Page 75 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Quantifiers 61
Once again, this is true, since the solution is y = 5 − 7 =−2. In fact, you
have probably realized by now that no matter what value we plug in for x, the
equation x + y = 5 will always have the solution y = 5 − x, so the statement
∃y(x + y = 5) will be true. Thus, the original statement ∀x∃y(x + y = 5) is
true.
On the other hand, the statement ∃y∀x(x + y = 5) means something entirely
different. This statement means that there is at least one value of y for which the
statement ∀x(x + y = 5) is true. Can we find such a value of y? Suppose, for
example, we try y = 4. Then we must determine whether or not the statement
∀x(x + 4 = 5) is true. This statement says that no matter what value we plug in
for x, the equation x + 4 = 5 holds, and this is clearly false. In fact, no value of
x other than x = 1 works in this equation. Thus, the statement ∀x(x + 4 = 5)
is false.
We have seen that when y = 4 the statement ∀x(x + y = 5) is false, but
maybe some other value of y will work. Remember, we are trying to determine
whether or not there is at least one value of y that works. Let’s try one more, say,
y = 9. Then we must consider the statement ∀x(x + 9 = 5), which says that no
matter what x is, the equation x + 9 = 5 holds. Once again this is clearly false,
since only x =−4 works in this equation. In fact, it should be clear by now
that no matter what value we plug in for y, the equation x + y = 5 will be true
for only one value of x, namely x = 5 − y, so the statement ∀x(x + y = 5)
will be false. Thus there are no values of y for which ∀x(x + y = 5) is true, so
the statement ∃y∀x(x + y = 5) is false.
Notice that we found that the statement ∀x∃y(x + y = 5) is true, but
∃y∀x(x + y = 5) is false. Apparently, the order of the quantifiers makes a
difference! What is responsible for this difference? The first statement says
that for every real number x, there is a real number y such that x + y = 5. For
example, when we tried x = 2 we found that y = 3 worked in the equation
x + y = 5, and with x = 7, y =−2 worked. Note that for different values of
x, we had to use different values of y to make the equation come out true. You
might think of this statement as saying that for each real number x there is a
corresponding real number y such that x + y = 5. On the other hand, when we
were analyzing the statement ∃y∀x(x + y = 5) we found ourselves searching
for a single value of y that made the equation x + y = 5 true for all values of x,
and this turned out to be impossible. For each value of x there is a correspond-
ing value of y that makes the equation true, but no single value of y works for
every x.
For another example, consider the statement ∀x∃yL(x, y), where the uni-
verse of discourse is the set of all people and L(x, y) means “x likes y.” This
statement says that for every person x, the statement ∃yL(x, y) is true. Now