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Are they only numbers that end with 9999... and 0000... after the dot or some other too? If so, can you give an example?

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In decimal base or for other bases? –  Sami Ben Romdhane Dec 17 '13 at 19:06
    
Yeah, in decimal base. –  jsoldi Dec 17 '13 at 19:07
    
They are the only numbers. –  Sami Ben Romdhane Dec 17 '13 at 19:09
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up vote 5 down vote accepted

Any rational number whose decimal expansion terminates, i.e., numbers of the form $$\dfrac{p}{2^m 5^n}$$ where $p,m,n \in \mathbb{Z}$ can have two different representations. For instance, $$\dfrac75 = 1.4 = 1.3\bar{9}$$

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Here 'terminates' refers to the same kind of reals I was referring to as ending with 0000... right? –  jsoldi Dec 17 '13 at 19:13
    
@jsoldi: that is correct. He is showing which they are. –  Ross Millikan Dec 17 '13 at 19:15
    
@jsoldi Yes, good question. Yes, true. –  user17762 Dec 17 '13 at 19:15
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The only real numbers with two decimal representations have their decimal representations agree up to some point, then one continues with $a999\ldots$ while the other with $b000\ldots$, where the digit $b$ is one more than the digit $a$. For example, $$5.5679999\ldots = 5.568000\ldots$$ Another example: $$34199.9999\ldots = 34200.0000\ldots$$

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