Archimedean equivalence. Proof of $ [g+h] geq min{ [g], [h] }$

0

$begingroup$

Prove that: $ [g+h] geq min{[g],[h]}$

Definitons:
Take any ordered abelian field $(G, leq)$. Two elements $g,h in G$ are archimedean equivalent if $exists n,n' in Bbb{N}$ : $n|g| geq |h|$ and $n'|h| geq |g|$.
$[g]$ denotes the equivalence class of $g$.

Attempt of proof:
Suppose that $ min{ [g], [h] }= [h]$

$[h] = {g: exists n_1, n_1' $such that$ n_1|h| geq |g| $and$ n_1'|g| geq |h|} $

$[h+g] ={ k: exists n_2, n_2' $such that$ n_2|g+h| geq |k| $and$ n_2'|k| geq |g+h| }$

I have to show that $k>g$ so I have to show that $ng<k$ for all $n in Bbb{N}$. From my assumption $[h] < [g]$ so it implies $|h| > |g|$ and $[g] neq[h]$. I am not sure what to do know. I have tried to multiply these inequalities by an inverse element for example of $n_1, n_1'$. I can also use the fact: $[g] geq [h]$ for all $h in G$ iff $g=0$.

real-analysis order-theory

share|cite|improve this question

edited Jan 22 at 17:32

Hikicianka

asked Jan 22 at 15:39

HikiciankaHikicianka

1498

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use $min$ for $min$.
  $endgroup$
  – Shaun
  Jan 22 at 15:46
  

  
  
  
  

add a comment | 

0

$begingroup$

Prove that: $ [g+h] geq min{[g],[h]}$

Definitons:
Take any ordered abelian field $(G, leq)$. Two elements $g,h in G$ are archimedean equivalent if $exists n,n' in Bbb{N}$ : $n|g| geq |h|$ and $n'|h| geq |g|$.
$[g]$ denotes the equivalence class of $g$.

Attempt of proof:
Suppose that $ min{ [g], [h] }= [h]$

$[h] = {g: exists n_1, n_1' $such that$ n_1|h| geq |g| $and$ n_1'|g| geq |h|} $

$[h+g] ={ k: exists n_2, n_2' $such that$ n_2|g+h| geq |k| $and$ n_2'|k| geq |g+h| }$

I have to show that $k>g$ so I have to show that $ng<k$ for all $n in Bbb{N}$. From my assumption $[h] < [g]$ so it implies $|h| > |g|$ and $[g] neq[h]$. I am not sure what to do know. I have tried to multiply these inequalities by an inverse element for example of $n_1, n_1'$. I can also use the fact: $[g] geq [h]$ for all $h in G$ iff $g=0$.

real-analysis order-theory

share|cite|improve this question

edited Jan 22 at 17:32

Hikicianka

asked Jan 22 at 15:39

HikiciankaHikicianka

1498

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use $min$ for $min$.
  $endgroup$
  – Shaun
  Jan 22 at 15:46
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

Prove that: $ [g+h] geq min{[g],[h]}$

Definitons:
Take any ordered abelian field $(G, leq)$. Two elements $g,h in G$ are archimedean equivalent if $exists n,n' in Bbb{N}$ : $n|g| geq |h|$ and $n'|h| geq |g|$.
$[g]$ denotes the equivalence class of $g$.

Attempt of proof:
Suppose that $ min{ [g], [h] }= [h]$

$[h] = {g: exists n_1, n_1' $such that$ n_1|h| geq |g| $and$ n_1'|g| geq |h|} $

$[h+g] ={ k: exists n_2, n_2' $such that$ n_2|g+h| geq |k| $and$ n_2'|k| geq |g+h| }$

I have to show that $k>g$ so I have to show that $ng<k$ for all $n in Bbb{N}$. From my assumption $[h] < [g]$ so it implies $|h| > |g|$ and $[g] neq[h]$. I am not sure what to do know. I have tried to multiply these inequalities by an inverse element for example of $n_1, n_1'$. I can also use the fact: $[g] geq [h]$ for all $h in G$ iff $g=0$.

real-analysis order-theory

share|cite|improve this question

edited Jan 22 at 17:32

Hikicianka

asked Jan 22 at 15:39

HikiciankaHikicianka

1498

$endgroup$

Prove that: $ [g+h] geq min{[g],[h]}$

Definitons:
Take any ordered abelian field $(G, leq)$. Two elements $g,h in G$ are archimedean equivalent if $exists n,n' in Bbb{N}$ : $n|g| geq |h|$ and $n'|h| geq |g|$.
$[g]$ denotes the equivalence class of $g$.

