Dedekind cut for 1/x
I’m kind of new to this concept and trying to get my head over it.
How do you construct a dedekind cut for $frac{1}{x}$ where $x$ is a positive real number?
elementary-number-theory
asked Nov 27 '18 at 21:54
sdkvsdkv
31
add a comment |
I’m kind of new to this concept and trying to get my head over it.
How do you construct a dedekind cut for $frac{1}{x}$ where $x$ is a positive real number?
elementary-number-theory
asked Nov 27 '18 at 21:54
sdkvsdkv
31
add a comment |
I’m kind of new to this concept and trying to get my head over it.
How do you construct a dedekind cut for $frac{1}{x}$ where $x$ is a positive real number?
elementary-number-theory
asked Nov 27 '18 at 21:54
sdkvsdkv
31
I’m kind of new to this concept and trying to get my head over it.
How do you construct a dedekind cut for $frac{1}{x}$ where $x$ is a positive real number?
elementary-number-theory
elementary-number-theory
asked Nov 27 '18 at 21:54
sdkvsdkv
31
asked Nov 27 '18 at 21:54
sdkvsdkv
31
asked Nov 27 '18 at 21:54
sdkvsdkv
31
asked Nov 27 '18 at 21:54
sdkvsdkv
31
asked Nov 27 '18 at 21:54
sdkvsdkv
31
31
add a comment |
add a comment |
1 Answer
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$newcommand{Q}{mathbb{Q}}$ A Dedekind cut is a partition of $Q$ satisfying certain additional nice properties. So, one way you could do this (somewhat trivially) is by taking $A = {a in Q mid a < frac{1}{x} }$ and $B = {b in Q mid b geq frac{1}{x} }$. Of course, this relies on the assumption that you already have $x$ defined.
answered Nov 27 '18 at 22:09
plattyplatty
3,370320
Okay, I think I get this! But what if I have a DC for x (something like $S= { r in mathbb{Q} |r<x }$, I guess?) and then try to construct a DC using $S$? I thought of something like $S'= { r in mathbb{Q} |rle 0$ or $frac{1}{r}<frac{1}{x}}$ but I'm not sure whether it works
– sdkv
Nov 28 '18 at 7:00
Sure, if you wanted to use $S$, I think $S' = { r in mathbb{Q} mid r leq 0 text{ or } forall s in S, s > 0 implies r < frac{1}{s}}$ should work.
– platty
Nov 28 '18 at 7:08
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$newcommand{Q}{mathbb{Q}}$ A Dedekind cut is a partition of $Q$ satisfying certain additional nice properties. So, one way you could do this (somewhat trivially) is by taking $A = {a in Q mid a < frac{1}{x} }$ and $B = {b in Q mid b geq frac{1}{x} }$. Of course, this relies on the assumption that you already have $x$ defined.
answered Nov 27 '18 at 22:09
plattyplatty
3,370320
Okay, I think I get this! But what if I have a DC for x (something like $S= { r in mathbb{Q} |r<x }$, I guess?) and then try to construct a DC using $S$? I thought of something like $S'= { r in mathbb{Q} |rle 0$ or $frac{1}{r}<frac{1}{x}}$ but I'm not sure whether it works
– sdkv
Nov 28 '18 at 7:00
Sure, if you wanted to use $S$, I think $S' = { r in mathbb{Q} mid r leq 0 text{ or } forall s in S, s > 0 implies r < frac{1}{s}}$ should work.
– platty
Nov 28 '18 at 7:08
add a comment |
$newcommand{Q}{mathbb{Q}}$ A Dedekind cut is a partition of $Q$ satisfying certain additional nice properties. So, one way you could do this (somewhat trivially) is by taking $A = {a in Q mid a < frac{1}{x} }$ and $B = {b in Q mid b geq frac{1}{x} }$. Of course, this relies on the assumption that you already have $x$ defined.
answered Nov 27 '18 at 22:09
plattyplatty
3,370320
Okay, I think I get this! But what if I have a DC for x (something like $S= { r in mathbb{Q} |r<x }$, I guess?) and then try to construct a DC using $S$? I thought of something like $S'= { r in mathbb{Q} |rle 0$ or $frac{1}{r}<frac{1}{x}}$ but I'm not sure whether it works
– sdkv
Nov 28 '18 at 7:00
Sure, if you wanted to use $S$, I think $S' = { r in mathbb{Q} mid r leq 0 text{ or } forall s in S, s > 0 implies r < frac{1}{s}}$ should work.
