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Sec 1.4 Nested Quantifiers
- A formula contains nested quantifiers when it contains at least one quantifier within the scope of another quantifier.
(ex) The sum of any two positive real numbers is positive.
∀x∀y ((x>0) ∧ (y>0)) → (x+y>0), D = reals
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(ex) For every integer m, there exists an integer n such that m<n (i.e., there is no greatest integer). ∀m∃n (m < n), D = integers
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(ex) There exists a positive integer m such that for every positive integer n, m is less than or equal to n (i.e., there is a least positive integer).
∃m∀n (m ≤ n), D = positive integers
While this is true of the positive integers (m=1), it is not true of the integers.
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(ex) There exists two integers m and n, such that m = -n. ∃m∃n (m = -n), D = integers While this is true of the integers (m=1 and n=-1), it is not true of the positive integers.
Problem-solving tips
- To prove ∀x∀y P(x,y) is true, must show P(x,y) is true for every combination of x and y in the domain. Use x and y to stand for arbitrary elements of the domain in the argument.
Problem-solving example 1
prove: ∀x∀y ((x>0) ∧ (y>0)) → (x+y>0), D = reals proof: if x≤0 or y≤0 then vacuously true. Let x be an arbitrary real such that x>0 and let y be an arbitrary real such that y>0. If we add y to each side of x>0, we get x+y>0+y. But since 0+y is equal to y which is greater than 0, we have x+y > 0+y > 0, so x+y > 0.
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Problem-solving tips
- To prove ∀x∃y P(x,y) is true, must show that for each x in the domain there is a y in the domain such that P(x,y) is true. Use x as an arbitrary element and then find a y for that x that makes P(x,y) true.
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Problem-solving example 2
prove: ∀m∃n (m < n), D = integers proof: let m be an arbitrary integer, then there exists an least one integer n, namely n = m+1, such that m<n.
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Problem-solving tips
- To prove ∃x∀y P(x,y) is true, must show that for some x in the domain, P(x,y) is true for every y in the domain. Having chosen an x, let y stand for an arbitrary element and show that P(x,y) is true.
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Problem-solving example 3
prove: ∃m∀n (m ≤ n), D = positive integers proof: let m=1. Then for every positive integer n, n has the property that m ≤ n.
Problem-solving tips
- To prove ∃x∃y P(x,y) is true, must find some x and some y in the domain which make P(x,y) true.
Problem-solving example 4
prove: ∃x∃y (x>1) ∧ (y>1) ∧ (xy=6), D = positive integers proof: let x = 2 and y = 3.
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Types of proofs for conditionals
1. direct proof – assume that p(x 1 , x 2 , ..., xn ) is true and then show that it follows that q(x 1 , x 2 , ..., xn ) is true.
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prove: for any integers m and n, if m is odd and n is even, then m+n is odd. proof: Since m is odd, there is an integer k 1 such that m = 2k 1 +1. Also, since n is even, there is an integer k 2 such that n = 2k 2. Therefore, m+n = (2k 1 +1) + 2k 2 = 2(k 1 + k 2 )
- Thus there is an integer k (namely, k = k 1 + k 2 ) such that m+n = 2k + 1. Therefore, m+n is odd.
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2. proof by contradiction (indirect proof) – prove p → q by assuming p and ¬q and deriving a contradiction (a proposition of the form r ∧ ¬r).
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prove: For all real numbers, x and y, if x+y ≥ 2 then either x ≥ 1 or y ≥ 1. proof: Let x and y be arbitrary real numbers and suppose ¬(x ≥ 1 ∨ y ≥ 1). But, ¬(x ≥ 1 ∨ y ≥ 1) ≡ ¬(x ≥ 1) ∧ ¬(y ≥ 1) ≡ (x < 1) ∧ (y < 1). Therefore, x+y < 1+ = 2. But this is a contradiction: (x+y < 2) ∧ (x+y ≥ 2). Therefore, (x ≥ 1 ∨ y ≥ 1) is true.
3. proof by contrapositive –
we prove p → q by assuming ¬q and deriving ¬p. In other words, we prove ¬q → ¬p by direct proof.
prove: for all integers m, if m 2 is odd, then m is odd. proof: The contrapositive is: if m is not odd then m^2 is not odd, or equivalently, if m is even then m^2 is even. So suppose m is even, then m=2k for some integer k. Now, m^2 = (2k)^2 = 2 2 k^2 = 2(2k^2 ). Therefore, m^2 is even. Therefore, we have proven the original conditional.
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4. proof by cases – used when the hypothesis naturally divides itself into various cases. Suppose we want to prove (p 1 ∨ p 2 ∨ ... ∨ pn ) → q, we can instead prove (p 1 → q) ∧ (p 2 → q) ∧ ... ∧ (p (^) n → q).
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prove: for every real number x, x ≤ |x|. proof: For every x, (x ≥ 0) ∨ (x < 0). Case 1: if x ≥ 0 then |x| = x by definition. Thus, |x| ≥ x. Case 2: if x < 0 then |x| = -x by definition. Since |x| = -x > 0 and 0 > x, |x| ≥ x. In either case, |x| ≥ x.
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- to prove theorems of the form:
p if and only if q we use the equivalence p ↔ q ≡ (p → q) ∧ (q → p).
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prove: for all integers n, n is odd iff n-1 is even. proof: (1) if n is odd then n = 2k+1 for some integer k. Now n-1 = (2k+1) – 1 = 2k. Therefore, n-1 is even. (2) if n-1 is even, then n-1 = 2k for some integer k. Now, n = 2k+1. Therefore, n is odd.
Deductive reasoning
- deductive reasoning – the process of drawing a conclusion from a set of propositions (the hypotheses).
- A (deductive) argument has the form:
if p 1 and p 2 and ... and pn , then q.
- Deductive arguments are also written as: p (^1) p (^2) ... p (^) n
∴ q
- and as: p 1 , p 2 , ... , pn / ∴ q.
