Is this number trancedental 0.110101000101000101... .?

Question

It's binary number, which has bits B_p set for each prime p.

(I do not know the answer)

Answer 1

0.110101000101000101... .
= 0.11 + 0.000101000101000101....

= 0.11 + 0.101 * 10^(-3) + 0.101 * 10^(-9) + 0.101 * 10^(-15) + ....

= 0.11 + sum of a GP with u_1 = 0.101 * 10^(-3) and r = 10^(-6)

So limiting sum = 11/100 + 0.101 * 10^(-3) / {1 - 10^(-6)}

= 11/100 + 0.101 * 10^(-3) * (10^6)/{10^6 - 1}
= 11/100 + 101/999999

= (10999989 + 10100)/99999900

= 11010089/99999900

This is certainly NOT transcendental but rational

zanti3 You say "We also know that the frequency of 1s will decrease as we get to the higher decimal points. "

THis is HUGE statement given that there is absolutely NO evidence that this will happen by the pattern so given!!

Answer 2

The number

1/2^2+1/2^3+1/2^5+1/2^7+1/2^11+1/2^13+1/2^17+.......= S

is irrational and also transcendental.

Irrational: Assume S is rational set S = a/b. We know that

prime gaps are arbitrarily large via a string of consecutive

composites n!+2,n!+3,......,n!+n etc so there are two

consecutive primes r and s which differ by at least

log2 +logb. Then

b*2^r*S = b*2^r(1/2^2+...+1/2^r)+b*2^r(1/2^s+.....) on rhs,

the first term is an integer and the second term is <1 since

b*2^r/2^s <1/2, and subsequent terms are,1/4,1/8,...

Yet b*2^r*S = a*2^r* an integer. So we have an integer

equals an integer plus, a real number less than 1. This

contradiction gives S irrational.

Transcendental: Suppose that S is algebraic so that we may

put S in a polynomial with integer coefficients to get zero:

F(S) = aS^n + bS^(n-1) + ... +yS +z = 0. We need a little

lemma: Gaps in powers of .1101010001010001010001000001... get
arbitrarily large ( that is # 0f zero's between the 1's).

Multiply F(S) by the number B*2^q where the gap between

q and the next prime r is so large that the non integer portions

of B*2^q* S^k, are all less than 1/ (a+b+c+...+z) which is

easy to do since a,b,c,..., z are constant integers. Then, as

before , a sum of integers plus a number less than one is

another integer (in this case,zero) which is a contradiction.

Therefore S is not algebraic and is transcendental.

Just prove the lemma!

Answer 3

Proving the transcendence of a given number frequently can be extraordinarily difficult to do, as it's required to show that the number cannot be the root of any algebraic polynomial equation with rational roots. There's already a long list of candidate numbers waiting to be proven transcendental or not, some waiting more than a century. It's a safe bet that your proposed number will be added to this list and stay there a very long time, and it's also a pretty good bet that it's transcendental, as algebraic numbers are actually in the minority, like how constructible regular polygons are in the minority. Because your number can be expressed as follows:

∑(1/2)^(p(n)-1)

where p(n) is the nth prime number, I can see a connection with the Riemann zeta function, so it should be interesting if this ever pops up in journals. In fact, let me keep looking for this series, I'll get back to you if I find anything.

Addendum: One place to look into is Louiville numbers, proven to be transcendental. I'll need more time to see if your number falls into this category.

Answer 4

If it were not transcendental, then the decimal expansion would eventually start repeating. Let's divide the given number by 10 so that each 10^(-p) digit is set to 1 for each prime p. (This division does not change the transcendentality.) Identify the repetition so that it starts with a 1; let -x be the position of the beginning of the repetition and y be the length of the repeated block of digits.
The digit at -(x+xy) wil be 1, indicating that (x+xy) is prime; but (x+xy)=x(y+1), so (x+xy) is not prime. Therefore, the number cannot end in a repeating block and is transcendental.

Answer 5

Well, the number is clearly irrational, since the prime numbers cannot come in any pattern that repeats to infinity. It is also clear that the irrational number cannot be normal, since there will be more 0s than 1s. We also know that the frequency of 1s will decrease as we get to the higher decimal points.

As for it being transcendental, the fact that the ratio of 0s to 1s decreases surely means this is the case. If the number is not transcendental, it would have to be some combination of nth roots (square roots, cube roots, etc.), and no nth roots behave like that. Now, I'll admit that proving this rigorously would be difficult.

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Wal C: I don't think you are reading this question correctly. There are an infinite number of 1 digits and an infinite number of 0 digits, since the number of prime and composite numbers are both infinite. However, the 0s will occur with greater frequency, since composite numbers occur more often than prime numbers. Also, the frequency of 1s WILL tend to decrease - this follows from the Prime Number Theorem, which states the density of primes approaches 1/ln(n) as n gets large. (This also makes sense intuitively - the bigger the number, the more potential factors it can have.)

* * * * *

Wow, knashha, your proof that this number is transcendental is way over my head but sure looks impressive. :) I have a proof in mind to show the number has to be irrational (a much easier task) that I'll outline here:

If the number is rational, then at some point in the decimal expansion it has to repeat indefinitely. Let n be the length of the repeating cycle. Let p be a prime number in the repeating cycle, i.e. one of the 1s in the repeating digits. (Since the number of primes is infinite, there has to be at least one.) Then, since the cycle repeats after every n digits, it follows that p + kn must also be prime for every positive integer k.

Now let q be a prime number that is not a factor of n. Since q and n are mutually prime, it follows that there will be a k < q for which q is a factor of p + kn. Therefore, the assumption that all the p + kn numbers are prime cannot be true, which means the number cannot be rational.

Answer 6

Proving a number is trancendental is a very difficult task and I think it is beyond the capabilities of most if not all Yahoo answerers.

Answer 7

It is an incomplete question or number so it is impossible to tell what the code means.
Binary Code most would not be in /10s , it could be, but the 0. throws it off.