Many statements in mathematics involve multple quantifiers. For example, βFor all real numbers \(x\text{,}\) there exists a real number \(y\) with \(x+y=0\text{.}\)β These statements, though frequent in math courses, represent some of the most complicated types of statements to understand. In this section we will try to understand the general structure of such statements.
Letβs look at the various ways we could have statements with two quantifiers. Since we have two quantifiers, we will have two variables. Thus, our predicate will now involve both variables. We can use the notation \(P(x, y)\) to mean a statement about \(x\) and \(y\text{.}\)
This means we can find a \(y\) in \(E\) that makes \(P(x, y)\) true for each \(x\) in \(D\text{.}\) In such statements \(y\) will often depend on \(x\text{.}\)
The order of the quantifiers matters. This means we can find a single\(x\) in \(D\) that makes \(P(x, y)\) true for every \(y\) in \(E\text{.}\) In such statements \(x\) does not depend on \(y\text{.}\) The same \(x\) must work for all the \(y\)βs.
Although we wonβt study the mathematical context of the following example in this class, it is a classic example in mathematics of a statement with multiple quantifiers and a conditional. It is so important that it is the subject of one set of math bike racks in front of Taylor Hall!
The definition of the limit of a function: For every \(\epsilon>0\text{,}\) there exists a \(\delta>0\) such that if \(|x-a|<\delta\) then \(|f(x)-L|<\epsilon\text{.}\) Or in symbols, \(\forall\epsilon>0, \exists\delta>0, |x-a|<\delta\rightarrow |f(x)-L|<\epsilon\text{.}\)
Now, we want to be able to negate statements with multiple quantifiers. There is nothing really new here, we just negate our quantified statements as we did for single quantifiers.
If you take Elementary Analysis, you will need to be able to negate the definition of the limit from ExampleΒ 3.3.2. Negate βFor every \(\epsilon>0\text{,}\) there exists a \(\delta>0\) such that if \(|x-a|<\delta\) then \(|f(x)-L|<\epsilon\text{.}\)β Hint: Our statement has the form \(\forall \epsilon, \exists \delta, P(\delta)\rightarrow Q(\epsilon).\) So the negation has the form \(\exists \epsilon, \forall \delta, P(\delta)\wedge \sim Q(\epsilon).\)
Let \(D=E=\{-2, -1, 0, 1, 2\}\text{.}\) Write the negations for each of the following statements and determine which is true, the given statement or the negation.
For each of the following, (1) rewrite the statement in English without the symbols \(\forall\) or \(\exists\) or variables. Then (2) indicate whether the statement is true or false.
For each of the following, (1) rewrite the statement in English without the symbols \(\forall\) or \(\exists\) or variables, trying to express your answer as simply as possible. Then (2) write the negation for each statement.
For each of the following, (1) write a new statement by interchanging the symbols \(\forall\) and \(\exists\text{.}\) Then (2) state which is true: the given statement, the interchanged statement, neither, or both.
For each of the following statements, (1) rewrite the statement formally using quantifiers and variables, and (2) write the negation for the statement.