Determining whether or not a set contains an element, and proving set equalities.
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I have two problems that are giving me a bit of a difficulty. I've already completed one of them, but I have a feeling I'm doing something wrong, and would greatly appreciate feedback. For the other one, I'm lost entirely.
For each of the following sets, determine whether 4 is an element of that set.
a) {x ∈ R | x is an integer greater than 4}
b) {x ∈ R | x is the square of an integer|
c) {4,{4}}
d) {{4},{{4}}}
e) {{4},{4,{4}}}
f) {{{4}}}
I answered as the following:
a. As x is greater than 4, 4 is not in the set.
b. If the integer is 2, the square of 2 is 4, meaning that 4 is an element of the set. (Truthfully, I'm on the fence about this. How do I know that the square is 2?)
c. The set contains the value 4, and the subset {4}. 4 is in the set.
d. The set contains the subsets {4} and {{4}}. 4 is not an element in the set. (I assume that {4} isn't the same as 4)
e. The set contains the subset {4} and {4,{4}}. 4 is an element of the subset, but not the set.
f. The set consists wholly of the subset {{{4}}}. 4 is not an element.
The second problem I'm lost on completely is
Prove or disprove:
A - (B n C) = (A-B) u (A-C)
I know that A is equivalent to (A u A), so I can go from
A- (B n C) = (A-B) u (A-C)
(A u A) - (B n C) = (A-B) u (A-C)
But from there, I'm lost. I don't even know how A-B or A-C would translate into anything else, so any assistance would be greatly, greatly appreciated.
discrete-mathematics
asked Oct 1 '14 at 0:17
FaraamFaraam
31
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add a comment |
$begingroup$
I have two problems that are giving me a bit of a difficulty. I've already completed one of them, but I have a feeling I'm doing something wrong, and would greatly appreciate feedback. For the other one, I'm lost entirely.
For each of the following sets, determine whether 4 is an element of that set.
a) {x ∈ R | x is an integer greater than 4}
b) {x ∈ R | x is the square of an integer|
c) {4,{4}}
d) {{4},{{4}}}
e) {{4},{4,{4}}}
f) {{{4}}}
I answered as the following:
a. As x is greater than 4, 4 is not in the set.
b. If the integer is 2, the square of 2 is 4, meaning that 4 is an element of the set. (Truthfully, I'm on the fence about this. How do I know that the square is 2?)
c. The set contains the value 4, and the subset {4}. 4 is in the set.
d. The set contains the subsets {4} and {{4}}. 4 is not an element in the set. (I assume that {4} isn't the same as 4)
e. The set contains the subset {4} and {4,{4}}. 4 is an element of the subset, but not the set.
f. The set consists wholly of the subset {{{4}}}. 4 is not an element.
The second problem I'm lost on completely is
Prove or disprove:
A - (B n C) = (A-B) u (A-C)
I know that A is equivalent to (A u A), so I can go from
A- (B n C) = (A-B) u (A-C)
(A u A) - (B n C) = (A-B) u (A-C)
But from there, I'm lost. I don't even know how A-B or A-C would translate into anything else, so any assistance would be greatly, greatly appreciated.
discrete-mathematics
asked Oct 1 '14 at 0:17
FaraamFaraam
31
$endgroup$
add a comment |
$begingroup$
I have two problems that are giving me a bit of a difficulty. I've already completed one of them, but I have a feeling I'm doing something wrong, and would greatly appreciate feedback. For the other one, I'm lost entirely.
For each of the following sets, determine whether 4 is an element of that set.
a) {x ∈ R | x is an integer greater than 4}
b) {x ∈ R | x is the square of an integer|
c) {4,{4}}
d) {{4},{{4}}}
e) {{4},{4,{4}}}
f) {{{4}}}
I answered as the following:
a. As x is greater than 4, 4 is not in the set.
b. If the integer is 2, the square of 2 is 4, meaning that 4 is an element of the set. (Truthfully, I'm on the fence about this. How do I know that the square is 2?)
c. The set contains the value 4, and the subset {4}. 4 is in the set.
d. The set contains the subsets {4} and {{4}}. 4 is not an element in the set. (I assume that {4} isn't the same as 4)
e. The set contains the subset {4} and {4,{4}}. 4 is an element of the subset, but not the set.
f. The set consists wholly of the subset {{{4}}}. 4 is not an element.
The second problem I'm lost on completely is
Prove or disprove:
A - (B n C) = (A-B) u (A-C)
I know that A is equivalent to (A u A), so I can go from
A- (B n C) = (A-B) u (A-C)
(A u A) - (B n C) = (A-B) u (A-C)
But from there, I'm lost. I don't even know how A-B or A-C would translate into anything else, so any assistance would be greatly, greatly appreciated.
discrete-mathematics
asked Oct 1 '14 at 0:17
FaraamFaraam
31
$endgroup$
I have two problems that are giving me a bit of a difficulty. I've already completed one of them, but I have a feeling I'm doing something wrong, and would greatly appreciate feedback. For the other one, I'm lost entirely.
For each of the following sets, determine whether 4 is an element of that set.
a) {x ∈ R | x is an integer greater than 4}
b) {x ∈ R | x is the square of an integer|
c) {4,{4}}
d) {{4},{{4}}}
e) {{4},{4,{4}}}
f) {{{4}}}
I answered as the following:
a. As x is greater than 4, 4 is not in the set.
b. If the integer is 2, the square of 2 is 4, meaning that 4 is an element of the set. (Truthfully, I'm on the fence about this. How do I know that the square is 2?)
c. The set contains the value 4, and the subset {4}. 4 is in the set.
d. The set contains the subsets {4} and {{4}}. 4 is not an element in the set. (I assume that {4} isn't the same as 4)
e. The set contains the subset {4} and {4,{4}}. 4 is an element of the subset, but not the set.
f. The set consists wholly of the subset {{{4}}}. 4 is not an element.
