Decimal representation of the set [0,1)
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I have encountered the next statement in statistics lecture (translated from german):
"From the analysis
you know that all but a countable number of $$w ∈ [0, 1)$$ represent a unique decimal representation
$$ω = 0.x_1x_2x_3...$$
for the countably many exceptions we choose that representation without period 9".
That sounds very counterintuitive and unprovable to me, is this true?
examples-counterexamples decimal-expansion
edited Dec 4 '18 at 14:26
José Carlos Santos
164k22131234
asked Dec 4 '18 at 14:15
AromaTheLoopAromaTheLoop
444
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add a comment |
$begingroup$
I have encountered the next statement in statistics lecture (translated from german):
"From the analysis
you know that all but a countable number of $$w ∈ [0, 1)$$ represent a unique decimal representation
$$ω = 0.x_1x_2x_3...$$
for the countably many exceptions we choose that representation without period 9".
That sounds very counterintuitive and unprovable to me, is this true?
examples-counterexamples decimal-expansion
edited Dec 4 '18 at 14:26
José Carlos Santos
164k22131234
asked Dec 4 '18 at 14:15
AromaTheLoopAromaTheLoop
444
$endgroup$
$begingroup$
"countable" means having the cardinality of $mathbb{N}$. in german: "abzählbar".
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– trancelocation
Dec 4 '18 at 14:19
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What exactly sounds counterintuitive and unprovable? The countability maybe?
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– drhab
Dec 4 '18 at 14:21
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@drhab José Carlos Santos answered my question below.
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:56
add a comment |
$begingroup$
I have encountered the next statement in statistics lecture (translated from german):
"From the analysis
you know that all but a countable number of $$w ∈ [0, 1)$$ represent a unique decimal representation
$$ω = 0.x_1x_2x_3...$$
for the countably many exceptions we choose that representation without period 9".
That sounds very counterintuitive and unprovable to me, is this true?
examples-counterexamples decimal-expansion
edited Dec 4 '18 at 14:26
José Carlos Santos
164k22131234
asked Dec 4 '18 at 14:15
AromaTheLoopAromaTheLoop
444
$endgroup$
I have encountered the next statement in statistics lecture (translated from german):
"From the analysis
you know that all but a countable number of $$w ∈ [0, 1)$$ represent a unique decimal representation
$$ω = 0.x_1x_2x_3...$$
for the countably many exceptions we choose that representation without period 9".
That sounds very counterintuitive and unprovable to me, is this true?
examples-counterexamples decimal-expansion
examples-counterexamples decimal-expansion
edited Dec 4 '18 at 14:26
José Carlos Santos
164k22131234
asked Dec 4 '18 at 14:15
AromaTheLoopAromaTheLoop
444
edited Dec 4 '18 at 14:26
José Carlos Santos
164k22131234
asked Dec 4 '18 at 14:15
AromaTheLoopAromaTheLoop
444
edited Dec 4 '18 at 14:26
José Carlos Santos
164k22131234
edited Dec 4 '18 at 14:26
José Carlos Santos
164k22131234
edited Dec 4 '18 at 14:26
José Carlos Santos
164k22131234
164k22131234
asked Dec 4 '18 at 14:15
AromaTheLoopAromaTheLoop
444
asked Dec 4 '18 at 14:15
AromaTheLoopAromaTheLoop
444
asked Dec 4 '18 at 14:15
AromaTheLoopAromaTheLoop
444
444
$begingroup$
"countable" means having the cardinality of $mathbb{N}$. in german: "abzählbar".
$endgroup$
– trancelocation
Dec 4 '18 at 14:19
$begingroup$
What exactly sounds counterintuitive and unprovable? The countability maybe?
$endgroup$
– drhab
Dec 4 '18 at 14:21
$begingroup$
@drhab José Carlos Santos answered my question below.
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:56
add a comment |
$begingroup$
"countable" means having the cardinality of $mathbb{N}$. in german: "abzählbar".
$endgroup$
– trancelocation
Dec 4 '18 at 14:19
$begingroup$
What exactly sounds counterintuitive and unprovable? The countability maybe?
$endgroup$
– drhab
Dec 4 '18 at 14:21
$begingroup$
@drhab José Carlos Santos answered my question below.
