Mixing
General question on $epsilon$ and choice of $a$ and $b$
Clash Royale CLAN TAG#URR8PPP
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Well I do have a small understanding issue here.
If you say that for every $epsilon > 0$ there are two values $a, b in mathbbR$ with $a neq b$ such that $ a - b < epsilon$. Thats not a problem there.
But when you say that there are two fixed values $x,y in mathbbR$ such that $x-y < epsilon$ for every $epsilon > 0$. How does this not imply that $x = y$ ?
I mean in the upper case you can choose $a = b+fracepsilon2$ or something. But in the second case you already got two fixed values which are so close together that the difference is smaller than every $epsilon > 0$. We had that in class and the lecturer told as it does not necessary mean that $x = y$.
calculus limits
edited Jul 31 at 14:23
Robert Soupe
9,94721947
asked Jul 31 at 11:34
Arjihad
1949
add a comment |Â
up vote
2
down vote
favorite
Well I do have a small understanding issue here.
If you say that for every $epsilon > 0$ there are two values $a, b in mathbbR$ with $a neq b$ such that $ a - b < epsilon$. Thats not a problem there.
But when you say that there are two fixed values $x,y in mathbbR$ such that $x-y < epsilon$ for every $epsilon > 0$. How does this not imply that $x = y$ ?
I mean in the upper case you can choose $a = b+fracepsilon2$ or something. But in the second case you already got two fixed values which are so close together that the difference is smaller than every $epsilon > 0$. We had that in class and the lecturer told as it does not necessary mean that $x = y$.
calculus limits
edited Jul 31 at 14:23
Robert Soupe
9,94721947
asked Jul 31 at 11:34
Arjihad
1949
Yes. I was already thinking about that but thats how we did it in class.
– Arjihad
Jul 31 at 11:39
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Well I do have a small understanding issue here.
If you say that for every $epsilon > 0$ there are two values $a, b in mathbbR$ with $a neq b$ such that $ a - b < epsilon$. Thats not a problem there.
But when you say that there are two fixed values $x,y in mathbbR$ such that $x-y < epsilon$ for every $epsilon > 0$. How does this not imply that $x = y$ ?
I mean in the upper case you can choose $a = b+fracepsilon2$ or something. But in the second case you already got two fixed values which are so close together that the difference is smaller than every $epsilon > 0$. We had that in class and the lecturer told as it does not necessary mean that $x = y$.
calculus limits
edited Jul 31 at 14:23
Robert Soupe
9,94721947
asked Jul 31 at 11:34
Arjihad
1949
Well I do have a small understanding issue here.
If you say that for every $epsilon > 0$ there are two values $a, b in mathbbR$ with $a neq b$ such that $ a - b < epsilon$. Thats not a problem there.
But when you say that there are two fixed values $x,y in mathbbR$ such that $x-y < epsilon$ for every $epsilon > 0$. How does this not imply that $x = y$ ?
I mean in the upper case you can choose $a = b+fracepsilon2$ or something. But in the second case you already got two fixed values which are so close together that the difference is smaller than every $epsilon > 0$. We had that in class and the lecturer told as it does not necessary mean that $x = y$.
calculus limits
edited Jul 31 at 14:23
Robert Soupe
9,94721947
asked Jul 31 at 11:34
Arjihad
1949
edited Jul 31 at 14:23
Robert Soupe
9,94721947
edited Jul 31 at 14:23
Robert Soupe
9,94721947
edited Jul 31 at 14:23
Robert Soupe
9,94721947
9,94721947
asked Jul 31 at 11:34
Arjihad
1949
asked Jul 31 at 11:34
Arjihad
1949
asked Jul 31 at 11:34
Arjihad
1949
1949
Yes. I was already thinking about that but thats how we did it in class.
– Arjihad
Jul 31 at 11:39
add a comment |Â
Yes. I was already thinking about that but thats how we did it in class.
– Arjihad
Jul 31 at 11:39
Yes. I was already thinking about that but thats how we did it in class.
– Arjihad
Jul 31 at 11:39
Yes. I was already thinking about that but thats how we did it in class.
– Arjihad
Jul 31 at 11:39
add a comment |Â
1 Answer
1
active
oldest
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up vote
3
down vote
If, for instance, $x=0$ and $y=1$, then it is true that, for every $varepsilon>0$, $x-y<varepsilon$. However, $xneq y$.
You would be right if the condition was $|x-y|<varepsilon$.
answered Jul 31 at 11:36
José Carlos Santos
112k1696172
Okay I see. What would be your answer if $x > y$ ?
– Arjihad
Jul 31 at 11:38
2
@Arjihad If $x>y$, then the assertion “for every $varepsilon>0$, $x-y<varepsilon$†is false.
– José Carlos Santos
Jul 31 at 11:39
Okay. Lets say we have $x > y$ such that $x-y < epsilon$ for every $epsilon > 0$. Does this imply that $x = y$ ?
– Arjihad
Jul 31 at 11:41
3
@Arjihad Yes, because if you havve two assertions $P$ and $Q$, and $P$ is false, then $Pimplies Q$, no matter what $Q$ is.
– José Carlos Santos
Jul 31 at 11:43
@Arjihad you may be thinking of requiring $x geq y$
– Sambo
Jul 31 at 12:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
If, for instance, $x=0$ and $y=1$, then it is true that, for every $varepsilon>0$, $x-y<varepsilon$. However, $xneq y$.
You would be right if the condition was $|x-y|<varepsilon$.
answered Jul 31 at 11:36
José Carlos Santos
112k1696172
Okay I see. What would be your answer if $x > y$ ?
– Arjihad
Jul 31 at 11:38
2
@Arjihad If $x>y$, then the assertion “for every $varepsilon>0$, $x-y<varepsilon$†is false.
