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Prove that $forall epsilon>0, epsilon>a implies 0 geq a$

0

1

$begingroup$

I am doing a course on basic real analysis in which firstly i am emphasising on real numbers. My book says that real number satisfies the following axioms.

---

1)Field Axiom
2)Extend Axiom
3)Order Axiom
4)Completeness axiom.

---

While doing the exercise of 3) I found following sets of questions.

$forall epsilon>0, epsilon>a implies 0 geq a$

$forall epsilon<0, epsilon<a implies a geq 0$

For the first part I think like this: however small is $epsilon$ if positive and as $epsilon>a$ so $a$ has to be negative.
But I am unable to write the formal proof.

real-analysis proof-writing real-numbers foundations

share|cite|improve this question

edited Mar 26 at 4:46

J. W. Tanner

4,7871420

asked Mar 25 at 21:14

M DesmondM Desmond

3228

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try the contrapositive: If $a>0$, can you find positive $epsilon$ that is not greater than $a$?
  $endgroup$
  – J. W. Tanner
  Mar 25 at 21:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i think $epsilon=fracan$ for any $n>1$ will work fine,but what about the 2nd question?
  $endgroup$
  – M Desmond
  Mar 25 at 21:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What about $fraca1000?$ Or other possibility.
  $endgroup$
  – mfl
  Mar 25 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think the 2nd question is correct as written. Take $a=-1$. There exists negative $epsilon$ (say $-2$) less than $-1$, but that does not mean $-1ge0$
  $endgroup$
  – J. W. Tanner
  Mar 25 at 21:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I do think so
  $endgroup$
  – J. W. Tanner
  Mar 25 at 22:11
  

  
  
  
  

 | 
show 1 more comment

0

1

$begingroup$

I am doing a course on basic real analysis in which firstly i am emphasising on real numbers. My book says that real number satisfies the following axioms.

---

1)Field Axiom
2)Extend Axiom
3)Order Axiom
4)Completeness axiom.

---

While doing the exercise of 3) I found following sets of questions.

$forall epsilon>0, epsilon>a implies 0 geq a$

$forall epsilon<0, epsilon<a implies a geq 0$

For the first part I think like this: however small is $epsilon$ if positive and as $epsilon>a$ so $a$ has to be negative.
But I am unable to write the formal proof.

real-analysis proof-writing real-numbers foundations

share|cite|improve this question

edited Mar 26 at 4:46

J. W. Tanner

4,7871420

asked Mar 25 at 21:14

M DesmondM Desmond

3228

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Try the contrapositive: If $a>0$, can you find positive $epsilon$ that is not greater than $a$?
  $endgroup$
  – J. W. Tanner
  Mar 25 at 21:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i think $epsilon=fracan$ for any $n>1$ will work fine,but what about the 2nd question?
  $endgroup$
  – M Desmond
  Mar 25 at 21:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What about $fraca1000?$ Or other possibility.
  $endgroup$
  – mfl
  Mar 25 at 21:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I don't think the 2nd question is correct as written. Take $a=-1$. There exists negative $epsilon$ (say $-2$) less than $-1$, but that does not mean $-1ge0$
  $endgroup$
  – J. W. Tanner
  Mar 25 at 21:45
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I do think so
  $endgroup$
  – J. W. Tanner
  Mar 25 at 22:11
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

I am doing a course on basic real analysis in which firstly i am emphasising on real numbers. My book says that real number satisfies the following axioms.

---

1)Field Axiom
2)Extend Axiom
3)Order Axiom
4)Completeness axiom.

---

While doing the exercise of 3) I found following sets of questions.

$forall epsilon>0, epsilon>a implies 0 geq a$

$forall epsilon<0, epsilon<a implies a geq 0$

For the first part I think like this: however small is $epsilon$ if positive and as $epsilon>a$ so $a$ has to be negative.
But I am unable to write the formal proof.

real-analysis proof-writing real-numbers foundations

share|cite|improve this question

edited Mar 26 at 4:46

J. W. Tanner

4,7871420

asked Mar 25 at 21:14

M DesmondM Desmond

3228

$endgroup$

share|cite|improve this question

edited Mar 26 at 4:46

J. W. Tanner

4,7871420

asked Mar 25 at 21:14

M DesmondM Desmond

3228

share|cite|improve this question

edited Mar 26 at 4:46

J. W. Tanner

4,7871420

asked Mar 25 at 21:14

M DesmondM Desmond

3228

edited Mar 26 at 4:46

J. W. Tanner

4,7871420

edited Mar 26 at 4:46

J. W. Tanner

4,7871420

asked Mar 25 at 21:14

M DesmondM Desmond

3228

asked Mar 25 at 21:14

M DesmondM Desmond

3228

2 Answers
2

active

oldest

votes

1

$begingroup$

If $a>0$, then take $epsilon=a/2,$ so $0<epsilon<a,$ so it is not true that for all $epsilon>0$, $epsilon>a$. The contrapositive of that is what you wanted for the first part.

