Proving $∀ε>0:x<y+epsilon ⇒ x<y$Proving limit using $epsilon - N$Contrapositive Epsilon-Delta Limits?An upper bound $u$ is the supremum of $A$ if and only if for all $epsilon > 0$ there is an $a in A$ such that $u-epsilon < a$Mathematical Rigor in Proving Limits by $epsilon-delta$ DefinitionLet $a,b in mathbbR^+$; prove that if $ab geqslant 1$, then $a geqslant 1$ or $b geqslant 1$Proving/disproving that √7 - √2 is irrational$Bbb Q$ is dense in $Bbb R$Let $a,b in mathbbR$ and suppose that for every $epsilon > 0$ we have $a ≥ b − ε$. Show that $a geq b $Proving $x$ between $alpha - epsilon$, if $alpha$ is supremumReal Analysis: Proving Continuity ($epsilon$ $delta$ argument)
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Proving $∀ε>0:x
Proving limit using $epsilon - N$Contrapositive Epsilon-Delta Limits?An upper bound $u$ is the supremum of $A$ if and only if for all $epsilon > 0$ there is an $a in A$ such that $u-epsilon < a$Mathematical Rigor in Proving Limits by $epsilon-delta$ DefinitionLet $a,b in mathbbR^+$; prove that if $ab geqslant 1$, then $a geqslant 1$ or $b geqslant 1$Proving/disproving that √7 - √2 is irrational$Bbb Q$ is dense in $Bbb R$Let $a,b in mathbbR$ and suppose that for every $epsilon > 0$ we have $a ≥ b − ε$. Show that $a geq b $Proving $x$ between $alpha - epsilon$, if $alpha$ is supremumReal Analysis: Proving Continuity ($epsilon$ $delta$ argument)
$begingroup$
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
$endgroup$
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
$endgroup$
5
$begingroup$
The conclusion should be $xleq y$.
$endgroup$
– gamma
Mar 10 at 20:13
$begingroup$
If $$epsilon=x-y$$ then your epsilon would be negative!
$endgroup$
– Dr. Sonnhard Graubner
Mar 10 at 20:13
$begingroup$
Why would it be negative if $x ge y$?
$endgroup$
– Ash
Mar 10 at 20:14
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
Mar 10 at 20:23
$begingroup$
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
Mar 10 at 20:32
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
$endgroup$
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
real-analysis proof-verification real-numbers
edited 2 days ago
Wrzlprmft
3,15111335
3,15111335
asked Mar 10 at 20:11
AshAsh
826
826
5
$begingroup$
The conclusion should be $xleq y$.
$endgroup$
– gamma
Mar 10 at 20:13
$begingroup$
If $$epsilon=x-y$$ then your epsilon would be negative!
$endgroup$
– Dr. Sonnhard Graubner
Mar 10 at 20:13
$begingroup$
Why would it be negative if $x ge y$?
$endgroup$
– Ash
Mar 10 at 20:14
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
Mar 10 at 20:23
$begingroup$
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
Mar 10 at 20:32
add a comment |
5
$begingroup$
The conclusion should be $xleq y$.
$endgroup$
– gamma
Mar 10 at 20:13
$begingroup$
If $$epsilon=x-y$$ then your epsilon would be negative!
$endgroup$
– Dr. Sonnhard Graubner
Mar 10 at 20:13
$begingroup$
Why would it be negative if $x ge y$?
$endgroup$
– Ash
Mar 10 at 20:14
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
Mar 10 at 20:23
$begingroup$
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
Mar 10 at 20:32
5
5
$begingroup$
The conclusion should be $xleq y$.
$endgroup$
– gamma
Mar 10 at 20:13
$begingroup$
The conclusion should be $xleq y$.
$endgroup$
– gamma
Mar 10 at 20:13
$begingroup$
If $$epsilon=x-y$$ then your epsilon would be negative!
$endgroup$
– Dr. Sonnhard Graubner
Mar 10 at 20:13
$begingroup$
If $$epsilon=x-y$$ then your epsilon would be negative!
$endgroup$
– Dr. Sonnhard Graubner
Mar 10 at 20:13
$begingroup$
Why would it be negative if $x ge y$?
$endgroup$
– Ash
Mar 10 at 20:14
$begingroup$
Why would it be negative if $x ge y$?
$endgroup$
– Ash
Mar 10 at 20:14
1
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
Mar 10 at 20:23
$begingroup$
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
Mar 10 at 20:23
$begingroup$
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
Mar 10 at 20:32
$begingroup$
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
Mar 10 at 20:32
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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 $x < y$.
$endgroup$
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 $x=y+varepsilon$.
$endgroup$
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.
$endgroup$
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
$begingroup$
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 $x < y$.
$endgroup$
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 $x < y$.
$endgroup$
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 $x < y$.
$endgroup$
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 $x < y$.
answered Mar 10 at 20:16
TheSilverDoeTheSilverDoe
3,837112
3,837112
add a comment |
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 $x=y+varepsilon$.
$endgroup$
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 $x=y+varepsilon$.
$endgroup$
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 $x=y+varepsilon$.
$endgroup$
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 $x=y+varepsilon$.
answered Mar 10 at 20:24
egregegreg
184k1486205
184k1486205
add a comment |
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.
$endgroup$
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.
$endgroup$
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.
$endgroup$
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.
answered Mar 10 at 20:31
fleabloodfleablood
72.2k22687
72.2k22687
add a comment |
add a comment |
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5
$begingroup$
The conclusion should be $xleq y$.
$endgroup$
– gamma
Mar 10 at 20:13
$begingroup$
If $$epsilon=x-y$$ then your epsilon would be negative!
$endgroup$
– Dr. Sonnhard Graubner
Mar 10 at 20:13
$begingroup$
Why would it be negative if $x ge y$?
$endgroup$
– Ash
Mar 10 at 20:14
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
Mar 10 at 20:23
$begingroup$
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
Mar 10 at 20:32