Pi question

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.

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

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).


355/113 is another good one.

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

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?

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

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…


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 = 22446688…/1133557799…

mr john

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

Au contraire. You simply have to know it to the right number of digits! :wink:

  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.

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!)[sup]4[/sup] (396)[sup]4n[/sup])))

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

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.

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

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!

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,

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.

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

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.


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

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

.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.


“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.


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 AN = BN.

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.


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

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