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Use of the term “whenever†in Logic
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How does the word "whenever" translate into logic? For instance:
$|f(x) - f(1)| geq epsilon$ whenever $|x-1| geq delta$
logic terminology logic-translation
edited 19 mins ago
Taroccoesbrocco
4,35461535
asked 1 hour ago
yoshi
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add a comment |Â
up vote
1
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How does the word "whenever" translate into logic? For instance:
$|f(x) - f(1)| geq epsilon$ whenever $|x-1| geq delta$
logic terminology logic-translation
edited 19 mins ago
Taroccoesbrocco
4,35461535
asked 1 hour ago
yoshi
1,018816
1
It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
– Stan Tendijck
1 hour ago
1
See my answer here.
– Taroccoesbrocco
56 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How does the word "whenever" translate into logic? For instance:
$|f(x) - f(1)| geq epsilon$ whenever $|x-1| geq delta$
logic terminology logic-translation
edited 19 mins ago
Taroccoesbrocco
4,35461535
asked 1 hour ago
yoshi
1,018816
How does the word "whenever" translate into logic? For instance:
$|f(x) - f(1)| geq epsilon$ whenever $|x-1| geq delta$
logic terminology logic-translation
logic terminology logic-translation
edited 19 mins ago
Taroccoesbrocco
4,35461535
asked 1 hour ago
yoshi
1,018816
edited 19 mins ago
Taroccoesbrocco
4,35461535
asked 1 hour ago
yoshi
1,018816
edited 19 mins ago
Taroccoesbrocco
4,35461535
edited 19 mins ago
Taroccoesbrocco
4,35461535
edited 19 mins ago
Taroccoesbrocco
4,35461535
4,35461535
asked 1 hour ago
yoshi
1,018816
asked 1 hour ago
yoshi
1,018816
asked 1 hour ago
yoshi
1,018816
1,018816
1
It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
– Stan Tendijck
1 hour ago
1
See my answer here.
– Taroccoesbrocco
56 mins ago
add a comment |Â
1
It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
– Stan Tendijck
1 hour ago
1
See my answer here.
– Taroccoesbrocco
56 mins ago
1
1
It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
– Stan Tendijck
1 hour ago
It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
– Stan Tendijck
1 hour ago
1
1
See my answer here.
– Taroccoesbrocco
56 mins ago
See my answer here.
– Taroccoesbrocco
56 mins ago
add a comment |Â
3 Answers
3
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oldest
votes
up vote
4
down vote
Whenever in this instant is used as the following :
$$textIfquad |x-1|geq delta quad textthen quad |f(x)-f(1)| geq epsilon $$
answered 1 hour ago
TheJedi
464
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TheJedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
3
down vote
"Whenever" can be a English representation of $Leftarrow$ or "is implied by".
The whole statement $|f(x) - f(1)| geq epsilon$ whenver $|x-1| geq delta$ is true when $|f(x) - f(1)| geq epsilon$ and $|x-1| geq delta$ are both true, or $|x-1| geq delta$ is false.
so: $|f(x) - f(1)| geq epsilon Leftarrow |x-1| geq delta$
answered 1 hour ago
James K
512212
add a comment |Â
up vote
1
down vote
$forall xbig[|x-1| geq delta rightarrow |f(x) - f(1)| geq epsilonbig]$
answered 58 mins ago
S.S.Danyal
1413
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Whenever in this instant is used as the following :
$$textIfquad |x-1|geq delta quad textthen quad |f(x)-f(1)| geq epsilon $$
answered 1 hour ago
TheJedi
464
New contributor
TheJedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
Whenever in this instant is used as the following :
$$textIfquad |x-1|geq delta quad textthen quad |f(x)-f(1)| geq epsilon $$
answered 1 hour ago
TheJedi
464
New contributor
TheJedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Whenever in this instant is used as the following :
$$textIfquad |x-1|geq delta quad textthen quad |f(x)-f(1)| geq epsilon $$
answered 1 hour ago
TheJedi
464
New contributor
TheJedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Whenever in this instant is used as the following :
$$textIfquad |x-1|geq delta quad textthen quad |f(x)-f(1)| geq epsilon $$
answered 1 hour ago
TheJedi
464
New contributor
TheJedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
TheJedi
464
New contributor
TheJedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 1 hour ago
TheJedi
464
answered 1 hour ago
TheJedi
464
464
New contributor
TheJedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
TheJedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
TheJedi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
add a comment |Â
up vote
3
down vote
"Whenever" can be a English representation of $Leftarrow$ or "is implied by".
