View Full Version : Pi question

SqrlCub

11-23-1999, 09:48 AM

I have two questions.

1. What two numbers do you divide to get pi? I thought it was something like 22/7 but when I did some calculater calculations it did not come out exactly right.

2. If we used another system of counting (not base 10) could we get pi to be a definite number? I think it would work in theory, but practice is something completely different.

Just some things that I think about.

HUGS!

Sqrl

------------------

Gasoline: As an accompaniement to cereal it made a refreshing change. Glen Baxter

RickG

11-23-1999, 10:09 AM

Pi is defined as the ratio of the circumference to the diameter of a circle. The rational number 22/7 is a useful approximation, as it is within 1% of the actual answer. However, pi is provably irrational (it cannot be represented as the ratio of two whole numbers, in any base system) and not algebraic (it is not the solution of any polynomial equation with rational coefficients). It is therefore a transcendental number, although meditating on it will not cause you to levitate, more's the pity).

Rick

Gilligan

11-23-1999, 11:40 AM

355/113 is another good one.

neuro-trash grrrl

11-23-1999, 11:42 AM

What do you mean, "definite number"? It already is a definite number, we'll just never be able to compute it exactly.

------------------

"That's entertainment!" —Vlad the Impaler

Keeves

11-23-1999, 11:56 AM

From context, I think the word "definite" (in the OP) was used to mean "writable with digits rather than letters, even if in a digit system other than base-ten".

In other words, if we used binary digits (you know, where forty-three is 101011) even on the right side of the "decimal" point, perhaps it would be writable with a repeating set of digits, or maybe even a finite set of digits.

I recall once seeing a formula for pi. It went something like this:

pi = 4/1 - 4/2 + 4/3 - 4/4 + 4/5 - 4/6 + 4/7 - 4/8 + 4/9 - 4/10 .....

Anyone remember what it is?

Liberal

11-23-1999, 01:29 PM

Arnold Winkelried

Is it alleged that, in base 6, .55555... is equal to 1?

------------------

"It is lucky for rulers that men do not think." — Adolf Hitler

Keeves

11-23-1999, 01:43 PM

Is it alleged that, in base 6, .55555... is equal to 1?I started figuring this out, and then I realized that (though you may not have intended it as such), it's a trick question!

The proper way to write one in base six is simply "1". Just like in any other base!

.55555... in base 6 approaches 1, in exactly the same way as .99999... approaches 1 in base 10!

SqrlCub

11-23-1999, 02:05 PM

Hey everyone, thanks for the input so far.

I was thinking that if we counted off in some odd based counting system perhaps base 353 or 22 or some such that Pi could occur naturally as one of the numbers. I know that the above systems are an over exxageration. But couldn't we do a Pi based counting system. We could have 1 = 3.14159247... (or whatever) and have 2 = twice that number. I believe Pi to the 10th digit can build a circle spreading the diameter of our solar system with only the bredth of an atom difference. Is that also true or would that need a significantly different number? When I was in school I always used the 22/7 method for Pi because I felt it would be more accurate, but recently I heard that it is not really the proper number. I understand that the mathemeticians that discovered Pi did something like put a two hexagons (or some other shape with easily measurable straight lines)on either side of the circumference of the circle and averaged their measurements. As people began needing more accurate measurements, they started putting more sides on the "hexagon".

OH Arnold, when you said, 2. An irrational number (one that cannot expressed as a fraction of two integers) is irrational in any base.

e.g.

1 / 3 = 0.2 in base 6...

It kind of answered what I was asking. Two tenths is much simpler than 0.333333333... or even than 1/3; however, I know a Pi based counting system would still be horribly inconvenient to use. It would make counting numbers very unweildy, if not irrational numbers themselves. I was kind of asking is it possible to devise a counting system based on irrational non-repeating numbers that would make enough logical sense to actually use? Or how about would it be possible to make an irrational based repeating or non-repeating counting system for theoretical usage.

Here is a brief example. Sorry I am not a mathemetician.

1 + 3 = 4 in base 10

but the same thing in a counting system where 1 = 1.5 could read completely different.

1.4a + 3a = 4.4a or when translated to base 10 it would really equal 6.1.

