Mixing
Every real number has a decimal representation.
$begingroup$
I was reading this answer explaining why all real number have a decimal representation. I think it is really a nice explanation but I don't really see were (I think it is a little hidden) we use the density of $mathbb Q$ into $mathbb R$ (which I think somehow we should use).
real-numbers decimal-expansion
edited Jan 22 at 17:17
Don Thousand
4,362734
asked Jan 22 at 17:05
roi_saumonroi_saumon
59438
$endgroup$
add a comment |
$begingroup$
I was reading this answer explaining why all real number have a decimal representation. I think it is really a nice explanation but I don't really see were (I think it is a little hidden) we use the density of $mathbb Q$ into $mathbb R$ (which I think somehow we should use).
real-numbers decimal-expansion
edited Jan 22 at 17:17
Don Thousand
4,362734
asked Jan 22 at 17:05
roi_saumonroi_saumon
59438
$endgroup$
add a comment |
$begingroup$
I was reading this answer explaining why all real number have a decimal representation. I think it is really a nice explanation but I don't really see were (I think it is a little hidden) we use the density of $mathbb Q$ into $mathbb R$ (which I think somehow we should use).
real-numbers decimal-expansion
edited Jan 22 at 17:17
Don Thousand
4,362734
asked Jan 22 at 17:05
roi_saumonroi_saumon
59438
$endgroup$
I was reading this answer explaining why all real number have a decimal representation. I think it is really a nice explanation but I don't really see were (I think it is a little hidden) we use the density of $mathbb Q$ into $mathbb R$ (which I think somehow we should use).
real-numbers decimal-expansion
real-numbers decimal-expansion
edited Jan 22 at 17:17
Don Thousand
4,362734
asked Jan 22 at 17:05
roi_saumonroi_saumon
59438
edited Jan 22 at 17:17
Don Thousand
4,362734
asked Jan 22 at 17:05
roi_saumonroi_saumon
59438
edited Jan 22 at 17:17
Don Thousand
4,362734
edited Jan 22 at 17:17
Don Thousand
4,362734
edited Jan 22 at 17:17
Don Thousand
4,362734
4,362734
asked Jan 22 at 17:05
roi_saumonroi_saumon
59438
asked Jan 22 at 17:05
roi_saumonroi_saumon
59438
asked Jan 22 at 17:05
roi_saumonroi_saumon
59438
59438
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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It seems to me the linked answer is virtually a proof that $mathbb Q$ is dense in $mathbb R$. It shows that a subset of $mathbb Q$—fractions with denominators which are powers of $10$—has elements arbitrarily close to any real number. Which makes the same true of $mathbb Q$, thereby proving that $mathbb Q$ is dense in $mathbb R$.
So the reason it doesn't use density of $mathbb Q$ in $mathbb R$ is that it proves it instead.
answered Jan 22 at 18:59
timtfjtimtfj
2,458420
$endgroup$
$begingroup$
Oh, l see. I could use the same reasonning for base b, where $b ge 2$ but instead of dividing the interval into $10$ pieces I divide in $b$ pieces right? For example for base $2$ I just choose if the number is in the left or right of the interval.
$endgroup$
– roi_saumon
Jan 22 at 20:35
1
$begingroup$
Yes, you'd be doing essentially the same thing but end up with a representation in base $b$.
$endgroup$
– timtfj
Jan 22 at 20:52
add a comment |
$begingroup$
It is used nowhere in that proof and I see no reason why it should.
answered Jan 22 at 17:15
José Carlos SantosJosé Carlos Santos
166k22132235
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It seems to me the linked answer is virtually a proof that $mathbb Q$ is dense in $mathbb R$. It shows that a subset of $mathbb Q$—fractions with denominators which are powers of $10$—has elements arbitrarily close to any real number. Which makes the same true of $mathbb Q$, thereby proving that $mathbb Q$ is dense in $mathbb R$.
So the reason it doesn't use density of $mathbb Q$ in $mathbb R$ is that it proves it instead.
answered Jan 22 at 18:59
timtfjtimtfj
2,458420
$endgroup$
$begingroup$
Oh, l see. I could use the same reasonning for base b, where $b ge 2$ but instead of dividing the interval into $10$ pieces I divide in $b$ pieces right? For example for base $2$ I just choose if the number is in the left or right of the interval.
$endgroup$
– roi_saumon
Jan 22 at 20:35
1
$begingroup$
Yes, you'd be doing essentially the same thing but end up with a representation in base $b$.