The whole statement $|f(x) - f(1)| geq epsilon$ whenver $|x-1| geq delta$ is true when $|f(x) - f(1)| geq epsilon$ and $|x-1| geq delta$ are both true, or $|x-1| geq delta$ is false.
so: $|f(x) - f(1)| geq epsilon Leftarrow |x-1| geq delta$
answered 1 hour ago
James K
512212
add a comment |Â
up vote
3
down vote
"Whenever" can be a English representation of $Leftarrow$ or "is implied by".
The whole statement $|f(x) - f(1)| geq epsilon$ whenver $|x-1| geq delta$ is true when $|f(x) - f(1)| geq epsilon$ and $|x-1| geq delta$ are both true, or $|x-1| geq delta$ is false.
so: $|f(x) - f(1)| geq epsilon Leftarrow |x-1| geq delta$
answered 1 hour ago
James K
512212
add a comment |Â
up vote
3
down vote
up vote
3
down vote
"Whenever" can be a English representation of $Leftarrow$ or "is implied by".
The whole statement $|f(x) - f(1)| geq epsilon$ whenver $|x-1| geq delta$ is true when $|f(x) - f(1)| geq epsilon$ and $|x-1| geq delta$ are both true, or $|x-1| geq delta$ is false.
so: $|f(x) - f(1)| geq epsilon Leftarrow |x-1| geq delta$
answered 1 hour ago
James K
512212
"Whenever" can be a English representation of $Leftarrow$ or "is implied by".
The whole statement $|f(x) - f(1)| geq epsilon$ whenver $|x-1| geq delta$ is true when $|f(x) - f(1)| geq epsilon$ and $|x-1| geq delta$ are both true, or $|x-1| geq delta$ is false.
so: $|f(x) - f(1)| geq epsilon Leftarrow |x-1| geq delta$
answered 1 hour ago
James K
512212
answered 1 hour ago
James K
512212
answered 1 hour ago
James K
512212
answered 1 hour ago
James K
512212
512212
add a comment |Â
add a comment |Â
up vote
1
down vote
$forall xbig[|x-1| geq delta rightarrow |f(x) - f(1)| geq epsilonbig]$
answered 58 mins ago
S.S.Danyal
1413
add a comment |Â
up vote
1
down vote
$forall xbig[|x-1| geq delta rightarrow |f(x) - f(1)| geq epsilonbig]$
answered 58 mins ago
S.S.Danyal
1413
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$forall xbig[|x-1| geq delta rightarrow |f(x) - f(1)| geq epsilonbig]$
answered 58 mins ago
S.S.Danyal
1413
$forall xbig[|x-1| geq delta rightarrow |f(x) - f(1)| geq epsilonbig]$
answered 58 mins ago
S.S.Danyal
1413
answered 58 mins ago
S.S.Danyal
1413
answered 58 mins ago
S.S.Danyal
1413
answered 58 mins ago
S.S.Danyal
1413
1413
add a comment |Â
add a comment |Â
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1
It means 'if'. So you can also read the statement as: If $|x-1|geqdelta$, then $|f(x)-f(1)|geqepsilon$. But it is nice to sometimes rephrase this to avoid repetition
– Stan Tendijck
1 hour ago
1
See my answer here.
– Taroccoesbrocco
56 mins ago