## Shifting Digits and a Point of View

*The number x _{n} is defined as the last digit in the decimal representation of the integer ⌊(√2)^{n}⌋ (n = 1, 2, ...). Determine whether the sequence x_{1}, x_{2}, ..., x_{n}, ... is periodic.* [Savchev & Andreescu, p. 146].

(⌊a⌋ is the whole part of number a which is more often at this site and elsewhere denoted [x].)

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

Observe that

y_{2n+1} | = (√2)^{2n+1} | |

= (√2)×2^{n}. |

Recollect that in the decimal system the multiplication of a real number by 10 causes the decimal point (or comma in some cultures) to shift one position to the right. The same happens in the binary system when a number is multiplied by 2. It follows that y_{2n+1} is a digit in the binary representation of √2 so that 0.y_{1}y_{3}y_{5}... is the binary representation of the fractional part of √2. Since √2 is irrational, the sequence y_{1}, y_{3}, y_{5}, ... is not periodic.

Let x_{n} = z_{n} (mod 2), where z_{n} is either 0 or 1, n = 1, 2, ... Clearly, _{2n+1} = y_{2n+1},_{2n+1}} is not periodic. We may now claim that neither is {z_{n}}. For if it were periodic with, say, period N, then it would be periodic with the period 2N as well. But this would imply that the sequence {y_{2n+1}} had a period of N, which is impossible. Therefore, the sequence of remainders {z_{n}} of the division of the terms {x_{n}} by 2 is not periodic. Thus {x_{n}} could not be periodic either.

### References

- S. Savchev, T. Andreescu,
*Mathematical Miniatures*, MAA, 2003

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

68353397