits, one after another, you will almost certainly not get a rational number. Since there's no particular reason that π must be rational, it would be a coincidence bordering on the miraculous if it were. It turns out that no such miracle occurs, and so π is irrational.

Is it coincidence that there are Pythagorean triples?

[gmalivuk]

No more than it's a coincidence that there are rational numbers. And similarly, a random set of three integers is highly unlikely to be a Pythagorean triple. (In the sense that, as n goes to infinity, the fraction of integer points in the box [-n,n]³ that are Pythagorean triples goes to zero.)

[ConMan]

Or to put it another way, if you're working with a particular type of construction in mind, don't be surprised that such a construction exists or that you can construct examples of that construction, but don't be fooled into thinking that that construction would be likely to come up if you were working with a broader group of numbers. So if you're doing standard geometry, then you're probably going to run into a bunch of nice integers (0, 1, things that are highly divisible, etc) and algebraic numbers (square roots, phi, etc) and a small subset of transcendental numbers (particularly ones closely related to pi) but in the grander scheme of things such numbers form a vast minority.

[PM 2Ring]

The Wallis product isn't a proof, per se, but it is suggestive that pi is irrational since it appears that it is not equal to any fraction with finite numerator and denominator, and it is a rather neat expression for pi, IMHO. To use the Wallis product to actually prove that pi's irrational you'd need to prove that it's equal to pi and that it can't be reduced to a fraction with finite numerator and denominator; I suspect that to do so would end up being equivalent to one of the more direct proofs already referenced in this thread.

[Proginoskes]

No more than it's a coincidence that there are rational numbers. And similarly, a random set of three integers is highly unlikely to be a Pythagorean triple. (In the sense that, as n goes to infinity, the fraction of integer points in the box [-n,n]³ that are Pythagorean triples goes to zero.)

Actually, it goes deeper than that. There are solutions to a^2 + b^2 = c^2 in positive integers, but no solutions to
a^3 + b^3 = c^3 in positive integers. It's not the existence of rational numbers; it's the fact that three of them "conspire" to make an equation like
a^2 + b^2 = c^2 possible.

[PM 2Ring]

Actually, it goes deeper than that. There are solutions to a^2 + b^2 = c^2 in positive integers, but no solutions to
a^3 + b^3 = c^3 in positive integers. It's not the existence of rational numbers; it's the fact that three of them "conspire" to make an equation like
a^2 + b^2 = c^2 possible.

As is well-known, Pythagorean triplets can easily be generated by:
a = u² - v²
b = 2uv
c = u² + v²
(FWIW, this relationship was known in ancient times). So is it a conspiracy between a, b and c, or should we say that the Pythagorean triplet is merely a simple consequence of an even deeper conspiracy between u and v?

We can go even deeper: Let z = u + iv, then z² = a + bi and c = |z²|. Thus the existence of Pythagorean triplets can be seen as a simple consequence of the multiplicative properties of the complex numbers (or more specifically, the Gaussian integers). I find it intriguing that to really understand the deeper properties of the integers you need to look at complex numbers.