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Proving a statement about real numbers

statement about properties of sequencesIssue with proof: Cauchy Completeness of Real NumbersProving Supremum of Product set of Nonnegative Real Numbers(Ir)rationality of Real NumbersProving no rational gaps between two rational numbers in QAm I proving this statement with itself?Proving/disproving that √7 - √2 is irrationalQuestion about 'A convergence sequence of real numbers is bounded.'Let $a,b in mathbb{R}$ and suppose that for every $epsilon > 0$ we have $a ≥ b − ε$. Show that $a geq b $General topology / real analysis - Suppose $S$ is a bounded and closed nonempty subset of real numbers. Prove sup S is in S

2

$begingroup$

If $x, y in Bbb R$ and $x<y+epsilon$ for every $epsilon >0$, then $x<y$.

Okay so I went about this by proving the contrapositive.

Proof: Let $x,yinBbb R$ and let $epsilon>0$. Suppose $xge y$. Take $epsilon = x-y$. This implies that $x=y+epsilon$, as needed.

Is this a valid proof or not?

real-analysis proof-verification real-numbers

share|cite|improve this question

edited 1 hour ago

J.G.

29.4k22846

asked 1 hour ago

AshAsh

234

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The conclusion should be $xleq y$.
  $endgroup$
  – gamma
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $$epsilon=x-y$$ then your epsilon would be negative!
  $endgroup$
  – Dr. Sonnhard Graubner
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why would it be negative if $x ge y$?
  $endgroup$
  – Ash
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
  $endgroup$
  – fleablood
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
  $endgroup$
  – fleablood
  1 hour ago
  

  
  
  
  

add a comment | 

2

$begingroup$

If $x, y in Bbb R$ and $x<y+epsilon$ for every $epsilon >0$, then $x<y$.

Okay so I went about this by proving the contrapositive.

Proof: Let $x,yinBbb R$ and let $epsilon>0$. Suppose $xge y$. Take $epsilon = x-y$. This implies that $x=y+epsilon$, as needed.

Is this a valid proof or not?

real-analysis proof-verification real-numbers

share|cite|improve this question

edited 1 hour ago

J.G.

29.4k22846

asked 1 hour ago

AshAsh

234

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The conclusion should be $xleq y$.
  $endgroup$
  – gamma
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $$epsilon=x-y$$ then your epsilon would be negative!
  $endgroup$
  – Dr. Sonnhard Graubner
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why would it be negative if $x ge y$?
  $endgroup$
  – Ash
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  1) The statment is false so your proof must be invalid 2). If $x = y$ then $epsilon = x-y = 0 not > 0$. And that's why the statement is false. If $x=y$ then $x=y < y + epsilon$ for all $epsilon > 0$ but $x not < y$. So the statement is false. What do you think the true statement would be? (Hint: You proof does prove the true statement.)
  $endgroup$
  – fleablood
  1 hour ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I do have to give you credit in that your proof is absolutely the correct idea and it WOULD have worked if the statement had been given properly.
  $endgroup$
  – fleablood
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

If $x, y in Bbb R$ and $x<y+epsilon$ for every $epsilon >0$, then $x<y$.

Okay so I went about this by proving the contrapositive.

Proof: Let $x,yinBbb R$ and let $epsilon>0$. Suppose $xge y$. Take $epsilon = x-y$. This implies that $x=y+epsilon$, as needed.

Is this a valid proof or not?

real-analysis proof-verification real-numbers

share|cite|improve this question

edited 1 hour ago

J.G.

29.4k22846

asked 1 hour ago

AshAsh

234

$endgroup$

share|cite|improve this question

edited 1 hour ago

J.G.

29.4k22846

asked 1 hour ago

AshAsh

234

share|cite|improve this question

edited 1 hour ago

J.G.

29.4k22846

asked 1 hour ago

AshAsh

234

edited 1 hour ago

J.G.

29.4k22846

edited 1 hour ago

J.G.

29.4k22846

asked 1 hour ago

AshAsh

234

asked 1 hour ago

AshAsh

234

3 Answers
3

active

oldest

votes

3

$begingroup$

The statement you want to prove is not true.

Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.

share|cite|improve this answer

answered 1 hour ago

TheSilverDoeTheSilverDoe

3,373112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're absolutely correct, thank you.
  $endgroup$
  – Ash
  1 hour ago
  

  
  
  
  

add a comment | 

1

$begingroup$

The proof is invalid for the simple fact that the statement is false, so you cannot prove it.

The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.

Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.

share|cite|improve this answer

answered 1 hour ago

egregegreg

184k1486205

$endgroup$

add a comment | 

0

$begingroup$

1) The statement is false.

Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.

2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.

Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.

3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.

4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.

share|cite|improve this answer

answered 1 hour ago

fleabloodfleablood

72.1k22687

$endgroup$

add a comment | 

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3

$begingroup$

The statement you want to prove is not true.

Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.

share|cite|improve this answer

answered 1 hour ago

TheSilverDoeTheSilverDoe

3,373112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're absolutely correct, thank you.
  $endgroup$
  – Ash
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

The statement you want to prove is not true.

Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.

share|cite|improve this answer

answered 1 hour ago

TheSilverDoeTheSilverDoe

3,373112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're absolutely correct, thank you.
  $endgroup$
  – Ash
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

The statement you want to prove is not true.

Take $x = y =0$. You have, for all $varepsilon > 0$, $x < y + varepsilon$ (because $varepsilon > 0$), but of course you don't have $x < y$.

share|cite|improve this answer

answered 1 hour ago

TheSilverDoeTheSilverDoe

3,373112

$endgroup$

share|cite|improve this answer

answered 1 hour ago

TheSilverDoeTheSilverDoe

3,373112

answered 1 hour ago

TheSilverDoeTheSilverDoe

3,373112

answered 1 hour ago

TheSilverDoeTheSilverDoe

3,373112

1

$begingroup$

The proof is invalid for the simple fact that the statement is false, so you cannot prove it.

The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.

Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.

share|cite|improve this answer

answered 1 hour ago

egregegreg

184k1486205

$endgroup$

add a comment | 

1

$begingroup$

The proof is invalid for the simple fact that the statement is false, so you cannot prove it.

The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.

Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.

share|cite|improve this answer

answered 1 hour ago

egregegreg

184k1486205

$endgroup$

add a comment | 

$begingroup$

The proof is invalid for the simple fact that the statement is false, so you cannot prove it.

The correct statement is “if, for all $varepsilon>0$, $x<y+varepsilon$, then $xle y$”.

Now your proof works! Suppose $x>y$ (that is, “not $xle y$”) and take $varepsilon=x-y$; then $varepsilon>0$ and $x=y+varepsilon$.

share|cite|improve this answer

answered 1 hour ago

egregegreg

184k1486205

$endgroup$

share|cite|improve this answer

answered 1 hour ago

egregegreg

184k1486205

answered 1 hour ago

egregegreg

184k1486205

answered 1 hour ago

egregegreg

184k1486205

0

$begingroup$

1) The statement is false.

Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.

2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.

Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.

3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.

4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.

share|cite|improve this answer

answered 1 hour ago

fleabloodfleablood

72.1k22687

$endgroup$

add a comment | 

0

$begingroup$

1) The statement is false.

Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.

2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.

Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.

3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.

4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.

share|cite|improve this answer

answered 1 hour ago

fleabloodfleablood

72.1k22687

$endgroup$

add a comment | 

$begingroup$

1) The statement is false.

Counter example: Let $x = y$. Then $x = y < y+epsilon$ for all $epsilon > 0$ but $x not < y$.

2) Your error is in claiming setting $epsilon = x - y$. If $x =y$ then that $epsilon not > 0$.

Your proof is doomed to fail for $x = y$ because the statement is false for $x =y$.

3) But you have proven that $x>y$ is impossible. That would mean if $epsilon = x - y > 0$ then $x = y + epsilon$ which contradicts our hypothesis that $x < y + epsilon$ for all $epsilon>0$.

4) So a TRUE statement would be: if $x < y + epsilon$ for all $epsilon > 0$ then $x le y$ and you successfully have proven that.

share|cite|improve this answer

answered 1 hour ago

fleabloodfleablood

72.1k22687

$endgroup$

share|cite|improve this answer

answered 1 hour ago

fleabloodfleablood

72.1k22687

answered 1 hour ago

fleabloodfleablood

72.1k22687

answered 1 hour ago

fleabloodfleablood

72.1k22687