Mixing
Binary relations on a set
$begingroup$
I have a homework problem that asks this...
a) List all the different binary relations on the set ${0,1}$
I assume that since the relation is not given then the answer must be the graph, or Cartesian product of the set. This only provides 4 sets. I should be getting this for a solution but I cannot see how to get there. Any explanation would be appreciated.
- $emptyset$
- ${(0, 0)}$
- ${(0, 1)}$
- ${(1, 0)}$
- ${(1, 1)}$
- ${(0, 0), (0, 1)}$
- ${(0, 0), (1, 0)}$
- ${(0, 0), (1, 1)}$
- ${(0, 1), (1, 0)}$
- ${(0, 1), (1, 1)}$
- ${(1, 0), (1, 1)}$
- ${(0, 0), (0, 1), (1, 0)}$
- ${(0, 0), (0, 1), (1, 1)}$
- ${(0, 0), (1, 0), (1, 1)}$
- ${(0, 1), (1, 0), (1, 1)}$
- ${(0, 0), (0, 1), (1, 0), (1, 1)}$
This is from an MIT OCW course which I am taking as an independent study as a high school junior.
relations
edited Apr 15 '14 at 4:07
Mark Fantini
4,86041936
asked Apr 15 '14 at 3:28
Liamdev631Liamdev631
1
$endgroup$
add a comment |
$begingroup$
I have a homework problem that asks this...
a) List all the different binary relations on the set ${0,1}$
I assume that since the relation is not given then the answer must be the graph, or Cartesian product of the set. This only provides 4 sets. I should be getting this for a solution but I cannot see how to get there. Any explanation would be appreciated.
- $emptyset$
- ${(0, 0)}$
- ${(0, 1)}$
- ${(1, 0)}$
- ${(1, 1)}$
- ${(0, 0), (0, 1)}$
- ${(0, 0), (1, 0)}$
- ${(0, 0), (1, 1)}$
- ${(0, 1), (1, 0)}$
- ${(0, 1), (1, 1)}$
- ${(1, 0), (1, 1)}$
- ${(0, 0), (0, 1), (1, 0)}$
- ${(0, 0), (0, 1), (1, 1)}$
- ${(0, 0), (1, 0), (1, 1)}$
- ${(0, 1), (1, 0), (1, 1)}$
- ${(0, 0), (0, 1), (1, 0), (1, 1)}$
This is from an MIT OCW course which I am taking as an independent study as a high school junior.
relations
edited Apr 15 '14 at 4:07
Mark Fantini
4,86041936
asked Apr 15 '14 at 3:28
Liamdev631Liamdev631
1
$endgroup$
2
$begingroup$
I think I found a solution. I guess I am supposed to list the subsets of AxA, which would give me 16 combinations.
$endgroup$
– Liamdev631
Apr 15 '14 at 3:41
add a comment |
$begingroup$
I have a homework problem that asks this...
a) List all the different binary relations on the set ${0,1}$
I assume that since the relation is not given then the answer must be the graph, or Cartesian product of the set. This only provides 4 sets. I should be getting this for a solution but I cannot see how to get there. Any explanation would be appreciated.
- $emptyset$
- ${(0, 0)}$
- ${(0, 1)}$
- ${(1, 0)}$
- ${(1, 1)}$
- ${(0, 0), (0, 1)}$
- ${(0, 0), (1, 0)}$
- ${(0, 0), (1, 1)}$
- ${(0, 1), (1, 0)}$
- ${(0, 1), (1, 1)}$
- ${(1, 0), (1, 1)}$
- ${(0, 0), (0, 1), (1, 0)}$
- ${(0, 0), (0, 1), (1, 1)}$
- ${(0, 0), (1, 0), (1, 1)}$
- ${(0, 1), (1, 0), (1, 1)}$
- ${(0, 0), (0, 1), (1, 0), (1, 1)}$
This is from an MIT OCW course which I am taking as an independent study as a high school junior.
relations
edited Apr 15 '14 at 4:07
Mark Fantini
4,86041936
asked Apr 15 '14 at 3:28
Liamdev631Liamdev631
1
$endgroup$
I have a homework problem that asks this...
a) List all the different binary relations on the set ${0,1}$
I assume that since the relation is not given then the answer must be the graph, or Cartesian product of the set. This only provides 4 sets. I should be getting this for a solution but I cannot see how to get there. Any explanation would be appreciated.
