Use of '.' in relations [closed]
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I am trying to understand what the following relation means:
$${(i, i · j) | i, j in mathbb N}$$
Any help is appreciated!
discrete-mathematics relations
closed as off-topic by amWhy, GNUSupporter 8964民主女神 地下教會, Ethan Bolker, Paul Frost, Gibbs Nov 23 at 17:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, GNUSupporter 8964民主女神 地下教會, Paul Frost, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-2
down vote
favorite
I am trying to understand what the following relation means:
$${(i, i · j) | i, j in mathbb N}$$
Any help is appreciated!
discrete-mathematics relations
closed as off-topic by amWhy, GNUSupporter 8964民主女神 地下教會, Ethan Bolker, Paul Frost, Gibbs Nov 23 at 17:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, GNUSupporter 8964民主女神 地下教會, Paul Frost, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
5
Multiplication would be the obvious guess.
– user3482749
Nov 23 at 11:35
Math needs context. Without it, it could mean just about anything...
– JuliusL33t
Nov 23 at 11:55
It is the (graph of the) function $mathbb Ntomathbb N$ prescribed by $dmapsto dcdot r$.
– drhab
Nov 23 at 12:04
The current version of the OP has nothing to do with "dot." I'm guessing something got lost in translation in the editing process...
– Eff
Nov 23 at 13:22
I'm voting to close this question because the OP's latest edit makes it meaningless.
– Ethan Bolker
Nov 23 at 13:23
add a comment |
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I am trying to understand what the following relation means:
$${(i, i · j) | i, j in mathbb N}$$
Any help is appreciated!
discrete-mathematics relations
I am trying to understand what the following relation means:
$${(i, i · j) | i, j in mathbb N}$$
Any help is appreciated!
discrete-mathematics relations
discrete-mathematics relations
edited Nov 23 at 13:24
Trevor Gunn
13.9k32045
13.9k32045
asked Nov 23 at 11:34
NorthernStars
13
13
closed as off-topic by amWhy, GNUSupporter 8964民主女神 地下教會, Ethan Bolker, Paul Frost, Gibbs Nov 23 at 17:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, GNUSupporter 8964民主女神 地下教會, Paul Frost, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, GNUSupporter 8964民主女神 地下教會, Ethan Bolker, Paul Frost, Gibbs Nov 23 at 17:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, GNUSupporter 8964民主女神 地下教會, Paul Frost, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
5
Multiplication would be the obvious guess.
– user3482749
Nov 23 at 11:35
Math needs context. Without it, it could mean just about anything...
– JuliusL33t
Nov 23 at 11:55
It is the (graph of the) function $mathbb Ntomathbb N$ prescribed by $dmapsto dcdot r$.
– drhab
Nov 23 at 12:04
The current version of the OP has nothing to do with "dot." I'm guessing something got lost in translation in the editing process...
– Eff
Nov 23 at 13:22
I'm voting to close this question because the OP's latest edit makes it meaningless.
– Ethan Bolker
Nov 23 at 13:23
add a comment |
5
Multiplication would be the obvious guess.
– user3482749
Nov 23 at 11:35
Math needs context. Without it, it could mean just about anything...
– JuliusL33t
Nov 23 at 11:55
It is the (graph of the) function $mathbb Ntomathbb N$ prescribed by $dmapsto dcdot r$.
– drhab
Nov 23 at 12:04
The current version of the OP has nothing to do with "dot." I'm guessing something got lost in translation in the editing process...
– Eff
Nov 23 at 13:22
I'm voting to close this question because the OP's latest edit makes it meaningless.
– Ethan Bolker
Nov 23 at 13:23
5
5
Multiplication would be the obvious guess.
– user3482749
Nov 23 at 11:35
Multiplication would be the obvious guess.
– user3482749
Nov 23 at 11:35
Math needs context. Without it, it could mean just about anything...
– JuliusL33t
Nov 23 at 11:55
Math needs context. Without it, it could mean just about anything...
– JuliusL33t
Nov 23 at 11:55
It is the (graph of the) function $mathbb Ntomathbb N$ prescribed by $dmapsto dcdot r$.