Attempt of proof:
Suppose that $ min{ [g], [h] }= [h]$

$[h] = {g: exists n_1, n_1' $such that$ n_1|h| geq |g| $and$ n_1'|g| geq |h|} $

$[h+g] ={ k: exists n_2, n_2' $such that$ n_2|g+h| geq |k| $and$ n_2'|k| geq |g+h| }$

I have to show that $k>g$ so I have to show that $ng<k$ for all $n in Bbb{N}$. From my assumption $[h] < [g]$ so it implies $|h| > |g|$ and $[g] neq[h]$. I am not sure what to do know. I have tried to multiply these inequalities by an inverse element for example of $n_1, n_1'$. I can also use the fact: $[g] geq [h]$ for all $h in G$ iff $g=0$.

real-analysis order-theory

real-analysis order-theory

share|cite|improve this question

edited Jan 22 at 17:32

Hikicianka

asked Jan 22 at 15:39

HikiciankaHikicianka

1498

share|cite|improve this question

edited Jan 22 at 17:32

Hikicianka

asked Jan 22 at 15:39

HikiciankaHikicianka

1498

share|cite|improve this question

share|cite|improve this question

edited Jan 22 at 17:32

Hikicianka

edited Jan 22 at 17:32

Hikicianka

edited Jan 22 at 17:32

Hikicianka

asked Jan 22 at 15:39

HikiciankaHikicianka

1498

asked Jan 22 at 15:39

HikiciankaHikicianka

1498

asked Jan 22 at 15:39

HikiciankaHikicianka

1498

1498

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use $min$ for $min$.
  $endgroup$
  – Shaun
  Jan 22 at 15:46
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Use $min$ for $min$.
  $endgroup$
  – Shaun
  Jan 22 at 15:46
  

  
  
  
  

$begingroup$
Use $min$ for $min$.
$endgroup$
– Shaun
Jan 22 at 15:46

$begingroup$
Use $min$ for $min$.
$endgroup$
– Shaun
Jan 22 at 15:46

add a comment | 

1 Answer
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$begingroup$

Assume that $ [g+h] < min([g],[h]) $. Let $ [g] = min([g], [h]) Longrightarrow [g] > [h] Longleftrightarrow forall_{n in mathbb{N}} ; |g| > n|h|$.

From assumption we have $ ; [g + h] < [g] Longrightarrow |g| < |g + h| ; text{and} ; [g + h] neq [g] $.

We know that
$$
|g| < |g + h| Longleftrightarrow forall_{n in mathbb{N}} ; n|g| < |g + h| leq |g| + |h| Longrightarrow (n - 1)|g| < |g + h| - |g| leq |h| \
Longrightarrow (n - 1)|g| < |h|
$$
So we have $$forall_{n in mathbb{N}} ; (n - 1)|g| < |h| quad wedge quad |g| > n|h| $$
Hence it doesn't true then $ [g+h] geq min([g],[h]) $. $square$

share|cite|improve this answer

answered Feb 6 at 19:39

nowek7nowek7

383

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add a comment | 

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1 Answer
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1 Answer
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active

oldest

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active

oldest

votes

1

$begingroup$

Assume that $ [g+h] < min([g],[h]) $. Let $ [g] = min([g], [h]) Longrightarrow [g] > [h] Longleftrightarrow forall_{n in mathbb{N}} ; |g| > n|h|$.

