Find a set of rational numbers where $sqrt3$ is the infimum. Prove itFind the limit if it exists of $S_n+1 = frac12(S_n +fracAS_n)$Infinite sums involving $pi$ and rational numbers.Troubles calculating a set infimumProof that the set of irrational numbers is dense in realsProof of the infinitude of rational and irrational numbersTest function to distinguish between irrational and rational numbersWhich set is more dense: set of irrational numbers or set of rational numbers?Approximating $sqrt2$ in rational numbersRudin exercise 2.18: perfect set with only irrational numbers.Trying to define the set of alternate derivatives associated with the set of periodic functions with some rational number periodInfimum of the set $xin mathbbQ;$?
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Find a set of rational numbers where $sqrt3$ is the infimum. Prove it
Find the limit if it exists of $S_n+1 = frac12(S_n +fracAS_n)$Infinite sums involving $pi$ and rational numbers.Troubles calculating a set infimumProof that the set of irrational numbers is dense in realsProof of the infinitude of rational and irrational numbersTest function to distinguish between irrational and rational numbersWhich set is more dense: set of irrational numbers or set of rational numbers?Approximating $sqrt2$ in rational numbersRudin exercise 2.18: perfect set with only irrational numbers.Trying to define the set of alternate derivatives associated with the set of periodic functions with some rational number periodInfimum of the set $xin mathbbQ;$?
$begingroup$
I thought about the expansion of $sqrt3$ as a series. But I didn't get anything useful. Also, I thought about a set with irrational numbers instead of rational numbers.
What is the general idea to attack this kind of problems? Since is the first one I'm doing of this type.
real-analysis calculus supremum-and-infimum
asked Mar 30 at 4:34
ClaraGarcíaClaraGarcía
12
$endgroup$
add a comment |
$begingroup$
I thought about the expansion of $sqrt3$ as a series. But I didn't get anything useful. Also, I thought about a set with irrational numbers instead of rational numbers.
What is the general idea to attack this kind of problems? Since is the first one I'm doing of this type.
real-analysis calculus supremum-and-infimum
asked Mar 30 at 4:34
ClaraGarcíaClaraGarcía
12
$endgroup$
$begingroup$
Hint: Consider $$(sqrt 3,2) cap mathbbQ$$
$endgroup$
– Story123
Mar 30 at 7:47
$begingroup$
As per this question, it is possible to build a sequence (and a set) for any $sqrtn$.
$endgroup$
– rtybase
Mar 30 at 9:26
add a comment |
$begingroup$
I thought about the expansion of $sqrt3$ as a series. But I didn't get anything useful. Also, I thought about a set with irrational numbers instead of rational numbers.
What is the general idea to attack this kind of problems? Since is the first one I'm doing of this type.
real-analysis calculus supremum-and-infimum
asked Mar 30 at 4:34
ClaraGarcíaClaraGarcía
12
$endgroup$
I thought about the expansion of $sqrt3$ as a series. But I didn't get anything useful. Also, I thought about a set with irrational numbers instead of rational numbers.
What is the general idea to attack this kind of problems? Since is the first one I'm doing of this type.
real-analysis calculus supremum-and-infimum
real-analysis calculus supremum-and-infimum
asked Mar 30 at 4:34
ClaraGarcíaClaraGarcía
12
asked Mar 30 at 4:34
ClaraGarcíaClaraGarcía
12
edited Apr 1 at 0:32
ClaraGarcía
asked Mar 30 at 4:34
ClaraGarcíaClaraGarcía
12
asked Mar 30 at 4:34
ClaraGarcíaClaraGarcía
12
asked Mar 30 at 4:34
ClaraGarcíaClaraGarcía
12
12
$begingroup$
Hint: Consider $$(sqrt 3,2) cap mathbbQ$$
$endgroup$
– Story123
Mar 30 at 7:47
$begingroup$
As per this question, it is possible to build a sequence (and a set) for any $sqrtn$.
$endgroup$
– rtybase
Mar 30 at 9:26
add a comment |
$begingroup$
Hint: Consider $$(sqrt 3,2) cap mathbbQ$$
$endgroup$
– Story123
Mar 30 at 7:47
$begingroup$
As per this question, it is possible to build a sequence (and a set) for any $sqrtn$.
