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These are no fractions, are they? (Posted on 2006-09-05) Difficulty: 3 of 5
You are probably quite certain that the following numbers are irrational. But can you prove it?

a = 0.149162536496481...

b = 0.2357111317192329...

c = 0.1248163264128256...

See The Solution Submitted by JLo    
Rating: 2.5000 (2 votes)

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Rephrasing of the problem | Comment 7 of 15 |
The relevant question seems to me to be: Is there a bounding value, so that all squares / primes / powers of two larger than that bound are "composed" of several smaller squares / primes / powers of 2 in order, where "composed" means concatenation of digits.

It seems extremely unlikely that that should be the case, but can one prove it?  Then these sequences of digits would be forever non-repeating, and the numbers therefore irrational.

My latest thought: the squares / primes / powers of 2 which appear after the start of perodicity and which have many more digits than the length of the period will themselves consist of periodical digits with the same period.  If the numbers are rational, there have to be such squares / primes / powers of 2 eventually, and exclusively.  It seems ludicrous, but how does one disprove it formally?


  Posted by vswitchs on 2006-09-06 13:15:23
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