Prove that a number is NOT periodic

Question

i'm reading "A concise introduction ti pure mathematics" by Liebeck and in the exercises of the second chapter i found this question:

"Show that the decimal expression for $\sqrt 2 $ is not periodic"

If i write $\sqrt 2$ in its decimal form, i should obtain something like:

$\sqrt 2 = {a_o}.{a_1}{a_2}{a_3}....{a_n}$

But how can i prove that there is not a string of periodic numbers in ${a_1}{a_2}{a_3}...{a_n}$?

Should i prove this by contradiction?

Thanks a lot for your help and excuse any grammatical mistakes i could have committed, English is not my born language.

If decimal expression is periodic, it's rational number. But $\sqrt2$ is not. — Michael Galuza, Jul 23 '15 at 17:13
Thanks a lot for your answer. Does that implies that every irrational number is non periodic? It is true that, if a number is rational, then it's decimal expression is periodic? — DUM00, Jul 23 '15 at 17:19
It's either periodic or terminating e.g. 1/10 = 0.1 in decimal. — Semiclassical, Jul 23 '15 at 17:55
@Semiclassical A terminating expansion is also a periodic expansion; just one in which every digit after a certain point is zero. (Or alternately, $b-1$ - e.g., $9$ in decimal expansions - but let's not get into that...) — Steven Stadnicki, Jul 23 '15 at 18:14

score 2 · Answer 1 · answered Jul 23 '15 at 18:07

Assume by contradiction a periodic representation of $\sqrt2$ on base $10$.

Let $A$ denote the digit-sequence in the integer part of $\sqrt2$.

Let $B$ denote the digit-sequence in the non-periodic prefix of the fractional part of $\sqrt2$.

Let $C$ denote the digit-sequence in the periodic remaining of the fractional part of $\sqrt2$.

Let $|A|$ denote the length of $A$, $|B|$ denote the length of $B$, and $|C|$ denote the length of $C$.

Then $\sqrt2=A+\dfrac{B}{10^{|B|}}+\dfrac{C}{10^{|B|+|C|}-1}$.

But ${A,B,C,|A|,|B|,|C|}\in\mathbb{N}\implies{A+\dfrac{B}{10^{|B|}}+\dfrac{C}{10^{|B|+|C|}-1}}\in\mathbb{Q}$.

This leads to the fact that $\sqrt2\in\mathbb{Q}$, which is false, hence the assumption is false.

Note that although the proof above is for base $10$, it can be used for any other integer base.

score 0 · Answer 2 · answered Jul 23 '15 at 18:08

If x has an expansion ,in any base B, which is eventually periodic, with period length P, then multiplying x by the Pth power of B and subtracting x, yields a rational value for (B**P - 1)x. Hence x is rational. Conversely,if x is rational, then computing its representation in base B by long division eventually gives a remainder that occurred before, as their are only B possible remainders; after that it repeats periodically.

Prove that a number is NOT periodic

2 Answers2