$endgroup$
– rtybase
Mar 30 at 9:26
$begingroup$
Hint: Consider $$(sqrt 3,2) cap mathbbQ$$
$endgroup$
– Story123
Mar 30 at 7:47
$begingroup$
Hint: Consider $$(sqrt 3,2) cap mathbbQ$$
$endgroup$
– Story123
Mar 30 at 7:47
$begingroup$
As per this question, it is possible to build a sequence (and a set) for any $sqrtn$.
$endgroup$
– rtybase
Mar 30 at 9:26
$begingroup$
As per this question, it is possible to build a sequence (and a set) for any $sqrtn$.
$endgroup$
– rtybase
Mar 30 at 9:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Apply Newton's Method to the function $f(x)=x^2-3$, starting with your favorite rational number that you are sure is $> sqrt3$.
answered Mar 30 at 4:39
Lee MosherLee Mosher
51.8k33889
$endgroup$
$begingroup$
Thanks, I will try it.
$endgroup$
– ClaraGarcía
Mar 30 at 5:07
add a comment |
$begingroup$
A simple example of a subset of $Bbb Q$ with infimum $sqrt3$, and which
does not presuppose existence of $sqrt3$ or of any irrational numbers is
$$ainBbb Q:a>0quadtextandquad a^2>3.$$
answered Mar 30 at 5:25
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
yep this is probably the simplest construction
$endgroup$
– qwr
Mar 30 at 5:32
add a comment |
$begingroup$
This isn't very educational, but $a_n = textceil(sqrt3 * 10^n)/10^n$ would work. So
$a_1 = textceil(1.73205... * 10)/10 = textceil(17.3205...)/10 = 18/10 = 1.8$
$a_2 = textceil(1.73205... * 100)/100 = textceil(173.205...)/100 = 174/100 = 1.74$
$a_3 = textceil(1.73205... * 1000)/1000 = textceil(1732.05...)/1000 = 1733/1000 = 1.733$
This would obviously work for any irrational number in the same way.
answered Mar 30 at 4:41
user2825632user2825632
2,6761617
$endgroup$
$begingroup$
Thanks!! Now, to prove it, it seems a little weird to use the definition of infimum. Let $S$ be the set, so for every $s in S$ we have: $s geq sqrt3 + varepsilon$ and now $a_n$ is our $s$, right?
$endgroup$
– ClaraGarcía
Mar 30 at 4:55
$begingroup$
Yep, that makes sense.
$endgroup$
– user2825632
Mar 30 at 5:01
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Apply Newton's Method to the function $f(x)=x^2-3$, starting with your favorite rational number that you are sure is $> sqrt3$.
answered Mar 30 at 4:39
Lee MosherLee Mosher
51.8k33889
$endgroup$
$begingroup$
Thanks, I will try it.
$endgroup$
– ClaraGarcía
Mar 30 at 5:07
add a comment |
$begingroup$
Hint: Apply Newton's Method to the function $f(x)=x^2-3$, starting with your favorite rational number that you are sure is $> sqrt3$.
answered Mar 30 at 4:39
Lee MosherLee Mosher
51.8k33889
$endgroup$
$begingroup$
Thanks, I will try it.
$endgroup$
– ClaraGarcía
Mar 30 at 5:07
add a comment |
$begingroup$
Hint: Apply Newton's Method to the function $f(x)=x^2-3$, starting with your favorite rational number that you are sure is $> sqrt3$.
answered Mar 30 at 4:39
Lee MosherLee Mosher
51.8k33889
$endgroup$
Hint: Apply Newton's Method to the function $f(x)=x^2-3$, starting with your favorite rational number that you are sure is $> sqrt3$.
answered Mar 30 at 4:39
Lee MosherLee Mosher
51.8k33889
answered Mar 30 at 4:39
Lee MosherLee Mosher
51.8k33889
answered Mar 30 at 4:39
Lee MosherLee Mosher
51.8k33889
answered Mar 30 at 4:39
Lee MosherLee Mosher
51.8k33889
51.8k33889
$begingroup$
Thanks, I will try it.