- $emptyset$
- ${(0, 0)}$
- ${(0, 1)}$
- ${(1, 0)}$
- ${(1, 1)}$
- ${(0, 0), (0, 1)}$
- ${(0, 0), (1, 0)}$
- ${(0, 0), (1, 1)}$
- ${(0, 1), (1, 0)}$
- ${(0, 1), (1, 1)}$
- ${(1, 0), (1, 1)}$
- ${(0, 0), (0, 1), (1, 0)}$
- ${(0, 0), (0, 1), (1, 1)}$
- ${(0, 0), (1, 0), (1, 1)}$
- ${(0, 1), (1, 0), (1, 1)}$
- ${(0, 0), (0, 1), (1, 0), (1, 1)}$
This is from an MIT OCW course which I am taking as an independent study as a high school junior.
relations
relations
edited Apr 15 '14 at 4:07
Mark Fantini
4,86041936
asked Apr 15 '14 at 3:28
Liamdev631Liamdev631
1
edited Apr 15 '14 at 4:07
Mark Fantini
4,86041936
asked Apr 15 '14 at 3:28
Liamdev631Liamdev631
1
edited Apr 15 '14 at 4:07
Mark Fantini
4,86041936
edited Apr 15 '14 at 4:07
Mark Fantini
4,86041936
edited Apr 15 '14 at 4:07
Mark Fantini
4,86041936
4,86041936
asked Apr 15 '14 at 3:28
Liamdev631Liamdev631
1
asked Apr 15 '14 at 3:28
Liamdev631Liamdev631
1
asked Apr 15 '14 at 3:28
Liamdev631Liamdev631
1
1
2
$begingroup$
I think I found a solution. I guess I am supposed to list the subsets of AxA, which would give me 16 combinations.
$endgroup$
– Liamdev631
Apr 15 '14 at 3:41
add a comment |
2
$begingroup$
I think I found a solution. I guess I am supposed to list the subsets of AxA, which would give me 16 combinations.
$endgroup$
– Liamdev631
Apr 15 '14 at 3:41
2
2
$begingroup$
I think I found a solution. I guess I am supposed to list the subsets of AxA, which would give me 16 combinations.
$endgroup$
– Liamdev631
Apr 15 '14 at 3:41
$begingroup$
I think I found a solution. I guess I am supposed to list the subsets of AxA, which would give me 16 combinations.
$endgroup$
– Liamdev631
Apr 15 '14 at 3:41
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You have four choices. $0$ can be related to $0$ or not. $0$ can be related to $1$ or not. $1$ can ... etc. until $1$ can be related to $1$ or not. The total number of options is therefore $$2*2*2*2=2^4=16$$
$endgroup$
$begingroup$
The question does not define what it considers a relation. Maybe does it mean aRb if a=b?
$endgroup$
– Liamdev631
Apr 15 '14 at 3:50
add a comment |
Your Answer
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1 Answer
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$begingroup$
You have four choices. $0$ can be related to $0$ or not. $0$ can be related to $1$ or not. $1$ can ... etc. until $1$ can be related to $1$ or not. The total number of options is therefore $$2*2*2*2=2^4=16$$
$endgroup$
$begingroup$
The question does not define what it considers a relation. Maybe does it mean aRb if a=b?
$endgroup$
– Liamdev631
Apr 15 '14 at 3:50
add a comment |
$begingroup$
You have four choices. $0$ can be related to $0$ or not. $0$ can be related to $1$ or not. $1$ can ... etc. until $1$ can be related to $1$ or not. The total number of options is therefore $$2*2*2*2=2^4=16$$
$endgroup$
$begingroup$
The question does not define what it considers a relation. Maybe does it mean aRb if a=b?
$endgroup$
– Liamdev631
Apr 15 '14 at 3:50
add a comment |
$begingroup$
You have four choices. $0$ can be related to $0$ or not. $0$ can be related to $1$ or not. $1$ can ... etc. until $1$ can be related to $1$ or not. The total number of options is therefore $$2*2*2*2=2^4=16$$
$endgroup$
You have four choices. $0$ can be related to $0$ or not. $0$ can be related to $1$ or not. $1$ can ... etc. until $1$ can be related to $1$ or not. The total number of options is therefore $$2*2*2*2=2^4=16$$
answered Apr 15 '14 at 3:35
user142299
$begingroup$
The question does not define what it considers a relation. Maybe does it mean aRb if a=b?
$endgroup$
– Liamdev631
Apr 15 '14 at 3:50
add a comment |
$begingroup$
The question does not define what it considers a relation. Maybe does it mean aRb if a=b?
$endgroup$
– Liamdev631
Apr 15 '14 at 3:50
$begingroup$
The question does not define what it considers a relation. Maybe does it mean aRb if a=b?
$endgroup$
– Liamdev631
Apr 15 '14 at 3:50
$begingroup$
The question does not define what it considers a relation. Maybe does it mean aRb if a=b?
$endgroup$
– Liamdev631
Apr 15 '14 at 3:50
add a comment |
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$begingroup$
I think I found a solution. I guess I am supposed to list the subsets of AxA, which would give me 16 combinations.
$endgroup$
– Liamdev631
Apr 15 '14 at 3:41