By means of an example, show that $P(A) + P(B) = 1$ does not mean that $B$ is the complement of $A$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)A, B, C and D, and we are trying to find the probability that exactly one event occurs.Infintely/Finitely Occurring Event Sequence in Terms of SubeventsThe letters A, E, I, P, Q, and R are arranged in a circle. Find the probability that at least 2 vowels are next to one another.expectation of Gamma distribution helpFinding independence of two random variablesFinding the probability of choosing six numbersSolving the probability of independent evnts without the complementFor events A, B from a sample space X, :Show there isn't the same probability for eventsExpected value of the mean of a Poisson distribution, where the mean has a Gamma distribution
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By means of an example, show that $P(A) + P(B) = 1$ does not mean that $B$ is the complement of $A$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)A, B, C and D, and we are trying to find the probability that exactly one event occurs.Infintely/Finitely Occurring Event Sequence in Terms of SubeventsThe letters A, E, I, P, Q, and R are arranged in a circle. Find the probability that at least 2 vowels are next to one another.expectation of Gamma distribution helpFinding independence of two random variablesFinding the probability of choosing six numbersSolving the probability of independent evnts without the complementFor events A, B from a sample space X, :Show there isn't the same probability for eventsExpected value of the mean of a Poisson distribution, where the mean has a Gamma distribution
$begingroup$
I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?
This is what I got to:
$P(A) + P(B) = 1$
$P(A) + P(A') = 1$
How could it be proven that $B$ isn't the complement of $A$?
Help would be greatly appreciated.
probability
edited Mar 24 at 18:22
Robert Howard
2,30331035
asked Mar 24 at 5:48
Gordon NgGordon Ng
342
$endgroup$
add a comment |
$begingroup$
I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?
This is what I got to:
$P(A) + P(B) = 1$
$P(A) + P(A') = 1$
How could it be proven that $B$ isn't the complement of $A$?
Help would be greatly appreciated.
probability
edited Mar 24 at 18:22
Robert Howard
2,30331035
asked Mar 24 at 5:48
Gordon NgGordon Ng
342
$endgroup$
6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51
7
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
Mar 24 at 10:01
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
Mar 24 at 18:23
add a comment |
$begingroup$
I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?
This is what I got to:
$P(A) + P(B) = 1$
$P(A) + P(A') = 1$
How could it be proven that $B$ isn't the complement of $A$?
Help would be greatly appreciated.
probability
edited Mar 24 at 18:22
Robert Howard
2,30331035
asked Mar 24 at 5:48
Gordon NgGordon Ng
342
$endgroup$
I'm in grade 10, and I've just started to learn about complementary events. I am rather perplexed with this question. Isn't this question kinda contradictory, since $P(A) + P(A') = 1$?
This is what I got to:
$P(A) + P(B) = 1$
$P(A) + P(A') = 1$
How could it be proven that $B$ isn't the complement of $A$?
Help would be greatly appreciated.
probability
probability
edited Mar 24 at 18:22
Robert Howard
2,30331035
asked Mar 24 at 5:48
Gordon NgGordon Ng
342
edited Mar 24 at 18:22
Robert Howard
2,30331035
asked Mar 24 at 5:48
Gordon NgGordon Ng
342
edited Mar 24 at 18:22
Robert Howard
2,30331035
edited Mar 24 at 18:22
Robert Howard
2,30331035
edited Mar 24 at 18:22
Robert Howard
2,30331035
2,30331035
asked Mar 24 at 5:48
Gordon NgGordon Ng
342
asked Mar 24 at 5:48
Gordon NgGordon Ng
342
asked Mar 24 at 5:48
Gordon NgGordon Ng
342
342
6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51
7
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
Mar 24 at 10:01
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
Mar 24 at 18:23
add a comment |
6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51
7
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
Mar 24 at 10:01
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
Mar 24 at 18:23
6
6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49
3
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51
7
7
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
Mar 24 at 10:01
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
Mar 24 at 10:01
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
Mar 24 at 18:23
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
Mar 24 at 18:23
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
Mar 24 at 18:29
add a comment |
$begingroup$
A counterexample
Take a normal 6-sided die.
Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6
Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6
P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1
But B is not the complement of A.
The complement of A is the event "roll any of the numbers 5 or 6".
By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.
Further
It also seems you have misunderstood the question.
You wrote How could it be proven that B isn't the complement of A?
This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.
answered Mar 24 at 11:28
user985366user985366
442310
$endgroup$
add a comment |
$begingroup$
As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.
answered Mar 24 at 5:55
TojrahTojrah
5246
$endgroup$
add a comment |
$begingroup$
From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.
answered Mar 24 at 11:29
user21820user21820
40.4k544163
$endgroup$
add a comment |
$begingroup$
Another example:
$$A = textGetting a head on coin A$$
$$B = textGetting a head on coin B$$
Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.
edited Mar 24 at 18:08
Robert Howard
2,30331035
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1214
$endgroup$
add a comment |
$begingroup$
Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus e$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.
