Negation of Statements (check answers...)
There exists x ∈ IR such that f(x) > 100.
Negation: f(x) <= 99 for all x ∈ IR (is this right?)
There exist a, b ∈ IN such that (a/b)^2 = 2.
(not sure about this one)
Everybody has one foot longer than the other.
negation: There is at least one person who has a foot shorter than the other. (is this right?)
Every parent over fifty has a child who is married.
negation: There exists one parent who has a child who is not married. (is this right?)
For all ε > 0 there exists n ∈ IN such that 1/n < ε.
(not sure about this one)
Negation: f(x) <= 99 for all x ∈ IR (is this right?)
There exist a, b ∈ IN such that (a/b)^2 = 2.
(not sure about this one)
Everybody has one foot longer than the other.
negation: There is at least one person who has a foot shorter than the other. (is this right?)
Every parent over fifty has a child who is married.
negation: There exists one parent who has a child who is not married. (is this right?)
For all ε > 0 there exists n ∈ IN such that 1/n < ε.
(not sure about this one)
Original post by E--
There exists x ∈ IR such that f(x) > 100.
Negation: f(x) <= 99 for all x ∈ IR (is this right?)
There exist a, b ∈ IN such that (a/b)^2 = 2.
(not sure about this one)
Everybody has one foot longer than the other.
negation: There is at least one person who has a foot shorter than the other. (is this right?)
Every parent over fifty has a child who is married.
negation: There exists one parent who has a child who is not married. (is this right?)
For all ε > 0 there exists n ∈ IN such that 1/n < ε.
(not sure about this one)
Negation: f(x) <= 99 for all x ∈ IR (is this right?)
There exist a, b ∈ IN such that (a/b)^2 = 2.
(not sure about this one)
Everybody has one foot longer than the other.
negation: There is at least one person who has a foot shorter than the other. (is this right?)
Every parent over fifty has a child who is married.
negation: There exists one parent who has a child who is not married. (is this right?)
For all ε > 0 there exists n ∈ IN such that 1/n < ε.
(not sure about this one)
1. Close but why go down to 99? The negation of f(x) > 100 is f(x) <= 100.
2. Start with negating the "there exists" part. In negation we always swap there exists and for all. As for (a/b)^2 = 2, what do you think the negation for this is?
3. This one is a little cheeky I think. If you have a foot shorter than the other, then you also have a foot longer than the other (just swap the order you're considering the feet in!). The only situation in which you don't have one foot longer than the other is if both your feet are the same length, right?
4. Close, but aren't you forgetting something about the parent?
5. Again, swap "for all" with "there exists" and vice versa. Then how do you negate an inequality?
(edited 5 years ago)
The first one should been;
Negation: f(x) <= 100 for all x ∈ IR
For the proposition: Every parent over fifty has a child who is married.
Let P be the set of all parents,
M be the set of all married adults.
Then the original proposition states: ∀x∈P, x>50→∃x'∈M You can think of x' as a child (element) out of all children the chosen parent has (out of the set of all children that the chosen parent has).
Therefore the negation is ∃x∈P, (x>50)Λ(∀x'∉M)
which can be read as; There exists a parent over 50 whose children are not married.Notice that if one can find a parent (who is over 50) with only one child and that is not married, then this can be given as a counter example to the original proposition to disprove it, so make sure that the negated proposition not only includes parents with more than one child, but also parent with one child.
Negation: f(x) <= 100 for all x ∈ IR
For the proposition: Every parent over fifty has a child who is married.
Let P be the set of all parents,
M be the set of all married adults.
Then the original proposition states: ∀x∈P, x>50→∃x'∈M You can think of x' as a child (element) out of all children the chosen parent has (out of the set of all children that the chosen parent has).
Therefore the negation is ∃x∈P, (x>50)Λ(∀x'∉M)
which can be read as; There exists a parent over 50 whose children are not married.Notice that if one can find a parent (who is over 50) with only one child and that is not married, then this can be given as a counter example to the original proposition to disprove it, so make sure that the negated proposition not only includes parents with more than one child, but also parent with one child.
(edited 5 years ago)
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