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Answer :
(a.) Example: ∃x∀y(x > y). True for Q, false for N.
(b.) Example: ∀x∃y(x + y = 0). True, but ∃x∀y(x + y = 0) is false.
(a.) An example of a formula that is false when quantifying over the natural numbers (N) but true when quantifying over the rational numbers (Q) is:
∃x∀y(x > y)
When quantifying over the natural numbers, this formula asserts the existence of a natural number x such that it is greater than all natural numbers y. This statement is false because there is no maximum natural number.
However, when quantifying over the rational numbers, this formula becomes true. The rational numbers include fractions, and for any rational number x, there exists a rational number y such that x is greater than y. This is because between any two rational numbers, there exists another rational number.
(b.) An example of a formula that is true with one ∀-quantifier and one ∃-quantifier but becomes false when the quantifiers are swapped is:
∀x∃y(x + y = 0)
When quantifying over the real numbers (R), this formula is true. It asserts that for any real number x, there exists a real number y such that their sum is zero. This is true since for every real number x, we can find its additive inverse, which sums to zero.
However, when the quantifiers are swapped, the formula ∃x∀y(x + y = 0) becomes false. This is because it asserts the existence of a real number x such that for all real numbers y, their sum is zero. In reality, there is no single real number that can satisfy this condition for all possible values of y.
For more questions on rational numbers
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