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:56
$begingroup$
"countable" means having the cardinality of $mathbb{N}$. in german: "abzählbar".
$endgroup$
– trancelocation
Dec 4 '18 at 14:19
$begingroup$
"countable" means having the cardinality of $mathbb{N}$. in german: "abzählbar".
$endgroup$
– trancelocation
Dec 4 '18 at 14:19
$begingroup$
What exactly sounds counterintuitive and unprovable? The countability maybe?
$endgroup$
– drhab
Dec 4 '18 at 14:21
$begingroup$
What exactly sounds counterintuitive and unprovable? The countability maybe?
$endgroup$
– drhab
Dec 4 '18 at 14:21
$begingroup$
@drhab José Carlos Santos answered my question below.
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:56
$begingroup$
@drhab José Carlos Santos answered my question below.
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:56
add a comment |
1 Answer
1
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oldest
votes
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This is true if “represent a” becomes “is represented by”. The exceptions are the numbers of $(0,1)$ which can be written with finitely many digits: $x=0.d_1d_2ldots d_{n-1}d_n$ (with $d_nin{1,2,3,4,5,6,7,8,9}$), because each such number can be also written as$$0.d_1d_2ldots d_{n-1}(d_n-1)999ldots$$For instance, $0.23=0.22999ldots$ All other numbers have a single decimal expansion.
answered Dec 4 '18 at 14:25
José Carlos SantosJosé Carlos Santos
164k22131234
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is true if “represent a” becomes “is represented by”. The exceptions are the numbers of $(0,1)$ which can be written with finitely many digits: $x=0.d_1d_2ldots d_{n-1}d_n$ (with $d_nin{1,2,3,4,5,6,7,8,9}$), because each such number can be also written as$$0.d_1d_2ldots d_{n-1}(d_n-1)999ldots$$For instance, $0.23=0.22999ldots$ All other numbers have a single decimal expansion.
answered Dec 4 '18 at 14:25
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
add a comment |
$begingroup$
This is true if “represent a” becomes “is represented by”. The exceptions are the numbers of $(0,1)$ which can be written with finitely many digits: $x=0.d_1d_2ldots d_{n-1}d_n$ (with $d_nin{1,2,3,4,5,6,7,8,9}$), because each such number can be also written as$$0.d_1d_2ldots d_{n-1}(d_n-1)999ldots$$For instance, $0.23=0.22999ldots$ All other numbers have a single decimal expansion.
answered Dec 4 '18 at 14:25
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
add a comment |
$begingroup$
This is true if “represent a” becomes “is represented by”. The exceptions are the numbers of $(0,1)$ which can be written with finitely many digits: $x=0.d_1d_2ldots d_{n-1}d_n$ (with $d_nin{1,2,3,4,5,6,7,8,9}$), because each such number can be also written as$$0.d_1d_2ldots d_{n-1}(d_n-1)999ldots$$For instance, $0.23=0.22999ldots$ All other numbers have a single decimal expansion.
answered Dec 4 '18 at 14:25
José Carlos SantosJosé Carlos Santos
164k22131234
$endgroup$
This is true if “represent a” becomes “is represented by”. The exceptions are the numbers of $(0,1)$ which can be written with finitely many digits: $x=0.d_1d_2ldots d_{n-1}d_n$ (with $d_nin{1,2,3,4,5,6,7,8,9}$), because each such number can be also written as$$0.d_1d_2ldots d_{n-1}(d_n-1)999ldots$$For instance, $0.23=0.22999ldots$ All other numbers have a single decimal expansion.
answered Dec 4 '18 at 14:25
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 4 '18 at 14:25
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 4 '18 at 14:25
José Carlos SantosJosé Carlos Santos
164k22131234
answered Dec 4 '18 at 14:25
José Carlos SantosJosé Carlos Santos
164k22131234
164k22131234
add a comment |
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$begingroup$
"countable" means having the cardinality of $mathbb{N}$. in german: "abzählbar".
$endgroup$
– trancelocation
Dec 4 '18 at 14:19
$begingroup$
What exactly sounds counterintuitive and unprovable? The countability maybe?
$endgroup$
– drhab
Dec 4 '18 at 14:21
$begingroup$
@drhab José Carlos Santos answered my question below.
$endgroup$
– AromaTheLoop
Dec 4 '18 at 17:56