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If I am asked to show that $\sqrt{2}$ does not have a periodic decimal expansion. Can I just prove that $\sqrt{2}$ is irrational , and since irrational numbers are don't have periodic decimal expansions then I am done? Thank you.

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  • $\begingroup$ Huh? What does it mean for a number to be periodic?? $\endgroup$ – bof Feb 19 '16 at 3:00
  • $\begingroup$ a periodic means after the decimal point number are repeating like:1.813813 $\endgroup$ – Abdulaziz Asiri Feb 19 '16 at 3:13
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Hint: If it's periodic, you could express it in terms of a geometric series with rational starting point and rational common factor.

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Suppose $\sqrt{2}$ has a periodic decimal expansion, i.e. it has the form $$\sqrt{2} = 1.a_1a_2 \cdots a_m \overline{d_1 d_2 \cdots d_n}$$ where the overline indicates the repeating digits. Then $$10^m \sqrt{2} = 1a_1a_2 \cdots a_m . \overline{d_1 d_2 \cdots d_n}$$ and $$10^{m+n} \sqrt{2} = 1a_1 a_2 \cdots a_m d_1 d_2 \cdots d_n . \overline{d_1 d_2 \cdots d_n}$$ Subtracting the first from the second yields $$(10^{m+n}-10^m)\sqrt{2} = 1a_1 a_2 \cdots a_m d_1 d_2 \cdots d_n - 1a_1a_2 \cdots a_m$$ and hence $$\sqrt{2} = \frac{1a_1 a_2 \cdots a_m d_1 d_2 \cdots d_n - 1a_1a_2 \cdots a_m}{10^{m+n}-10^m}$$ This implies that $\sqrt{2}$ is rational, which is known to be false.

This argument generalises to show that all irrational numbers have nonperiodic expansions (in any number base).

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