Prove that a number is NOT periodic 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.
 A: 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.
A: 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.