(Where 1a = 1.5 and 2a = 3 3a = 4.5 and so forth)I know this is a cheesy example but why isn't it possible to have a counting system where 1Pi + 3Pi = 4Pi? I suppose that you could use an irrational base for the given number rather than the traditional counting numbers; however the whole numbers would be all screwed up. Am I just thinking completely screwy?

Thanks for all the help beforehand.

HUGS!

Sqrl

------------------

Gasoline: As an accompaniement to cereal it made a refreshing change. Glen Baxter

Arnold Winkelried

11-23-1999, 02:30 PM

Libertarian, as Keeves said,

0.5555... in base 6 would be

(5/6) + (5/36) + (5/216) + (5/1296) ...

which will be a number very close to one.

SqrlCub

My math days are behind me now, but every time I've seen a different base used, it was an integer base.

I suppose you could suggest a counting method where, for example

a = pi

b = 2 * pi

c = 3 * pi

etc...

but I fail to see the advantages. How would you write the number 1? the number 2?

Since integers are the "natural" counting method and used much more often than pi, all you would do is make the notation of one number very easy, and make the notation of any other number very difficult. And the exact value of the base of your counting system would still not be known.

------------------

La franchise ne consiste pas à dire tout ce que l'on pense, mais à penser tout ce que l'on dit.

H. de Livry

RickG

11-23-1999, 03:04 PM

.55555... (base 6) is exactly equal to 1, in the sense that the infinite series represented by the repeating decimal (5/6 + 5 36 +5/216 + ...) sums to 1. It is true that any partial sum of the series is less than 1, but the partial sums approach 1 as the number of terms increases.

For the base ten case, the proof is as follows:

Let x = 0.9999999...

Then 10*x = 9.99999...

Subtracting x from both sides, we have

9*x = 9, thus x = 1, QED.

This is a standard method for summing convergent series. The demonstration for the base 6 case is left as an exercise for the reader.

Rick

Padeye

11-23-1999, 03:47 PM

Going to binary (Base Two) wouldn't work, but if one were to use a Base Pi counting system, "pi" would be written "10"...

"Base pi" doesn't help either because the ratio remains the same. You've done no different than drawing a cirle with a circumference that is an integer. The diameter is only expressable as an irrational number.

Liberal

11-23-1999, 04:07 PM

RickG

I have a counter-proof (inductive) that (in base 10) .99999... is not equal to 1. I would appreciate your bringing your opinion to bear on it.

Let me present first your (i.e., the classic)deductive proof just a bit more formally, plus one classic inductive proof, and then my counter-proof. Naturally, if my proof is sound, and it contradicts the classic proofs, then I will have to demonstrate something specific that is wrong with both of them.

Classic Deductive Proof

Axiom 1: For every real number, A, A = A.

Axiom 2: If A = B, then A*N = B*N.

Axiom 3: If A = B, then A-N = B-N.

Axiom 4: A = .99999...

Premise 1: 10A = 9.99999...

Premise 2: 10A-A = 9.99999...-A

Premise 3: 9A = 9.99999... - .99999...

Premise 4: 9A = 9

Premise 5: 9A/9 = 9/9

Premise 6: A = 1

Conclusion: Since A = A, by Axiom 1, and A = 1, by Premise 6, and A = .99999..., by Axiom 4, then 1 = .99999...

Classic Inductive Proof

Axiom 1: For every real number A, A = A.

Axiom 2: If A > B and B > C, then A > C.

Observation 1: If .99999... is not equal to 1, then there must be some number, A, between .99999... and 1, such that .99999... < A < 1.

Conclusion: Since no such number exists, 1 must be equal to .99999...

Inductive Counter-proof

Axiom 1: For every real number, A, A = A.

Axiom 2: If A = B, then A-B = B-A = 0.

Axiom 3: A = 1

Axiom 4: B = .99999...

Observation 1: 1 - .9 = .1

Observation 2: 1 - .99 = .01

Observation 3: 1 - .999 = .001

Observation 4: 1 - .9999 = .0001

Observation 5: 1 - .99999 = .00001

Observation 6: In general, it appears that 1-.9(N digits) = .0(N-1 digits)...1

Observation 7: Though the trailing 1 of .0(N-1 digits)...1 moves further and further to the right with each additional digit in .9(N digits), and is therefore of less and less magnitude, it does not ever disappear, and therefore never attains the value of 0.