– José Carlos Santos
Jul 31 at 11:39
Okay. Lets say we have $x > y$ such that $x-y < epsilon$ for every $epsilon > 0$. Does this imply that $x = y$ ?
– Arjihad
Jul 31 at 11:41
3
@Arjihad Yes, because if you havve two assertions $P$ and $Q$, and $P$ is false, then $Pimplies Q$, no matter what $Q$ is.
– José Carlos Santos
Jul 31 at 11:43
@Arjihad you may be thinking of requiring $x geq y$
– Sambo
Jul 31 at 12:37
add a comment |Â
up vote
3
down vote
If, for instance, $x=0$ and $y=1$, then it is true that, for every $varepsilon>0$, $x-y<varepsilon$. However, $xneq y$.
You would be right if the condition was $|x-y|<varepsilon$.
answered Jul 31 at 11:36
José Carlos Santos
112k1696172
Okay I see. What would be your answer if $x > y$ ?
– Arjihad
Jul 31 at 11:38
2
@Arjihad If $x>y$, then the assertion “for every $varepsilon>0$, $x-y<varepsilon$†is false.
– José Carlos Santos
Jul 31 at 11:39
Okay. Lets say we have $x > y$ such that $x-y < epsilon$ for every $epsilon > 0$. Does this imply that $x = y$ ?
– Arjihad
Jul 31 at 11:41
3
@Arjihad Yes, because if you havve two assertions $P$ and $Q$, and $P$ is false, then $Pimplies Q$, no matter what $Q$ is.
– José Carlos Santos
Jul 31 at 11:43
@Arjihad you may be thinking of requiring $x geq y$
– Sambo
Jul 31 at 12:37
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If, for instance, $x=0$ and $y=1$, then it is true that, for every $varepsilon>0$, $x-y<varepsilon$. However, $xneq y$.
You would be right if the condition was $|x-y|<varepsilon$.
answered Jul 31 at 11:36
José Carlos Santos
112k1696172
If, for instance, $x=0$ and $y=1$, then it is true that, for every $varepsilon>0$, $x-y<varepsilon$. However, $xneq y$.
You would be right if the condition was $|x-y|<varepsilon$.
answered Jul 31 at 11:36
José Carlos Santos
112k1696172
answered Jul 31 at 11:36
José Carlos Santos
112k1696172
answered Jul 31 at 11:36
José Carlos Santos
112k1696172
answered Jul 31 at 11:36
José Carlos Santos
112k1696172
112k1696172
Okay I see. What would be your answer if $x > y$ ?
– Arjihad
Jul 31 at 11:38
2
@Arjihad If $x>y$, then the assertion “for every $varepsilon>0$, $x-y<varepsilon$†is false.
– José Carlos Santos
Jul 31 at 11:39
Okay. Lets say we have $x > y$ such that $x-y < epsilon$ for every $epsilon > 0$. Does this imply that $x = y$ ?
– Arjihad
Jul 31 at 11:41
3
@Arjihad Yes, because if you havve two assertions $P$ and $Q$, and $P$ is false, then $Pimplies Q$, no matter what $Q$ is.
– José Carlos Santos
Jul 31 at 11:43
@Arjihad you may be thinking of requiring $x geq y$
– Sambo
Jul 31 at 12:37
add a comment |Â
Okay I see. What would be your answer if $x > y$ ?
– Arjihad
Jul 31 at 11:38
2
@Arjihad If $x>y$, then the assertion “for every $varepsilon>0$, $x-y<varepsilon$†is false.
– José Carlos Santos
Jul 31 at 11:39
Okay. Lets say we have $x > y$ such that $x-y < epsilon$ for every $epsilon > 0$. Does this imply that $x = y$ ?
– Arjihad
Jul 31 at 11:41
3
@Arjihad Yes, because if you havve two assertions $P$ and $Q$, and $P$ is false, then $Pimplies Q$, no matter what $Q$ is.
– José Carlos Santos
Jul 31 at 11:43
@Arjihad you may be thinking of requiring $x geq y$
– Sambo
Jul 31 at 12:37
Okay I see. What would be your answer if $x > y$ ?
– Arjihad
Jul 31 at 11:38
Okay I see. What would be your answer if $x > y$ ?
– Arjihad
Jul 31 at 11:38
2
2
@Arjihad If $x>y$, then the assertion “for every $varepsilon>0$, $x-y<varepsilon$†is false.
– José Carlos Santos
Jul 31 at 11:39
@Arjihad If $x>y$, then the assertion “for every $varepsilon>0$, $x-y<varepsilon$†is false.
– José Carlos Santos
Jul 31 at 11:39
Okay. Lets say we have $x > y$ such that $x-y < epsilon$ for every $epsilon > 0$. Does this imply that $x = y$ ?
– Arjihad
Jul 31 at 11:41
Okay. Lets say we have $x > y$ such that $x-y < epsilon$ for every $epsilon > 0$. Does this imply that $x = y$ ?
– Arjihad
Jul 31 at 11:41
3
3
@Arjihad Yes, because if you havve two assertions $P$ and $Q$, and $P$ is false, then $Pimplies Q$, no matter what $Q$ is.
– José Carlos Santos
Jul 31 at 11:43
@Arjihad Yes, because if you havve two assertions $P$ and $Q$, and $P$ is false, then $Pimplies Q$, no matter what $Q$ is.
– José Carlos Santos
Jul 31 at 11:43
@Arjihad you may be thinking of requiring $x geq y$
– Sambo
Jul 31 at 12:37
@Arjihad you may be thinking of requiring $x geq y$
– Sambo
Jul 31 at 12:37
add a comment |Â
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Yes. I was already thinking about that but thats how we did it in class.
– Arjihad
Jul 31 at 11:39