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## Could the MRB constant be rational?

Maple

The best definition for the MRB constant that I know of is found at

Up until last week I'd been operating under the assumption that the MRB constant is irrational.

It is the sum of irrational numbers; in fact it is a sum of an infinite number of irrational numbers.  However, the sum of irrational numbers, whether it is two or infinity of them, is not necessarily irrational. For instance, if you had an infinite number of irrational numbers and half of them were the sqrt(2) and the other half were –sqrt(2), their sum would not have to be irrational, would it?

Recall that a number is rational if it can be expressed as a ratio of two integers, p and q. Its decimal expansion will either terminate or become repeating. I've computed 299,998 digits of the MRB constant and have found neither termination point, nor found the digits to be repeating. You can check them for yourself at http://marvinrayburns.com/299998mrb.txt . So to determine if it was rational you would have to have integers,p1,q1,p,q with q1 and q!=0 such that p1/q1=sum((-1)^n*(n^(1/n)-1),n=1..infinity) with p1 and q1 being coprime, which implies that with p and q being coprime:

p/q= sum((-1)^n*(n^(1/n)),n=1..infinity). Multiplying both sides by q we get

p= q*sum((-1)^n*(n^(1/n)),n=1..infinity). By an assosiate law,

p=sum((-1)^n*q*(n^(1/n)),n=1..infinity) Expressing n and q by the same exponent we get

p=sum((-1)^n*(n*q^n)^(1/n),n=1..infinity). When we expand the series we see that

p=-1^(1/1)*(q^1)^(1/1)+2^(1/2)*(q^2)^(1/2)-3^(1/3)*(q^3)^(1/3)+4^(1/4)*(q^4)^(1/4)-... . Since (A^B)^C=(A^C)^B when they are all positive:

p=-1^(1/1)*(q^(1/1))^1+2^(1/2)*(q^(1/2))^2-3^(1/3)*(q^(1/3))^3+4^(1/4)*(q^(1/4))^4... . Accordingly

______________________Second noticed mistake is here.______________________________

p=-1^(1/1)*a1^1+2^(1/2)*a2^2-3^(1/3)*a3^3+4^(1/4)*a4^4-..., where ax=(q^(1/x))^x., and assuming q and x are positive  (q^(1/x))^x =q. Thus

______________________First noticed mistake is here.______________________________

p=-1^(1/1)*q^1+2^(1/2)*q^2-3^(1/3)*q^3+4^(1/4)*q^4-,,,. Since 1^(1/1) =1,

p=-1q^1+2^(1/2)*q^2-3^(1/3)*q^3+4^(1/4)*q^4-,,,. . Factoring out a q we get

p=q*(-1+2^(1/2)*q-3^(1/3)*q^2+4^(1/4)*q^3-,,,).  Dividing both sides by q we get

p/q=-1+2^(1/2)*q-3^(1/3)*q^2+4^(1/4)*q^3-,,,. Converting back to a series we get

p/q= sum((-1)^n*(n^(1/n)*q),n=1..infinity). Hence

p/q= q*sum((-1)^n*(n^(1/n)),n=1..infinity). But remember that   [1]

p/q= sum((-1)^n*(n^(1/n)),n=1..infinity).                                   [2]

subtracting [2]  from [1] gives

0=(q-1)*sum((-1)^n*(n^(1/n)),n=1..infinity). Dividing by the sum of the series we see that

0=q-1, and 1=q. Thus q=1, and since q=1 and equation [2],

p=sum((-1)^n*(n^(1/n)),n=1..infinity) which is not an integer.  QED

I bet I made a few mistakes here. If you see them before I do, go ahead and say so.

So I will stick with:

It seems vey difficult to show that p and q can be integers, but probably just as hard to prove that they can't be integers; don't you think so?

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