Conclusion: Since 1-.99999... > 0, then by Axiom 2, .99999... does not equal 1.

Note that if Observation 6 is proved, the entire proof becomes deductive.

Summary

I have named the number representing 1-.99999... "eta". But before I present its properties, I am beholden to show flaws in the classic proofs that 1 does equal .99999....

Flaw 1

The flaw in the classic deductive proof occurs early, in Premise 1. The step ignores the fact that, for N 9s to the right of the decimal, there will be N-1 9s after multiplying by 10. For example, .99999*10 = 9.9999. It does not equal 9.99999.

Flaw 2

The conclusion assumes only one possible interpretation for there being no number between .99999... and 1, because the assumption is made that the set of real numbers is infinitely large, an assumption not presented as an axiom and not proved.

(Yes, I am aware of Cantor's work, but save that for another post.)

The existence of eta implies the existence of a smallest non-zero real number. In other words, eta is exactly the distance between .99999... and 1 on the real number line, or, put another way, eta is beside zero on the real number line.

We can use the same extrapolation process to discover another number, the multiplicative inverse of eta, and therefore the largest number in magnitude, thus: 1/.9 = 10; 1/.99 = 100; 1/.999 = 1000; etc; therefore, 1/.99999(N digits) = 10(N digits). That number, I named "omega".

There are also some practical arithmetic applications for the same extrapolation process, as solutions to heretofore perplexing problems. For example, what is 4*.77777...?

Well, 4*.7 = 2.8; 4*.77 = 3.08; 4*.777 = 3.108; 4*.7777 = 3.1108; 4 * .77777 = 3.11108; therefore, 4*.7 = 3.1(N digits)07.

What do you think?

------------------

"It is lucky for rulers that men do not think." — Adolf Hitler

Arnold Winkelried

11-23-1999, 04:21 PM

The inductive counter-proof is incorrect because in writing 0.999... we are assuming an "infinite" number of 9s following the decimal point, so there would be no remainder for 1 - 0.99999.

------------------

La franchise ne consiste pas à dire tout ce que l'on pense, mais à penser tout ce que l'on dit.

H. de Livry

Liberal

11-23-1999, 04:27 PM

Arnold Winkelried

[quote]...we are assuming an "infinite" number of 9s following the decimal point...[/b]

But that criticism, though respectfully appreciated, bears the same flaw as with the classic inductive proof that my counter-proof counters. Such an assumption is a priori, and must be presented as axiomatic. One thing that has always bothered me about transfinitists is their inconsistent insistence that (1) infinity is not a number, but (2) there are an infinite number of so-and-so's (9s, digits, whatever).

------------------

"It is lucky for rulers that men do not think." — Adolf Hitler

Cabbage

11-23-1999, 04:56 PM

Infinity is a number(s) (I'll call the infinity we're talking about omega). The problem with the argument:

1-.9999 is always greater than zero for a finite number of 9's, therefore 1-.999... is greater than zero,

is flawed because induction shows that 1-.999... is greater than zero for any finite number of nines--induction does not make the step to showing that this is true for an infinite number of nines.

Consider the following argument:

1 is finite.

If n is finite, n+1 is also finite.

Therefore, the set of natural numbers is finite.

This seems to be equivalent to the flaw in the proof that .9999... is not 1--it works as long as we have finitely many 9's, but not infinitely many 9's. Induction doesn't make that step to omega.

{:-Df

11-23-1999, 05:06 PM

Padeye:

"Base pi" doesn't help either because the ratio remains the same. You've done no different than drawing a cirle with a circumference that is an integer. The diameter is only expressable as an irrational number.

Of course it's no different. (And as I understated, counting integer quantities, such as a baker's dozen (13) in Base pi would remain "a bit harder.") But as SqrlCub and Arnold Wrinkle-eyed later discussed, this was just the kind of thing SqrlCub was looking for. Useless (or, "usefulness unknown"), but easy to write.

The only thing I'd quibble is that, using the Arabic numerals and numbering convention with which we are all familiar, "pi" should be written 10, not 1. (Base ten, ten is written 10; Base six, six is written 10; etc. - and all deriving necessarily from the set of whole numbers.) If "pi" is 1 (the first, er..."whole" component of the series) AND also the base we use for describing larger numbers - then that's Base One(?), in which you can't calculate or even count anything. (I guess 1=>1, 2=>11, 3=>111...and that's not a "counting" system.)

