Lets say we have irrational numbers &#x03B1;<!-- α --> 1 </msub> , . .

racodelitusmn

racodelitusmn

Answered question

2022-07-06

Lets say we have irrational numbers α 1 , . . . , α n in the interval ( 0 , 1 ). Represent each αi as a binary expansion 0. a i 1 a i 2 . . . where each a a i j { 0 , 1 }. Define the "dovetail" of the α i to be the number with binary expansion 0. a 1 1 a 2 1 a 3 1 . . . a n 1 a 1 2 a 2 2 a 3 2 . . . a n 2 a 1 3 a 2 3 a 3 3 . . .
So that the dovetail is irrational?

Answer & Explanation

isscacabby17

isscacabby17

Beginner2022-07-07Added 13 answers

It's never rational. This follows from the fact that a real number is rational if and only if its binary expansion is eventually periodic (possibly with endless 0s).

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