$endgroup$
– ClaraGarcía
Mar 30 at 5:07
add a comment |
$begingroup$
Thanks, I will try it.
$endgroup$
– ClaraGarcía
Mar 30 at 5:07
$begingroup$
Thanks, I will try it.
$endgroup$
– ClaraGarcía
Mar 30 at 5:07
$begingroup$
Thanks, I will try it.
$endgroup$
– ClaraGarcía
Mar 30 at 5:07
add a comment |
$begingroup$
A simple example of a subset of $Bbb Q$ with infimum $sqrt3$, and which
does not presuppose existence of $sqrt3$ or of any irrational numbers is
$$ainBbb Q:a>0quadtextandquad a^2>3.$$
answered Mar 30 at 5:25
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
yep this is probably the simplest construction
$endgroup$
– qwr
Mar 30 at 5:32
add a comment |
$begingroup$
A simple example of a subset of $Bbb Q$ with infimum $sqrt3$, and which
does not presuppose existence of $sqrt3$ or of any irrational numbers is
$$ainBbb Q:a>0quadtextandquad a^2>3.$$
answered Mar 30 at 5:25
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
$begingroup$
yep this is probably the simplest construction
$endgroup$
– qwr
Mar 30 at 5:32
add a comment |
$begingroup$
A simple example of a subset of $Bbb Q$ with infimum $sqrt3$, and which
does not presuppose existence of $sqrt3$ or of any irrational numbers is
$$ainBbb Q:a>0quadtextandquad a^2>3.$$
answered Mar 30 at 5:25
Lord Shark the UnknownLord Shark the Unknown
108k1162136
$endgroup$
A simple example of a subset of $Bbb Q$ with infimum $sqrt3$, and which
does not presuppose existence of $sqrt3$ or of any irrational numbers is
$$ainBbb Q:a>0quadtextandquad a^2>3.$$
answered Mar 30 at 5:25
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 30 at 5:25
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 30 at 5:25
Lord Shark the UnknownLord Shark the Unknown
108k1162136
answered Mar 30 at 5:25
Lord Shark the UnknownLord Shark the Unknown
108k1162136
108k1162136
$begingroup$
yep this is probably the simplest construction
$endgroup$
– qwr
Mar 30 at 5:32
add a comment |
$begingroup$
yep this is probably the simplest construction
$endgroup$
– qwr
Mar 30 at 5:32
$begingroup$
yep this is probably the simplest construction
$endgroup$
– qwr
Mar 30 at 5:32
$begingroup$
yep this is probably the simplest construction
$endgroup$
– qwr
Mar 30 at 5:32
add a comment |
$begingroup$
This isn't very educational, but $a_n = textceil(sqrt3 * 10^n)/10^n$ would work. So
$a_1 = textceil(1.73205... * 10)/10 = textceil(17.3205...)/10 = 18/10 = 1.8$
$a_2 = textceil(1.73205... * 100)/100 = textceil(173.205...)/100 = 174/100 = 1.74$
$a_3 = textceil(1.73205... * 1000)/1000 = textceil(1732.05...)/1000 = 1733/1000 = 1.733$
This would obviously work for any irrational number in the same way.
answered Mar 30 at 4:41
user2825632user2825632
2,6761617
$endgroup$
$begingroup$
Thanks!! Now, to prove it, it seems a little weird to use the definition of infimum. Let $S$ be the set, so for every $s in S$ we have: $s geq sqrt3 + varepsilon$ and now $a_n$ is our $s$, right?
$endgroup$
– ClaraGarcía
Mar 30 at 4:55
$begingroup$
Yep, that makes sense.
$endgroup$
– user2825632
Mar 30 at 5:01
add a comment |
$begingroup$
This isn't very educational, but $a_n = textceil(sqrt3 * 10^n)/10^n$ would work. So
$a_1 = textceil(1.73205... * 10)/10 = textceil(17.3205...)/10 = 18/10 = 1.8$
$a_2 = textceil(1.73205... * 100)/100 = textceil(173.205...)/100 = 174/100 = 1.74$
$a_3 = textceil(1.73205... * 1000)/1000 = textceil(1732.05...)/1000 = 1733/1000 = 1.733$
This would obviously work for any irrational number in the same way.
answered Mar 30 at 4:41
user2825632user2825632
2,6761617
$endgroup$
$begingroup$
Thanks!! Now, to prove it, it seems a little weird to use the definition of infimum. Let $S$ be the set, so for every $s in S$ we have: $s geq sqrt3 + varepsilon$ and now $a_n$ is our $s$, right?
