Consider the universe N of all integers along with the following open statement:

What happens when we quantify the m variable? With the existential quantifier, we get

∀m [p(m, n)]

which can be translated to “For every integer m, we have that m + n = 0.” Is this true? We would need to know what value n holds to be certain. We can’t just pick any value of n. The above statement would be false when m = 1 and n = 2. We just can’t be sure. As such, ∀m [p(m, n)] does not have a truth value.

Now consider the statement

∃m [p(m, n)]

which can be translated to “There exists an integer m such that m + n = 0.” Again, until a value for n is known, we can’t be sure. Of course, we could figure out what value n would need to be in order for m + n = 0. For example, when m = 5, we could make n = -5. But again, the value of n is unknown. As such, the statement ∃m [p(m, n)] is open, and has no definite truth value.

A basic law of arithmetic is the commutative law, where, in a sum or product, the two constituent parts can be swapped without affecting the result. In other words, for any two real numbers x and y, we have that

x + y = y + x.

Consider the universe 𝒰 of all real numbers. We can express the commutative laws of arithmetic using quantifiers like this:

∀x∀y [x + y = y + x]
∀x∀y [x * y = y * x]

The order of the quantified variables does not matter, so we could also express the commutative laws of arithmetic like this:

∀y∀x [x + y = y + x]
∀y∀x [x * y = y * x]

Consider the universe N consisting of all integers.

Consider the statement

“The integer 100 is the sum of two perfect squares.”

The presence of the definite article the suggests that the existential quantifier is the implied quantifier. As such, we could express the statement using quantifiers like this:

∃m∃n [100 = m² + n²].

Again, the order of the quantification does not matter in this case, so we could also write

∃n∃m [100 = m² + n²].

One more way to write this is to specify which universe each variable belongs to using the ∈ symbol discussed on this page:

∃m∈N ∃n∈N [100 = m² + n²].

Consider the universe of all real numbers, which we’ll denote R.

When dealing with real numbers, another commonly used arithmetic law is the distribution of multiplication over addition. Typically, this is expressed as follows:

“For every real number x, y, and z, we have that x(y + z) = xy + xz.”

Here, the quantifier being used is the universal quantifier, and it is being applied to all variables, so we can concisely represent this statement with the following proposition:

∀x ∀y ∀z [x(y + z) = xy + xz].

Using the short-hand notation discussed above, we could also write this as

∀x, y, z [x(y + z) = xy + xz].

Consider the universe consisting of all real numbers.

It is known that every number has an additive inverse. What this basically means is this:

“For every real number x, there exists some other number y where x + y = 0.”

Notice that we used the phrases “For every….” and “There exists….”, which suggests both a universal quantifier and an existential quantifier. The phrase “For every…” comes first, suggesting it is the first quantifier, and so the existential quantifier comes second. As such we can rewrite the above statement as

∀x ∃y [x + y = 0].

Since x is universally quantified, we can pick arbitrary values to test and see if this works.

If x = 3008, then we know that y = -3008 will work.

Similarly, taking x = -π², we can set y = π².

Indeed, as soon as a value of x is selected, we can just take y = -x, because since x is a real number, so is -x. Thus, no matter what real number we pick for x, we can always find an appropriate value for y.

Let’s reconsider Example 1.10.5 regarding additive inverses.

Suppose we swapped quantifiers like we did when we examined repeated quantifiers. In doing so, we form the following statement:

∃y ∀x [x + y = 0].

This proposition can be translated into English as “There exists some real number y, such that for every real number x, we have that x + y = 0.”

Here, the proposition is asserting the existence of some such real number. Then, once that real number is determined, supposedly, every real number added to it will yield a sum of 0.

We know that y can’t be 17, because 3 + 17 ≠ 0 (here, y = 17 and x = 3.)

Similarly, we know that y can’t be -0.745 because 20 – 0.745 ≠ 0 (here, y = -0.745 and x = 20.)

As a matter of fact, no such value of y can exist, because as soon as we pick a value for y, there are infinitely many ways to pick the value of x (since x is universally quantified, the statement asserts that every value of x will work, which so far is not the case.) Since there are infinitely many values to substitute in for x, there are infinitely many sums.

Out of those infinitely many sums, only one sum will be 0. That happens when we pick x to be equal to -y. Of course, since x is universally quantified, we could also pick the real number (-y + 1), which will yield a sum of 1, and not 0.

Surprisingly, if this statement were true, every single real number would need to be equal to 0, an obviously ridiculous scenario!

Suppose we have universe 𝒰, and that the open statements a(x, y), b(x, y), and c(x, y) are defined on that universe.

We want to negate the following quantified statement:

∃x ∀y [a(x, y) ∧ b(x, y) → c(x, y)].

Before we start, we can make the following short-hand substitutions.

a(x, y): a
b(x, y): b
c(x, y): c

We do this to make the proposition a little less messy:

∃x ∀y [(a ∧ b) → c].

Now we perform the negation:

Substituting a(x, y), b(x, y), c(x, y) back in for a, b, and c respectively, we get the negation of

∃x ∀y [a(x, y) ∧ b(x, y) → c(x, y)]

is the statement

∀x ∃y [a(x, y) ∧ b(x, y) ∧ ¬c(x, y)].