– drhab
Nov 23 at 12:04
It is the (graph of the) function $mathbb Ntomathbb N$ prescribed by $dmapsto dcdot r$.
– drhab
Nov 23 at 12:04
The current version of the OP has nothing to do with "dot." I'm guessing something got lost in translation in the editing process...
– Eff
Nov 23 at 13:22
The current version of the OP has nothing to do with "dot." I'm guessing something got lost in translation in the editing process...
– Eff
Nov 23 at 13:22
I'm voting to close this question because the OP's latest edit makes it meaningless.
– Ethan Bolker
Nov 23 at 13:23
I'm voting to close this question because the OP's latest edit makes it meaningless.
– Ethan Bolker
Nov 23 at 13:23
add a comment |
2 Answers
2
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If $i,j in mathbb N$, then $(i,i cdot j)$ is the pair with $i$ in the first coordinate and the product $i cdot j$ in the second coordinate, hence
$(i,i cdot j) in mathbb N times mathbb N$ .
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(Working from the original version rather than the current which makes no sense) this is a way of writing down the relation "divides". $i$ divides $icdot j$ (that is $itimes j$) for each natural number $j$, and it divides no other natural numbers. So this relation contains all the pairs of that form.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If $i,j in mathbb N$, then $(i,i cdot j)$ is the pair with $i$ in the first coordinate and the product $i cdot j$ in the second coordinate, hence
$(i,i cdot j) in mathbb N times mathbb N$ .
add a comment |
up vote
1
down vote
If $i,j in mathbb N$, then $(i,i cdot j)$ is the pair with $i$ in the first coordinate and the product $i cdot j$ in the second coordinate, hence
$(i,i cdot j) in mathbb N times mathbb N$ .
add a comment |
up vote
1
down vote
up vote
1
down vote
If $i,j in mathbb N$, then $(i,i cdot j)$ is the pair with $i$ in the first coordinate and the product $i cdot j$ in the second coordinate, hence
$(i,i cdot j) in mathbb N times mathbb N$ .
If $i,j in mathbb N$, then $(i,i cdot j)$ is the pair with $i$ in the first coordinate and the product $i cdot j$ in the second coordinate, hence
$(i,i cdot j) in mathbb N times mathbb N$ .
answered Nov 23 at 11:38
Fred
43k1643
43k1643
add a comment |
add a comment |
up vote
0
down vote
(Working from the original version rather than the current which makes no sense) this is a way of writing down the relation "divides". $i$ divides $icdot j$ (that is $itimes j$) for each natural number $j$, and it divides no other natural numbers. So this relation contains all the pairs of that form.
add a comment |
up vote
0
down vote
(Working from the original version rather than the current which makes no sense) this is a way of writing down the relation "divides". $i$ divides $icdot j$ (that is $itimes j$) for each natural number $j$, and it divides no other natural numbers. So this relation contains all the pairs of that form.
add a comment |
up vote
0
down vote
up vote
0
down vote
(Working from the original version rather than the current which makes no sense) this is a way of writing down the relation "divides". $i$ divides $icdot j$ (that is $itimes j$) for each natural number $j$, and it divides no other natural numbers. So this relation contains all the pairs of that form.
(Working from the original version rather than the current which makes no sense) this is a way of writing down the relation "divides". $i$ divides $icdot j$ (that is $itimes j$) for each natural number $j$, and it divides no other natural numbers. So this relation contains all the pairs of that form.
answered Nov 23 at 13:25
Especially Lime
21.2k22655
21.2k22655
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add a comment |
5
Multiplication would be the obvious guess.
– user3482749
Nov 23 at 11:35
Math needs context. Without it, it could mean just about anything...
– JuliusL33t
Nov 23 at 11:55
It is the (graph of the) function $mathbb Ntomathbb N$ prescribed by $dmapsto dcdot r$.
– drhab
Nov 23 at 12:04
The current version of the OP has nothing to do with "dot." I'm guessing something got lost in translation in the editing process...
– Eff
Nov 23 at 13:22
I'm voting to close this question because the OP's latest edit makes it meaningless.
– Ethan Bolker
Nov 23 at 13:23