Dividing pi into a series to make Base pi countable is an arbitrary task (I vote for "bits," in Base ten "1/8" pi), thus revealing the whole thing to be the nonsense (irrational sense that I can't compute?) that it is.

TheDude

11-23-1999, 06:07 PM

The existence of eta implies the existence of a smallest non-zero real number.

So in order to disprove your hypothesis, all we must do is prove that no such thing exists.

We will do this by assuming it to be true and finding a contradiction (reductio ad absurdum).

Assume that the smallest positive non-zero real number eta exists.

Divide eta by two to obtain eta/2=zeta.

Zeta is not zero, because if it were, eta would also be zero (because eta=zeta*2).

But zeta is positive because eta is.

And furthermore, zeta is smaller than eta by the following proof:

zeta>0

zeta+zeta>0+zeta

eta/2+eta/2>zeta

eta>zeta.

So zeta is a smaller positive non-zero real number than eta.

But we assumed that eta was the smallest such number.

So the smallest positive non-zero real number must not exist.

I won't get into the proof, but it can be proven that between any two arbitrarily close real numbers, there exists an infinite number of real numbers. Which is a consequence of the fact that there are more real numbers than integers (that is, the real numbers are not countable).

So, in conclusion, 0.999999.... (with an infinite number of nines) is identical to 1 because the limit of 1-sum(9*10^(-n)) as n approaches infinity is 0.

TheDude

TheDude

11-23-1999, 06:15 PM

the assumption is made that the set of real numbers is infinitely large, an assumption not presented as an axiom and not proved.

This is not an assumption but a basic outgrowth of the definition of the real numbers. It is not an axiom and can be proven if you so desire, but it will involve me busting out my Real Analysis notes from college (without a doubt the hardest class I ever took in college or grad school).

TheDude

andros

11-23-1999, 06:28 PM

libertarian sed:

The flaw in the classic deductive proof occurs early, in Premise 1. The step ignores

the fact that, for N 9s to the right of the decimal, there will be N-1 9s after

multiplying by 10. For example, .99999*10 = 9.9999. It does not equal 9.99999.

Then asserted:

One thing that has always bothered me

about transfinitists is their inconsistent insistence that (1) infinity is not a number

So, it's your belief that infinity is a number? In the top "flaw," you claim that if there are n 9s after the decimal, there must be n-1 after multiplying by 10.

That can only be true if n is finite. It is not (that's what the elipses mean in .999...). [infinity]/[any real number]=[infinity].

So, if n is infinite, 10*n is also infinite.

-andros-

Lumpy

11-23-1999, 07:05 PM

Why .99999.... Equals 1:

1/9 = .11111....

2/9 = .22222....

3/9 = .33333....

4/9 = .44444....

5/9 = .55555....

6/9 = .66666....

7/9 = .77777....

8/9 = .88888....

9/9 = .99999....

Or you could say, 3/9 = .33333.... , so (3/9) *3 = .99999.....

When I began to work on the question does .99999.... equal 1, I believed that it didn't but I wrote a proof that says it does. For the following proof, please realize that the sum of r^n from n=0 to infinity is 1/(1-r) where r < 1. Anyone who has studied Calculus will back me up on this. Also I don't know if theres a way to type the summation symbol, so please bear with me.

.999999999.... =

.9(1.1111111.....) =

(9/10)(1 + 1/10 + 1/100 + 1/1000....) =

(9/10)(sum from n=0 to infinity of (1/10)^n) =

(9/10)(1/(1-(1/10)))=

(9/10)(1/(9/10))=

(9/10)(10/9)=

1

Since we are dealing with a number with infinite decimal places, everyone should remember when dealing with infinity, what seems logical is not always correct. For instance the question of what is infinity divided by infinity?

Some would argue that anything divided by itself is 1. This is logical.

Others would argue that infinity divided by anything is still infinity. Logical.

Finally, others might argue that anything divided by infinity is 0. Also Logical.

What about infinity minus infinity?

Infinity is not a number, but an idea and does not follow the rules of numbers. Thats why functions are never considered at infinity, but rather as they approach infinity.