The second problem I'm lost on completely is
Prove or disprove:
A - (B n C) = (A-B) u (A-C)
I know that A is equivalent to (A u A), so I can go from
A- (B n C) = (A-B) u (A-C)
(A u A) - (B n C) = (A-B) u (A-C)
But from there, I'm lost. I don't even know how A-B or A-C would translate into anything else, so any assistance would be greatly, greatly appreciated.
discrete-mathematics
discrete-mathematics
asked Oct 1 '14 at 0:17
FaraamFaraam
31
asked Oct 1 '14 at 0:17
FaraamFaraam
31
asked Oct 1 '14 at 0:17
FaraamFaraam
31
asked Oct 1 '14 at 0:17
FaraamFaraam
31
asked Oct 1 '14 at 0:17
FaraamFaraam
31
31
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2 Answers
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Your logic in 1 is fine. For part 2, to prove two sets are equal, show they are subsets of each other.
So, assume $xin A-(Bcap C)$ Xo, $xin A$ and $xnotin (Bcap C)$ Thus, we have $xnotin B$ or $xnotin C$ along with $xin A$. thus, we have $xin A-B$ or $xin A-C$
Hence, $xin (A-B)cup(A-C)$
To argue the reverse, just start with an element of the right hand side and follow the same logic backward to the left hand side
answered Oct 1 '14 at 1:04
AlanAlan
8,58221636
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For your set equality,
draw a Venn diagram with
circles for A, B, and C.
You can then visually see
what the two sides are
and why they are equal.
answered Oct 1 '14 at 1:29
marty cohenmarty cohen
73.3k549128
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2 Answers
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2 Answers
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$begingroup$
Your logic in 1 is fine. For part 2, to prove two sets are equal, show they are subsets of each other.
So, assume $xin A-(Bcap C)$ Xo, $xin A$ and $xnotin (Bcap C)$ Thus, we have $xnotin B$ or $xnotin C$ along with $xin A$. thus, we have $xin A-B$ or $xin A-C$
Hence, $xin (A-B)cup(A-C)$
To argue the reverse, just start with an element of the right hand side and follow the same logic backward to the left hand side
answered Oct 1 '14 at 1:04
AlanAlan
8,58221636
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add a comment |
$begingroup$
Your logic in 1 is fine. For part 2, to prove two sets are equal, show they are subsets of each other.
So, assume $xin A-(Bcap C)$ Xo, $xin A$ and $xnotin (Bcap C)$ Thus, we have $xnotin B$ or $xnotin C$ along with $xin A$. thus, we have $xin A-B$ or $xin A-C$
Hence, $xin (A-B)cup(A-C)$
To argue the reverse, just start with an element of the right hand side and follow the same logic backward to the left hand side
answered Oct 1 '14 at 1:04
AlanAlan
8,58221636
$endgroup$
add a comment |
$begingroup$
Your logic in 1 is fine. For part 2, to prove two sets are equal, show they are subsets of each other.
So, assume $xin A-(Bcap C)$ Xo, $xin A$ and $xnotin (Bcap C)$ Thus, we have $xnotin B$ or $xnotin C$ along with $xin A$. thus, we have $xin A-B$ or $xin A-C$
Hence, $xin (A-B)cup(A-C)$
To argue the reverse, just start with an element of the right hand side and follow the same logic backward to the left hand side
answered Oct 1 '14 at 1:04
AlanAlan
8,58221636
$endgroup$
Your logic in 1 is fine. For part 2, to prove two sets are equal, show they are subsets of each other.
So, assume $xin A-(Bcap C)$ Xo, $xin A$ and $xnotin (Bcap C)$ Thus, we have $xnotin B$ or $xnotin C$ along with $xin A$. thus, we have $xin A-B$ or $xin A-C$
Hence, $xin (A-B)cup(A-C)$
To argue the reverse, just start with an element of the right hand side and follow the same logic backward to the left hand side
answered Oct 1 '14 at 1:04
AlanAlan
8,58221636
answered Oct 1 '14 at 1:04
AlanAlan
8,58221636
answered Oct 1 '14 at 1:04
AlanAlan
8,58221636
answered Oct 1 '14 at 1:04
AlanAlan
8,58221636
8,58221636
add a comment |
add a comment |
$begingroup$
For your set equality,
draw a Venn diagram with
circles for A, B, and C.
You can then visually see
what the two sides are
and why they are equal.
answered Oct 1 '14 at 1:29
marty cohenmarty cohen
73.3k549128
$endgroup$
add a comment |
$begingroup$
For your set equality,
draw a Venn diagram with
circles for A, B, and C.
You can then visually see
what the two sides are
and why they are equal.
answered Oct 1 '14 at 1:29
marty cohenmarty cohen
73.3k549128
$endgroup$
add a comment |
$begingroup$
For your set equality,
draw a Venn diagram with
circles for A, B, and C.
You can then visually see
what the two sides are
and why they are equal.
answered Oct 1 '14 at 1:29
marty cohenmarty cohen
73.3k549128
$endgroup$
For your set equality,
draw a Venn diagram with
circles for A, B, and C.
You can then visually see
what the two sides are
and why they are equal.
answered Oct 1 '14 at 1:29
marty cohenmarty cohen
73.3k549128
answered Oct 1 '14 at 1:29
marty cohenmarty cohen
73.3k549128
answered Oct 1 '14 at 1:29
marty cohenmarty cohen
73.3k549128
answered Oct 1 '14 at 1:29
marty cohenmarty cohen
73.3k549128
73.3k549128
add a comment |
add a comment |
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