If $a<0$, then take $epsilon=a/2$ so $0>epsilon>a$ so it is not true that for all $epsilon<0$, $epsilon<a$. The contrapositive of that is that if for all $epsilon<0$, $epsilon<a$, then $age0$.

share|cite|improve this answer

answered Mar 25 at 21:54

J. W. TannerJ. W. Tanner

4,7871420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir tell me one thing , what is the negation of $exists epsilon < 0 , epsilon < a$ ? Is it $forall epsilon > 0, epsilon > a$ ?
  $endgroup$
  – M Desmond
  Mar 25 at 22:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The negation of that is $forallepsilonbf <$$0, epsilonge a$
  $endgroup$
  – J. W. Tanner
  Mar 25 at 22:27
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I can't see your axioms, nor can I parse the strange order of your highlighted line, but I'd prove the title theorem's contrapositive. If $a> 0$ (which by order, I guess, is equivalent to $0notge a$) then the positive choice $epsilon:=a/2$ contradicts $epsilon>a$ (by whatever axiomatic proof you'd manage of $a>0implies a>a/2>0$).

share|cite|improve this answer

answered Mar 25 at 21:19

J.G.J.G.

33.5k23252

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please see the link i have provided below for axioms,and see the order axioms in that link.
  $endgroup$
  – M Desmond
  Mar 25 at 21:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MDesmond Your link gave me a 404 error, but in any case your question ought to be self-contained.
  $endgroup$
  – J.G.
  Mar 25 at 21:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry for that, also i am not sure about the $exists$ in 2nd statement, please let me know what do you think ?( Should it be $forall$) ( the print actually disappeared)
  $endgroup$
  – M Desmond
  Mar 25 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @J.W.Tanner Thanks for fixing the link. It seems to only state the first two axioms. I think it would be better to write $forallepsilon>0 (epsilon >a)implies 0ge a$ and $existsepsilon<0(epsilon<a)implies age 0$.
  $endgroup$
  – J.G.
  Mar 25 at 21:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe different devices of mine handle the link differently. I agree the second $exists$ should be $forall$. Please take my previous comment as advice on standard ordering of symbols rather on which statements hold.
  $endgroup$
  – J.G.
  Mar 25 at 22:00
  

  
  
  
  

 | 
show 4 more comments

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1

$begingroup$

If $a>0$, then take $epsilon=a/2,$ so $0<epsilon<a,$ so it is not true that for all $epsilon>0$, $epsilon>a$. The contrapositive of that is what you wanted for the first part.

If $a<0$, then take $epsilon=a/2$ so $0>epsilon>a$ so it is not true that for all $epsilon<0$, $epsilon<a$. The contrapositive of that is that if for all $epsilon<0$, $epsilon<a$, then $age0$.

share|cite|improve this answer

answered Mar 25 at 21:54

J. W. TannerJ. W. Tanner

4,7871420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir tell me one thing , what is the negation of $exists epsilon < 0 , epsilon < a$ ? Is it $forall epsilon > 0, epsilon > a$ ?
  $endgroup$
  – M Desmond
  Mar 25 at 22:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The negation of that is $forallepsilonbf <$$0, epsilonge a$
  $endgroup$
  – J. W. Tanner
  Mar 25 at 22:27
  
  

  
  
  
  

add a comment | 

1

$begingroup$

If $a>0$, then take $epsilon=a/2,$ so $0<epsilon<a,$ so it is not true that for all $epsilon>0$, $epsilon>a$. The contrapositive of that is what you wanted for the first part.

If $a<0$, then take $epsilon=a/2$ so $0>epsilon>a$ so it is not true that for all $epsilon<0$, $epsilon<a$. The contrapositive of that is that if for all $epsilon<0$, $epsilon<a$, then $age0$.

share|cite|improve this answer

answered Mar 25 at 21:54

J. W. TannerJ. W. Tanner

4,7871420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sir tell me one thing , what is the negation of $exists epsilon < 0 , epsilon < a$ ? Is it $forall epsilon > 0, epsilon > a$ ?
  $endgroup$
  – M Desmond
  Mar 25 at 22:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The negation of that is $forallepsilonbf <$$0, epsilonge a$
  $endgroup$
  – J. W. Tanner
  Mar 25 at 22:27
  
  

  
  
  
  

add a comment | 

$begingroup$

If $a>0$, then take $epsilon=a/2,$ so $0<epsilon<a,$ so it is not true that for all $epsilon>0$, $epsilon>a$. The contrapositive of that is what you wanted for the first part.

