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 x₁, x₂, ..., 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].)

Solution

Observe that

	y_2n+1	= (√2)²ⁿ⁺¹
		= (√2)×2ⁿ.

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.y₁y₃y₅... is the binary representation of the fractional part of √2. Since √2 is irrational, the sequence y₁, y₃, y₅, ... is not periodic.

Let x_n = z_n (mod 2), where z_n is either 0 or 1, n = 1, 2, ... Clearly, z_2n+1 = y_2n+1, and we just proved that the sequence {y_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

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