Heres another puzzler. The area under the curve of 1/(2^x) from 1 to infinity is actually finite.

mr john

11-24-1999, 12:02 AM

You may never levitate but if you meditate on pi long enough you will find you can win any debate using circular logic. You'll never get to the 'end' of pi. it is a beastly infinite ratio like 66.66666666....

------------------

"Pardon me while I have a strange interlude."-Marx

I am so suggestible. Now I'm hungry for pie. Guess I'd better avoid all commercials. Dont know about that last number in the above post(666!) arg!hee hee

{:-Df

11-24-1999, 12:12 AM

If we used another system of counting (not base 10) could we get pi to be a definite number? I think it would work in theory, but practice is something completely different.

Going to binary (Base Two) wouldn't work, but if one were to use a Base Pi counting system, "pi" would be written "10"...

Of course, currently simple tasks like counting a baker's dozen doughnuts would be a bit harder...

Liberal

11-24-1999, 12:16 AM

Keeves

You might be talking about the formula first presented by John Wallis in 1655 in Arithmetica Infinitorum. It was the first time an expression for pi was show as a product of rational numbers!

pi/2 = 2*2*4*4*6*6*8*8.../1*1*3*3*5*5*7*7*9*9...

[b]mr john[b]

You'll never get to the 'end' of pi. it is a beastly infinite ratio like 66.66666666....

Well, sort of, except that pi is transcendent. None of its digits ever repeat in any pattern.

------------------

"It is lucky for rulers that men do not think." — Adolf Hitler

Polycarp

11-24-1999, 12:19 AM

It is therefore a transcendental number, although meditating on it will not cause you to levitate, more's the pity.

Au contraire. You simply have to know it to the right number of digits! ;)

Arnold Winkelried

11-24-1999, 12:23 AM

1. There are many fractions that represent an approximation of pi. For mental arithmetic, one commonly used value is 22/7. Another, closer fraction would be 355/113. Nowadays since many digits of pi are known, fractions aren't used very much.

2. An irrational number (one that cannot expressed as a fraction of two integers) is irrational in any base.

e.g.

1 / 3 = 0.2 in base 6

1 / 3 = 0.333333... in base 10

1/3 has an infinite number of digits (after the decimal point) in one base, doesn't in another base. But that is only true for fractions.

There are many infinite series for calculating pi.

One that is easy to remember is the Leibniz series,

pi / 4 = 1 - (1/3) + (1/5) - (1/7) + (1/9) ...

One currently used by supercomputers to calculate the value of pi (according to an article I read) was devised by the indian mathematician Ramanujan.

1 / <font face="Symbol">p</font> = sqrt (8) / 9801 * (sum from n = 0 to infinity (((4n)! (1103 + 26390n)) / ((n!)4 (396)4n)))

------------------

La franchise ne consiste pas à dire tout ce que l'on pense, mais à penser tout ce que l'on dit.

H. de Livry

Arnold Winkelried

11-24-1999, 12:25 AM

Maybe my point (2) above wasn't too clear. Since pi is an irrational number (i.e. cannot be represented exactly by a fraction), then it doesn't matter what base you chose to write it in, it will always have an infinite number of digits after the decimal point.

SqrlCub

11-24-1999, 08:23 AM

This helps explain it better to me. My main problem with a number that has an infinite non-repeating series after it is that any product of that given multiplicant will also have an infinite series of numbers after it. I know what I was asking isn't really practical but IMO I thought it would get rid of the infinite series in some way. Numbers are screwy that way.

Thanks for all the insight so far. I look forward to reading more in this topic.

HUGS!

Sqrl

------------------

Gasoline: As an accompaniement to cereal it made a refreshing change. Glen Baxter

Cabbage

11-24-1999, 01:26 PM

Actually, just because one (or both) of the factors in a product has a nonrepeating decimal expansion doesn't mean the product will. For example, both pi and 1/pi have nonrepeating decimal expansions, but they multiply to 1. There's an infinite number of ways this can happen, and it's not always easy to tell what kinds of arithmetic combinations of irrational/transcendental numbers will produce another irrational/transcendental.

vBulletin® v3.8.7, Copyright ©2000-2017, vBulletin Solutions, Inc.