If $a<0$, then take $epsilon=a/2$ so $0>epsilon>a$ so it is not true that for all $epsilon<0$, $epsilon<a$. The contrapositive of that is that if for all $epsilon<0$, $epsilon<a$, then $age0$.

share|cite|improve this answer

answered Mar 25 at 21:54

J. W. TannerJ. W. Tanner

4,7871420

$endgroup$

share|cite|improve this answer

answered Mar 25 at 21:54

J. W. TannerJ. W. Tanner

4,7871420

answered Mar 25 at 21:54

J. W. TannerJ. W. Tanner

4,7871420

answered Mar 25 at 21:54

J. W. TannerJ. W. Tanner

4,7871420

0

$begingroup$

I can't see your axioms, nor can I parse the strange order of your highlighted line, but I'd prove the title theorem's contrapositive. If $a> 0$ (which by order, I guess, is equivalent to $0notge a$) then the positive choice $epsilon:=a/2$ contradicts $epsilon>a$ (by whatever axiomatic proof you'd manage of $a>0implies a>a/2>0$).

share|cite|improve this answer

answered Mar 25 at 21:19

J.G.J.G.

33.5k23252

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please see the link i have provided below for axioms,and see the order axioms in that link.
  $endgroup$
  – M Desmond
  Mar 25 at 21:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MDesmond Your link gave me a 404 error, but in any case your question ought to be self-contained.
  $endgroup$
  – J.G.
  Mar 25 at 21:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry for that, also i am not sure about the $exists$ in 2nd statement, please let me know what do you think ?( Should it be $forall$) ( the print actually disappeared)
  $endgroup$
  – M Desmond
  Mar 25 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @J.W.Tanner Thanks for fixing the link. It seems to only state the first two axioms. I think it would be better to write $forallepsilon>0 (epsilon >a)implies 0ge a$ and $existsepsilon<0(epsilon<a)implies age 0$.
  $endgroup$
  – J.G.
  Mar 25 at 21:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe different devices of mine handle the link differently. I agree the second $exists$ should be $forall$. Please take my previous comment as advice on standard ordering of symbols rather on which statements hold.
  $endgroup$
  – J.G.
  Mar 25 at 22:00
  

  
  
  
  

 | 
show 4 more comments

0

$begingroup$

I can't see your axioms, nor can I parse the strange order of your highlighted line, but I'd prove the title theorem's contrapositive. If $a> 0$ (which by order, I guess, is equivalent to $0notge a$) then the positive choice $epsilon:=a/2$ contradicts $epsilon>a$ (by whatever axiomatic proof you'd manage of $a>0implies a>a/2>0$).

share|cite|improve this answer

answered Mar 25 at 21:19

J.G.J.G.

33.5k23252

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Please see the link i have provided below for axioms,and see the order axioms in that link.
  $endgroup$
  – M Desmond
  Mar 25 at 21:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MDesmond Your link gave me a 404 error, but in any case your question ought to be self-contained.
  $endgroup$
  – J.G.
  Mar 25 at 21:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am sorry for that, also i am not sure about the $exists$ in 2nd statement, please let me know what do you think ?( Should it be $forall$) ( the print actually disappeared)
  $endgroup$
  – M Desmond
  Mar 25 at 21:46
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @J.W.Tanner Thanks for fixing the link. It seems to only state the first two axioms. I think it would be better to write $forallepsilon>0 (epsilon >a)implies 0ge a$ and $existsepsilon<0(epsilon<a)implies age 0$.
  $endgroup$
  – J.G.
  Mar 25 at 21:52
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe different devices of mine handle the link differently. I agree the second $exists$ should be $forall$. Please take my previous comment as advice on standard ordering of symbols rather on which statements hold.
  $endgroup$
  – J.G.
  Mar 25 at 22:00
  

  
  
  
  

 | 
show 4 more comments

$begingroup$

I can't see your axioms, nor can I parse the strange order of your highlighted line, but I'd prove the title theorem's contrapositive. If $a> 0$ (which by order, I guess, is equivalent to $0notge a$) then the positive choice $epsilon:=a/2$ contradicts $epsilon>a$ (by whatever axiomatic proof you'd manage of $a>0implies a>a/2>0$).

share|cite|improve this answer

answered Mar 25 at 21:19

J.G.J.G.

33.5k23252

$endgroup$

share|cite|improve this answer

answered Mar 25 at 21:19

J.G.J.G.

33.5k23252

answered Mar 25 at 21:19

J.G.J.G.

33.5k23252

answered Mar 25 at 21:19

J.G.J.G.

33.5k23252