Since $S = A cup B cup e$ we have:
$$
P(A) + P(B) + P(e) = 1
$$
but also
$$
P(e) = 0
$$
since it is finite, thus:
$$
P(A) + P(B) = 1
$$
For example $S = mathbbN$, $e=0$, $A=x mid x in mathbbN_0 wedge x text is even$ and $B=x mid x in mathbbN_0 wedge x text is odd$.
answered Mar 24 at 16:28
BakuriuBakuriu
12116
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
Mar 24 at 18:29
add a comment |
$begingroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
Mar 24 at 18:29
add a comment |
$begingroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
$endgroup$
Take any event of probability $frac 1 2$ and take $B=A$.
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
answered Mar 24 at 5:54
Kavi Rama MurthyKavi Rama Murthy
76.6k53471
76.6k53471
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
Mar 24 at 18:29
add a comment |
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
Mar 24 at 18:29
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
Mar 24 at 18:29
$begingroup$
And if it helps to give a more concrete example, we can say: I'm going to flip a coin just once. Let $A$ be the event that the flip will be heads, and let $B$ also be the event that the flip will be heads.
$endgroup$
– aschepler
Mar 24 at 18:29
add a comment |
$begingroup$
A counterexample
Take a normal 6-sided die.
Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6
Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6
P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1
But B is not the complement of A.
The complement of A is the event "roll any of the numbers 5 or 6".
By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.
Further
It also seems you have misunderstood the question.
You wrote How could it be proven that B isn't the complement of A?
This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.
answered Mar 24 at 11:28
user985366user985366
442310
$endgroup$
add a comment |
$begingroup$
A counterexample
Take a normal 6-sided die.
Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6
Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6
P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1
But B is not the complement of A.
The complement of A is the event "roll any of the numbers 5 or 6".
By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.
Further
It also seems you have misunderstood the question.
You wrote How could it be proven that B isn't the complement of A?
This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.
answered Mar 24 at 11:28
user985366user985366
442310
$endgroup$
add a comment |
$begingroup$
A counterexample
Take a normal 6-sided die.
Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6
Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6
P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1
But B is not the complement of A.
The complement of A is the event "roll any of the numbers 5 or 6".
By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.
Further
It also seems you have misunderstood the question.
You wrote How could it be proven that B isn't the complement of A?
This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.
answered Mar 24 at 11:28
user985366user985366
442310
$endgroup$
A counterexample
Take a normal 6-sided die.
Let event A be "roll any of the numbers 1, 2, 3 or 4". P(A) = 4/6
Let event B be "roll any of the numbers 1 or 2". P(B) = 2/6
P(A) + P(B) = 4/6 + 2/6 = 6/6 = 1
But B is not the complement of A.
The complement of A is the event "roll any of the numbers 5 or 6".
By this example, we've shown that P(A) + P(B) = 1 does not imply that A and B are complements.
Further
It also seems you have misunderstood the question.
You wrote How could it be proven that B isn't the complement of A?
This is not what you need to prove, and you cannot prove it just by knowing the probabilities. What you need to show is that it isn't always the case. You can show this by giving 1 counterexample, as above.
answered Mar 24 at 11:28
user985366user985366
442310
answered Mar 24 at 11:28
user985366user985366
442310
answered Mar 24 at 11:28
user985366user985366
442310
answered Mar 24 at 11:28
user985366user985366
442310
442310
add a comment |
add a comment |
$begingroup$
As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.
answered Mar 24 at 5:55
TojrahTojrah
5246
$endgroup$
add a comment |
$begingroup$
As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.
answered Mar 24 at 5:55
TojrahTojrah
5246
$endgroup$
add a comment |
$begingroup$
As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.
answered Mar 24 at 5:55
TojrahTojrah
5246
$endgroup$
As commented by @amd you can have many counter examples such as in throwing of dice. You can define any two events A and B. Such that n(A) + n(B)=6. So P(A)+P(B)=1. But A and B need not be necessarily disjoint sets.
answered Mar 24 at 5:55
TojrahTojrah
5246
answered Mar 24 at 5:55
TojrahTojrah
5246
answered Mar 24 at 5:55
TojrahTojrah
5246
answered Mar 24 at 5:55
TojrahTojrah
5246
5246
add a comment |
add a comment |
$begingroup$
From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.
answered Mar 24 at 11:29
user21820user21820
40.4k544163
$endgroup$
add a comment |
$begingroup$
From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.
answered Mar 24 at 11:29
user21820user21820
40.4k544163
$endgroup$
add a comment |
$begingroup$
From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.
answered Mar 24 at 11:29
user21820user21820
40.4k544163
$endgroup$
From the two statements you obtained (correctly), you can further obtain $P(A') = P(B)$. But that does not imply $A'=B$. Just as $x^2 = y^2$ for reals $x,y$ does not imply $x = y$.
