- #1
TylerH
- 729
- 0
I know .999... = 1. I'm just arguing against this method of proof.
A common proof I see that [itex].999 \ldots = 1[/itex] is that [itex]sup\{.9, .99, .999, \ldots \} = 1[/itex], but this is only true if you assume [itex].999 \ldots \ge 1[/itex]. If you assume, as most argue, that [itex].999 \ldots < 1[/itex], then [itex]sup \{.9, .99, .999, \ldots \} = .999 \ldots < 1[/itex]. Of course, by assuming [itex].999 \ldots < 1[/itex], you get the nonsense expected at the end of a proof by contradiction, but you still have to proof that [itex].999 \ldots < 1[/itex] is nonsense by proving [itex].999 \ldots = 1[/itex]. Therefore, the supremum method is useless.
Is my logic correct?
A common proof I see that [itex].999 \ldots = 1[/itex] is that [itex]sup\{.9, .99, .999, \ldots \} = 1[/itex], but this is only true if you assume [itex].999 \ldots \ge 1[/itex]. If you assume, as most argue, that [itex].999 \ldots < 1[/itex], then [itex]sup \{.9, .99, .999, \ldots \} = .999 \ldots < 1[/itex]. Of course, by assuming [itex].999 \ldots < 1[/itex], you get the nonsense expected at the end of a proof by contradiction, but you still have to proof that [itex].999 \ldots < 1[/itex] is nonsense by proving [itex].999 \ldots = 1[/itex]. Therefore, the supremum method is useless.
Is my logic correct?
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