What's wrong with my method that gives $0.\bar9 = 1$?

Asked Dec 05 '13 at 18:21

Active Dec 05 '13 at 18:21

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This is what I learned :

$x = 0.\bar3$ (I)

$10x=3.\bar3$ (II)

(II)-(I) : $9x = 3 => x = 1/3.$ And It's OK.

But now, I think this similar calculation gives me wrong answer:

$x = 0.\bar9$ (I)

$10x=9.\bar9$ (II)

(II)-(I) : $9x = 9 => x = 1$

Why does: $0.\bar9 = 1 $ ?

asked Dec 05 '13 at 18:21

Emadpres

1

This is true, take a look at http://math.stackexchange.com/questions/11/does-99999-1 . – Stefan Dec 05 '13 at 18:25
@Stefan Thanks ! I got the point ! – Emadpres Dec 05 '13 at 18:30

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