answered Mar 24 at 11:29
user21820user21820
40.4k544163
answered Mar 24 at 11:29
user21820user21820
40.4k544163
answered Mar 24 at 11:29
user21820user21820
40.4k544163
answered Mar 24 at 11:29
user21820user21820
40.4k544163
40.4k544163
add a comment |
add a comment |
$begingroup$
Another example:
$$A = textGetting a head on coin A$$
$$B = textGetting a head on coin B$$
Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.
edited Mar 24 at 18:08
Robert Howard
2,30331035
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1214
$endgroup$
add a comment |
$begingroup$
Another example:
$$A = textGetting a head on coin A$$
$$B = textGetting a head on coin B$$
Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.
edited Mar 24 at 18:08
Robert Howard
2,30331035
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1214
$endgroup$
add a comment |
$begingroup$
Another example:
$$A = textGetting a head on coin A$$
$$B = textGetting a head on coin B$$
Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.
edited Mar 24 at 18:08
Robert Howard
2,30331035
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1214
$endgroup$
Another example:
$$A = textGetting a head on coin A$$
$$B = textGetting a head on coin B$$
Both have probability $0.5$, so $P(A)+P(B)=1$, but you could get a head on either, both or neither.
edited Mar 24 at 18:08
Robert Howard
2,30331035
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1214
edited Mar 24 at 18:08
Robert Howard
2,30331035
edited Mar 24 at 18:08
Robert Howard
2,30331035
edited Mar 24 at 18:08
Robert Howard
2,30331035
2,30331035
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1214
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1214
answered Mar 24 at 17:57
Artemis FowlArtemis Fowl
1214
1214
add a comment |
add a comment |
$begingroup$
Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus e$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.
Since $S = A cup B cup e$ we have:
$$
P(A) + P(B) + P(e) = 1
$$
but also
$$
P(e) = 0
$$
since it is finite, thus:
$$
P(A) + P(B) = 1
$$
For example $S = mathbbN$, $e=0$, $A=x mid x in mathbbN_0 wedge x text is even$ and $B=x mid x in mathbbN_0 wedge x text is odd$.
answered Mar 24 at 16:28
BakuriuBakuriu
12116
$endgroup$
add a comment |
$begingroup$
Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus e$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.
Since $S = A cup B cup e$ we have:
$$
P(A) + P(B) + P(e) = 1
$$
but also
$$
P(e) = 0
$$
since it is finite, thus:
$$
P(A) + P(B) = 1
$$
For example $S = mathbbN$, $e=0$, $A=x mid x in mathbbN_0 wedge x text is even$ and $B=x mid x in mathbbN_0 wedge x text is odd$.
answered Mar 24 at 16:28
BakuriuBakuriu
12116
$endgroup$
add a comment |
$begingroup$
Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus e$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.
Since $S = A cup B cup e$ we have:
$$
P(A) + P(B) + P(e) = 1
$$
but also
$$
P(e) = 0
$$
since it is finite, thus:
$$
P(A) + P(B) = 1
$$
For example $S = mathbbN$, $e=0$, $A=x mid x in mathbbN_0 wedge x text is even$ and $B=x mid x in mathbbN_0 wedge x text is odd$.
answered Mar 24 at 16:28
BakuriuBakuriu
12116
$endgroup$
Let $S$ be any infinite set and let $e in S$. Now let $A$ and $B$ be a partition of $S setminus e$ with both $A$ and $B$ infinite. Let $P(C)$ be the probability of drawing randomly an element of $S$ that is in the set $C$.
Since $S = A cup B cup e$ we have:
$$
P(A) + P(B) + P(e) = 1
$$
but also
$$
P(e) = 0
$$
since it is finite, thus:
$$
P(A) + P(B) = 1
$$
For example $S = mathbbN$, $e=0$, $A=x mid x in mathbbN_0 wedge x text is even$ and $B=x mid x in mathbbN_0 wedge x text is odd$.
answered Mar 24 at 16:28
BakuriuBakuriu
12116
answered Mar 24 at 16:28
BakuriuBakuriu
12116
answered Mar 24 at 16:28
BakuriuBakuriu
12116
answered Mar 24 at 16:28
BakuriuBakuriu
12116
12116
add a comment |
add a comment |
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6
$begingroup$
I believe a better statement would be "show that $B$ is not necessarily the complement of $A$".
$endgroup$
– Paras Khosla
Mar 24 at 5:49
3
$begingroup$
A common way to show that something doesn’t hold is to come up with a specific counterexample.
$endgroup$
– amd
Mar 24 at 5:51
7
$begingroup$
If you flip a coin and get tails, does it mean that the next person you'll meet cannot be a girl?
$endgroup$
– Eric Duminil
Mar 24 at 10:01
$begingroup$
Welcome to MSE; this is a good first question! Here's a brief tutorial on how we format math on this site, like I did in my edit to your question: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Robert Howard
Mar 24 at 18:23