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Why is it true that all irrational numbers are non-terminating/non-repeating decimals?

By definition, an irrational number is one that can't be expressed as a ratio of integers.

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    $\begingroup$ Unclear what the question is: are you asking why they are non-terminating, why they are non-repeating, or why they can't be expressed as a ratio of integers? $\endgroup$ – user147263 Nov 29 '15 at 20:54
  • $\begingroup$ Duplicate of How to know a irrationals never repeat? $\endgroup$ – user147263 Nov 29 '15 at 20:55
  • $\begingroup$ All of the above. $\endgroup$ – Dwayne Rousseau Nov 29 '15 at 20:55
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    $\begingroup$ Still unclear, because one of those things is a definition of irrational number. And too broad. $\endgroup$ – user147263 Nov 29 '15 at 20:55
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    $\begingroup$ Possible duplicate of How to know that irrational numbers never repeat? $\endgroup$ – Nicholas Pipitone Nov 29 '15 at 20:56
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The definition: a number is irrational if and only if it's not rational, i.e. it can't be expressed as a ratio of two integers. This answers one part of your question.

The other part: I'll prove the contrapositive. If $x$ has a repeating decimal expansion (this includes terminating decimal expansions), then $x$ is rational.

Proof: If $x$ has a repeating decimal expansion, then it can always be written in the following form:

Let $c,b$ be non-negative integers and $a_i\in\{0,1,2,\ldots,9\}$ and $t$ is the number of digits of $b$. $$x=\overline{c.ba_1a_2\ldots a_ka_1a_2\ldots a_ka_1a_2\ldots}$$ $$10^tx=\overline{cb.a_1a_2\ldots a_ka_2a_2\ldots a_ka_1a_2\ldots}$$ $$10^{kt}x=\overline{cba_1a_2\ldots a_k.a_1a_2\ldots a_ka_1a_2\ldots}$$ $$10^{kt}x-10^{t}x=\overline{cba_1a_2\ldots a_k}-\overline{cb}$$ $$x=\frac{\overline{cba_1a_2\ldots a_k}-\overline{cb}}{10^{kt}-10^t}$$

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  • $\begingroup$ Thanks, user236182; I now understand the reasoning. $\endgroup$ – Dwayne Rousseau Nov 29 '15 at 23:50
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  • If the decimal expansion of a number $x$ is terminating, with $n$ digits after the decimal point, say, then $10^nx$ is an integer $m$ (the decimal expansion is shifted by $n$ places to the left, hence has nothing after the decimal point) so that $x =\frac{m}{10^n}$ is a fraction of integers, aka. rational number.
  • If the decimal expansion of a number $x$ is eventually repeating, with a epriod of length $n\ge1$, say, then $10^nx$ has a decimal expansion that matches that of $x$ beyond some point, so that in computing $10^nx-x$ everything cancels beyond some point, i.e., $10^nx-x$ is a rational $\frac ab$ per first bullet point. Solving for $x$ we find $x=\frac{a}{(10^n-1)b}$, which is again a fraction of integers.
  • If $x$ is the ratio/fraction of integers, then by definition it is rational.

Hence for an irrational number (by definition a number that is not a rational number) none of the three options above can occur.

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  • $\begingroup$ Thanks, Hagen von Eitzen for this, for I get it now. $\endgroup$ – Dwayne Rousseau Nov 29 '15 at 21:24

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