$endgroup$
– timtfj
Jan 22 at 20:52
add a comment |
$begingroup$
It seems to me the linked answer is virtually a proof that $mathbb Q$ is dense in $mathbb R$. It shows that a subset of $mathbb Q$—fractions with denominators which are powers of $10$—has elements arbitrarily close to any real number. Which makes the same true of $mathbb Q$, thereby proving that $mathbb Q$ is dense in $mathbb R$.
So the reason it doesn't use density of $mathbb Q$ in $mathbb R$ is that it proves it instead.
answered Jan 22 at 18:59
timtfjtimtfj
2,458420
$endgroup$
$begingroup$
Oh, l see. I could use the same reasonning for base b, where $b ge 2$ but instead of dividing the interval into $10$ pieces I divide in $b$ pieces right? For example for base $2$ I just choose if the number is in the left or right of the interval.
$endgroup$
– roi_saumon
Jan 22 at 20:35
1
$begingroup$
Yes, you'd be doing essentially the same thing but end up with a representation in base $b$.
$endgroup$
– timtfj
Jan 22 at 20:52
add a comment |
$begingroup$
It seems to me the linked answer is virtually a proof that $mathbb Q$ is dense in $mathbb R$. It shows that a subset of $mathbb Q$—fractions with denominators which are powers of $10$—has elements arbitrarily close to any real number. Which makes the same true of $mathbb Q$, thereby proving that $mathbb Q$ is dense in $mathbb R$.
So the reason it doesn't use density of $mathbb Q$ in $mathbb R$ is that it proves it instead.
answered Jan 22 at 18:59
timtfjtimtfj
2,458420
$endgroup$
It seems to me the linked answer is virtually a proof that $mathbb Q$ is dense in $mathbb R$. It shows that a subset of $mathbb Q$—fractions with denominators which are powers of $10$—has elements arbitrarily close to any real number. Which makes the same true of $mathbb Q$, thereby proving that $mathbb Q$ is dense in $mathbb R$.
So the reason it doesn't use density of $mathbb Q$ in $mathbb R$ is that it proves it instead.
answered Jan 22 at 18:59
timtfjtimtfj
2,458420
answered Jan 22 at 18:59
timtfjtimtfj
2,458420
answered Jan 22 at 18:59
timtfjtimtfj
2,458420
answered Jan 22 at 18:59
timtfjtimtfj
2,458420
2,458420
$begingroup$
Oh, l see. I could use the same reasonning for base b, where $b ge 2$ but instead of dividing the interval into $10$ pieces I divide in $b$ pieces right? For example for base $2$ I just choose if the number is in the left or right of the interval.
$endgroup$
– roi_saumon
Jan 22 at 20:35
1
$begingroup$
Yes, you'd be doing essentially the same thing but end up with a representation in base $b$.
$endgroup$
– timtfj
Jan 22 at 20:52
add a comment |
$begingroup$
Oh, l see. I could use the same reasonning for base b, where $b ge 2$ but instead of dividing the interval into $10$ pieces I divide in $b$ pieces right? For example for base $2$ I just choose if the number is in the left or right of the interval.
$endgroup$
– roi_saumon
Jan 22 at 20:35
1
$begingroup$
Yes, you'd be doing essentially the same thing but end up with a representation in base $b$.
$endgroup$
– timtfj
Jan 22 at 20:52
$begingroup$
Oh, l see. I could use the same reasonning for base b, where $b ge 2$ but instead of dividing the interval into $10$ pieces I divide in $b$ pieces right? For example for base $2$ I just choose if the number is in the left or right of the interval.
$endgroup$
– roi_saumon
Jan 22 at 20:35
$begingroup$
Oh, l see. I could use the same reasonning for base b, where $b ge 2$ but instead of dividing the interval into $10$ pieces I divide in $b$ pieces right? For example for base $2$ I just choose if the number is in the left or right of the interval.
$endgroup$
– roi_saumon
Jan 22 at 20:35
1
1
$begingroup$
Yes, you'd be doing essentially the same thing but end up with a representation in base $b$.
$endgroup$
– timtfj
Jan 22 at 20:52
$begingroup$
Yes, you'd be doing essentially the same thing but end up with a representation in base $b$.
$endgroup$
– timtfj
Jan 22 at 20:52
add a comment |
$begingroup$
It is used nowhere in that proof and I see no reason why it should.
answered Jan 22 at 17:15
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
It is used nowhere in that proof and I see no reason why it should.
answered Jan 22 at 17:15
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
add a comment |
$begingroup$
It is used nowhere in that proof and I see no reason why it should.
answered Jan 22 at 17:15
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
It is used nowhere in that proof and I see no reason why it should.
answered Jan 22 at 17:15
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 17:15
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 17:15
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 17:15
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
add a comment |
add a comment |
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