– platty
Nov 28 '18 at 7:08
add a comment |
$newcommand{Q}{mathbb{Q}}$ A Dedekind cut is a partition of $Q$ satisfying certain additional nice properties. So, one way you could do this (somewhat trivially) is by taking $A = {a in Q mid a < frac{1}{x} }$ and $B = {b in Q mid b geq frac{1}{x} }$. Of course, this relies on the assumption that you already have $x$ defined.
answered Nov 27 '18 at 22:09
plattyplatty
3,370320
$newcommand{Q}{mathbb{Q}}$ A Dedekind cut is a partition of $Q$ satisfying certain additional nice properties. So, one way you could do this (somewhat trivially) is by taking $A = {a in Q mid a < frac{1}{x} }$ and $B = {b in Q mid b geq frac{1}{x} }$. Of course, this relies on the assumption that you already have $x$ defined.
answered Nov 27 '18 at 22:09
plattyplatty
3,370320
answered Nov 27 '18 at 22:09
plattyplatty
3,370320
answered Nov 27 '18 at 22:09
plattyplatty
3,370320
answered Nov 27 '18 at 22:09
plattyplatty
3,370320
3,370320
Okay, I think I get this! But what if I have a DC for x (something like $S= { r in mathbb{Q} |r<x }$, I guess?) and then try to construct a DC using $S$? I thought of something like $S'= { r in mathbb{Q} |rle 0$ or $frac{1}{r}<frac{1}{x}}$ but I'm not sure whether it works
– sdkv
Nov 28 '18 at 7:00
Sure, if you wanted to use $S$, I think $S' = { r in mathbb{Q} mid r leq 0 text{ or } forall s in S, s > 0 implies r < frac{1}{s}}$ should work.
– platty
Nov 28 '18 at 7:08
add a comment |
Okay, I think I get this! But what if I have a DC for x (something like $S= { r in mathbb{Q} |r<x }$, I guess?) and then try to construct a DC using $S$? I thought of something like $S'= { r in mathbb{Q} |rle 0$ or $frac{1}{r}<frac{1}{x}}$ but I'm not sure whether it works
– sdkv
Nov 28 '18 at 7:00
Sure, if you wanted to use $S$, I think $S' = { r in mathbb{Q} mid r leq 0 text{ or } forall s in S, s > 0 implies r < frac{1}{s}}$ should work.
– platty
Nov 28 '18 at 7:08
Okay, I think I get this! But what if I have a DC for x (something like $S= { r in mathbb{Q} |r<x }$, I guess?) and then try to construct a DC using $S$? I thought of something like $S'= { r in mathbb{Q} |rle 0$ or $frac{1}{r}<frac{1}{x}}$ but I'm not sure whether it works
– sdkv
Nov 28 '18 at 7:00
Okay, I think I get this! But what if I have a DC for x (something like $S= { r in mathbb{Q} |r<x }$, I guess?) and then try to construct a DC using $S$? I thought of something like $S'= { r in mathbb{Q} |rle 0$ or $frac{1}{r}<frac{1}{x}}$ but I'm not sure whether it works
– sdkv
Nov 28 '18 at 7:00
Sure, if you wanted to use $S$, I think $S' = { r in mathbb{Q} mid r leq 0 text{ or } forall s in S, s > 0 implies r < frac{1}{s}}$ should work.
– platty
Nov 28 '18 at 7:08
Sure, if you wanted to use $S$, I think $S' = { r in mathbb{Q} mid r leq 0 text{ or } forall s in S, s > 0 implies r < frac{1}{s}}$ should work.
– platty
Nov 28 '18 at 7:08
add a comment |
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