$endgroup$
– ClaraGarcía
Mar 30 at 4:55
$begingroup$
Yep, that makes sense.
$endgroup$
– user2825632
Mar 30 at 5:01
add a comment |
$begingroup$
This isn't very educational, but $a_n = textceil(sqrt3 * 10^n)/10^n$ would work. So
$a_1 = textceil(1.73205... * 10)/10 = textceil(17.3205...)/10 = 18/10 = 1.8$
$a_2 = textceil(1.73205... * 100)/100 = textceil(173.205...)/100 = 174/100 = 1.74$
$a_3 = textceil(1.73205... * 1000)/1000 = textceil(1732.05...)/1000 = 1733/1000 = 1.733$
This would obviously work for any irrational number in the same way.
answered Mar 30 at 4:41
user2825632user2825632
2,6761617
$endgroup$
This isn't very educational, but $a_n = textceil(sqrt3 * 10^n)/10^n$ would work. So
$a_1 = textceil(1.73205... * 10)/10 = textceil(17.3205...)/10 = 18/10 = 1.8$
$a_2 = textceil(1.73205... * 100)/100 = textceil(173.205...)/100 = 174/100 = 1.74$
$a_3 = textceil(1.73205... * 1000)/1000 = textceil(1732.05...)/1000 = 1733/1000 = 1.733$
This would obviously work for any irrational number in the same way.
answered Mar 30 at 4:41
user2825632user2825632
2,6761617
edited Mar 30 at 4:44
answered Mar 30 at 4:41
user2825632user2825632
2,6761617
answered Mar 30 at 4:41
user2825632user2825632
2,6761617
answered Mar 30 at 4:41
user2825632user2825632
2,6761617
2,6761617
$begingroup$
Thanks!! Now, to prove it, it seems a little weird to use the definition of infimum. Let $S$ be the set, so for every $s in S$ we have: $s geq sqrt3 + varepsilon$ and now $a_n$ is our $s$, right?
$endgroup$
– ClaraGarcía
Mar 30 at 4:55
$begingroup$
Yep, that makes sense.
$endgroup$
– user2825632
Mar 30 at 5:01
add a comment |
$begingroup$
Thanks!! Now, to prove it, it seems a little weird to use the definition of infimum. Let $S$ be the set, so for every $s in S$ we have: $s geq sqrt3 + varepsilon$ and now $a_n$ is our $s$, right?
$endgroup$
– ClaraGarcía
Mar 30 at 4:55
$begingroup$
Yep, that makes sense.
$endgroup$
– user2825632
Mar 30 at 5:01
$begingroup$
Thanks!! Now, to prove it, it seems a little weird to use the definition of infimum. Let $S$ be the set, so for every $s in S$ we have: $s geq sqrt3 + varepsilon$ and now $a_n$ is our $s$, right?
$endgroup$
– ClaraGarcía
Mar 30 at 4:55
$begingroup$
Thanks!! Now, to prove it, it seems a little weird to use the definition of infimum. Let $S$ be the set, so for every $s in S$ we have: $s geq sqrt3 + varepsilon$ and now $a_n$ is our $s$, right?
$endgroup$
– ClaraGarcía
Mar 30 at 4:55
$begingroup$
Yep, that makes sense.
$endgroup$
– user2825632
Mar 30 at 5:01
$begingroup$
Yep, that makes sense.
$endgroup$
– user2825632
Mar 30 at 5:01
add a comment |
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$begingroup$
Hint: Consider $$(sqrt 3,2) cap mathbbQ$$
$endgroup$
– Story123
Mar 30 at 7:47
$begingroup$
As per this question, it is possible to build a sequence (and a set) for any $sqrtn$.
$endgroup$
– rtybase
Mar 30 at 9:26