One or more counterexamples defeat “every”
The first exercise asks for the negation of “P(n) holds for every integer n.” The predicate P(n) becomes true or false when n is supplied. Holding for every integer means there is no integer where it fails. Its negation is therefore “There exists an integer n for which P(n) is false.” The domain remains the integers before and after negation.
Let P(n) mean “n is even.” The integer 1 is a counterexample, so “Every integer is even” is false. Yet 0 is even, so “P(n) is false for every integer” goes too far. The integer 3 is another counterexample, so “exactly one” is also too strong. Existence means at least one, without restricting the number to one. “P(n) is true for some integer” is not the negation either: if P(n) is n² ≥ 0, that statement and the original universal claim are both true.
Negating “and” does not require both to fail
In the second exercise, A and B each stand for a statement. “A and B” is true only when both are true. Its negation says that at least one is false: “not A or not B.” Both may be false, but the cases with just one false must also be included.
For an integer n, let A mean “n > 0” and B mean “n is even,” and take n = 3. A is true and B is false, so “A and B” is false. Its negation is true, while “not A and not B” is false. List the assignments as (true, true), (true, false), (false, true), (false, false). The conjunction is true only in the first row; its negation is true in the other three.
Mathematical “or” includes both true
The third exercise asks which cases “A or B” includes. The or used here is true when at least one statement is true. With the preceding predicates at n = 2, the integer is both positive and even. Both statements are true, and this case is included in “A or B.” To require exactly one, state that extra restriction explicitly.
“A or B” is false only when both statements are false. Its negation is therefore “not A and not B.” Compare this with the negation of and: negate each statement and exchange and with or. These two equivalences are De Morgan’s laws. Check that the results agree in all four truth assignments instead of relying only on a word-swapping mnemonic.
Preserve both the scope and the domain
Negating the whole statement “A and B” also differs from writing “not A and B.” Negating A alone misses the case where A is true and B is false. Put parentheses around the entire statement being negated before checking its connective. Similarly, the negation of “n > 0” is “n ≤ 0,” including zero, rather than just “n < 0.”
Write down the domain as well. Among integers, “not even” means “odd,” but changing the domain to all real numbers does not preserve that replacement. Also, knowing P(n) is true for one n differs from knowing it is true for every integer. Do not silently exchange “some” and “every” while forming a negation.
Rework the three exercises with a domain note and four rows
For paper practice, write “domain = integers” and “every → at least one counterexample” beside the first exercise. Use the even-integer predicate to check why “all fail” and “exactly one fails” are too strong. For the next two exercises, list all four truth assignments for A and B, enter the results for and and or, then reverse each truth value. In the order (true, true), (true, false), (false, true), (false, false), the negation of and is false, true, true, true; the negation of or is false, false, false, true.
To review substituting values for a variable, revisit the prerequisite “Expressions: give the unknown a role.” To move from defining conditions to counting outcomes, continue to the related “Arrangements and selections: count each result once.” The practice preset selects standard, high-school-level questions on logic and applications, including other matching questions beyond these three.