From assumption we have $ ; [g + h] < [g] Longrightarrow |g| < |g + h| ; text{and} ; [g + h] neq [g] $.

We know that
$$
|g| < |g + h| Longleftrightarrow forall_{n in mathbb{N}} ; n|g| < |g + h| leq |g| + |h| Longrightarrow (n - 1)|g| < |g + h| - |g| leq |h| \
Longrightarrow (n - 1)|g| < |h|
$$
So we have $$forall_{n in mathbb{N}} ; (n - 1)|g| < |h| quad wedge quad |g| > n|h| $$
Hence it doesn't true then $ [g+h] geq min([g],[h]) $. $square$

share|cite|improve this answer

answered Feb 6 at 19:39

nowek7nowek7

383

$endgroup$

add a comment | 

1

$begingroup$

Assume that $ [g+h] < min([g],[h]) $. Let $ [g] = min([g], [h]) Longrightarrow [g] > [h] Longleftrightarrow forall_{n in mathbb{N}} ; |g| > n|h|$.

From assumption we have $ ; [g + h] < [g] Longrightarrow |g| < |g + h| ; text{and} ; [g + h] neq [g] $.

We know that
$$
|g| < |g + h| Longleftrightarrow forall_{n in mathbb{N}} ; n|g| < |g + h| leq |g| + |h| Longrightarrow (n - 1)|g| < |g + h| - |g| leq |h| \
Longrightarrow (n - 1)|g| < |h|
$$
So we have $$forall_{n in mathbb{N}} ; (n - 1)|g| < |h| quad wedge quad |g| > n|h| $$
Hence it doesn't true then $ [g+h] geq min([g],[h]) $. $square$

share|cite|improve this answer

answered Feb 6 at 19:39

nowek7nowek7

383

$endgroup$

add a comment | 

1

1

1

$begingroup$

Assume that $ [g+h] < min([g],[h]) $. Let $ [g] = min([g], [h]) Longrightarrow [g] > [h] Longleftrightarrow forall_{n in mathbb{N}} ; |g| > n|h|$.

From assumption we have $ ; [g + h] < [g] Longrightarrow |g| < |g + h| ; text{and} ; [g + h] neq [g] $.

We know that
$$
|g| < |g + h| Longleftrightarrow forall_{n in mathbb{N}} ; n|g| < |g + h| leq |g| + |h| Longrightarrow (n - 1)|g| < |g + h| - |g| leq |h| \
Longrightarrow (n - 1)|g| < |h|
$$
So we have $$forall_{n in mathbb{N}} ; (n - 1)|g| < |h| quad wedge quad |g| > n|h| $$
Hence it doesn't true then $ [g+h] geq min([g],[h]) $. $square$

share|cite|improve this answer

answered Feb 6 at 19:39

nowek7nowek7

383

$endgroup$

Assume that $ [g+h] < min([g],[h]) $. Let $ [g] = min([g], [h]) Longrightarrow [g] > [h] Longleftrightarrow forall_{n in mathbb{N}} ; |g| > n|h|$.

From assumption we have $ ; [g + h] < [g] Longrightarrow |g| < |g + h| ; text{and} ; [g + h] neq [g] $.

We know that
$$
|g| < |g + h| Longleftrightarrow forall_{n in mathbb{N}} ; n|g| < |g + h| leq |g| + |h| Longrightarrow (n - 1)|g| < |g + h| - |g| leq |h| \
Longrightarrow (n - 1)|g| < |h|
$$
So we have $$forall_{n in mathbb{N}} ; (n - 1)|g| < |h| quad wedge quad |g| > n|h| $$
Hence it doesn't true then $ [g+h] geq min([g],[h]) $. $square$

share|cite|improve this answer

answered Feb 6 at 19:39

nowek7nowek7

383

share|cite|improve this answer

share|cite|improve this answer

answered Feb 6 at 19:39

nowek7nowek7

383

answered Feb 6 at 19:39

nowek7nowek7

383

answered Feb 6 at 19:39

nowek7nowek7

383

383

add a comment | 

add a comment | 

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