|
To state the obvious, the three numbers a, b, and c are generated by concatenating the series of squares, primes and powers of two. Then we precede the whole thing with "0." to obtain a decimal number. In order to prove that none of these numbers is rational, we must show that the decimal expansion is not periodic.
First of all, have a look at Art M's posts for a nice, elementary proof covering all three numbers using a unified approach.
Below an alternative proof:
a) The series of squares also includes the squares of 10, 100, 1000,... Therefore the decimal expansion of a contains arbitrarily long sequences of zeros and cannot be periodic.
b) A bit of a tough one. I must admit that I have no proof that works without some "hard" theorems about prime numbers. The "easiest" that will do the trick for us is Bertrand's Postulate, the one Art M also used in his proof. There is a rather short proof by Erdos available for those who are interested. Bertrand's Postulate states that there is always a prime number between the positive integers n and 2n. From Bertrand's Postulate follows immediately that there is always a prime number with exactly k digits for every k>=1; this weaker statement is what we'll use.
So let's assume b gets periodic at some point with period length l. Because of our weakend version of Bertrand's Postulate, we know that there must be prime numbers with exactly l, 2l, 3l... digits. When these primes get large enough, we also know that they lie completey within the "repeating region" of our decimal number b. Let's say the prime p with (k+1)*l digits is one of those. Then we have p = d + d*10^l + d*10^(2l) + ... + d*10^(kl) for some integer d with l digits. Obviously d divides p and p cannot be a prime number, which contradicts our assumption, hence the decimal expansion of b cannot be periodic.
c) It can be shown that every positive integer represents the starting digits of a power of two. For a proof you can have a look at the solution to "Evil Powers". Therefore there are powers of two starting with digits 101, 1001, 1001,... which again makes it impossible for the series of power-of-two-digits to ever become periodic. Not the most elementary proof maybe, but a quick one for me to write ;-)
|
