Negation of a statement

  • #1

Homework Statement


Why is the negation of: Atleast one hole contains atleast n+1 pigeons where n is a natural number

There are less than n+1 pigeons per hole?

Wouldn't it be: There are less than one hole that contains less than n+1 pigeons?

Wouldn't the negation of there exists a hole that contains atleast n+1 pigeons yield

all holes contain less than n+1 pigeons?

Why is the existential quantifier assumed for the phrase: "Atleast one hole" in the first part of the sentence but not for the "atleast n+1 pigeons"?
 
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Answers and Replies

  • #2
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Why is the negation of: At least one hole contains at least n+1 pigeons where n is a natural number
Let's make it formal and set ##N(L)## the number of pigeons in the hole ##L##.
Then the statement says: ##\exists L\, : \,|N(L)|\geq n+1##
which has the negation: ##\forall L \, : \,|N(L)| < n+1##
or in words
There are less than n+1 pigeons per hole?
 
  • #3
Let's make it formal and set ##N(L)## the number of pigeons in the hole ##L##.
Then the statement says: ##\exists L\, : \,|N(L)|\geq n+1##
which has the negation: ##\forall L \, : \,|N(L)| < n+1##
or in words
What about my other comments? because thats exactly what I'm confused about, why is atleast turned into an existential quantifier in the beginning of the sentence but not in the other part of the sentence?
 
  • #4
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11,697
What about my other comments? because that's exactly what I'm confused about, why is at least turned into an existential quantifier in the beginning of the sentence but not in the other part of the sentence?
"At least one hole contains at least n+1 pigeons."

  • at least one hole = subject = exists, for otherwise we wouldn't talk about it; it specifies the subject we are talking about and which is not the empty set, i.e. it is necessary to exist = ##\exists\, L##
  • contains = predicate = announces a property, i.e. something can be said about the hole = ##":"##
  • at least n+1 pigeons = object = the something which can be said about the hole are the number of pigeons in it, i.e. the hole has n+1 pigeons = ##N(L)>n##

Thus we have ##\exists L\, \, : \, N(L) > n##

We can also say: ##\{\,L\,|\,N(L)>n\,\}\neq \emptyset## and then the negation is, that this set is empty. If the set is empty, then it's complement is the entire space, which are all holes in this case. So ##\{\,L\,|\,N(L)>n\,\}= \emptyset \Longrightarrow \{\,L\,|\,N(L)\leq n\,\} = \{\,L\,\}##

We can also say: ##\exists L\, : \,L \wedge N(L)>n## with the negation ##\forall L\, : \,\lnot L \vee N(L)\leq n##, which means, either it isn't a hole, or in case it is, there are at most ##n## pigeons in it.
 

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