1.5 Indirect Proof and 1.6 Proofs Involving Quantifiers

Let m and n be integers. Then m and n have the same parity iff m^2 + n^2 is even.

1) Suppose m and n have the same parity. We have two cases. a} If both m and n are even b} If both m and n are odd 2) Suppose m^2 + n^2 is even. To show that n has the same parity as m, we use some previous examples and exercises about even and odd integers. Again we have two cases. a} If m is even, then is even. b} If m is odd, then is odd.

Proof of (∃x)P(x)

1) in constructive proof we actually name an object a in the universe such that P(a) is true, which directly verifies that the truth set of P(x) is nonempty 2) is to show that there must be some object for which is true, without ever actually producing a particular object 3)

prove √2 is irrational

Assume that √2 is a rational number. <We assume ∼P.> Then for √2 = a / b some integers a and b, where b not equal to 0 and a and b have no common factors. <The statement ment Q is "a and b have no common factors."> From √2 = a/ b we have 2 = a^2/ b^2 which implies that 2b^2 = a^ 2. Therefore is even and so a is even. (Recall the example we proved on page 40.) It follows that there exists an integer k such that and therefore 2b^2 = a^2 = (2k)^2 = 4k^ 2 b2 = 2k2, which shows b^2 is even. Therefore b is even. Since both a and b are even, a and b do have a common factor of 2. < We have deduced the statement ∼Q.> This is a contradiction. We conclude that is √2 irrational.

For given integers x and y, give a direct proof, a proof by contraposition, and a proof by contradiction of the following statement: If x and y are odd integers, then xy is odd.

Direct Proof. Assume x is odd and y is odd. Then integers m and n exist so that x = 2m + 1 and y = 2n + 1. Proof by Contraposition. xy is even --> (x is even V y is even). Proof by Contradiction. Suppose that the statement "If x and y are odd integers, then xy is odd" is false. Then x is odd and y is odd, and xy is not odd. Since xy is not odd, xy is even. Therefore 2 divides xy. Then by Euclid's Lemma, 2 divides x or 2 divides y. Thus either x is even or y is even. But x is odd and y is odd. This is a contradiction. We conclude that if x and y are odd integers, then xy is odd.

Prove that for every natural number n, 4n^2 − 6.8n + 2.88 > 0.

Proof. The statement has the form (∀x)P(x), where the universe is N and P(x) is 4n^2 − 6.8n + 2.88 > 0. Let n be a natural number. Then n ≥ 1, so n − .8 and n − .9 are both positive. Therefore, 4(n − .8)(n − .9) = 4n2 − 6.8n + 2.88 is positive. We conclude that 4n2 − 6.8n + 2.88 > 0 for all natural numbers n.

PROOF OF (∃!x)P(x)

Proof: (i) Prove that is (∃x)P(x) true. Use any method. (ii) Prove that (∀y)(∀z)[P(y) ∧ P( z)⇒y = z] Assume that y and z are objects in the universe such that P(y) and P( z) are true. ... Therefore, y = z. From (i) and (ii) conclude that ( E!x)P(x) is true. see exampleon page 55

Proofs of biconditional sentences

P⇐⇒Q equivalent to (P⇒Q) ∧ (Q⇒P). TWO-PART PROOF OF P⇐⇒Q Proof. (i) Show P⇒Q. (ii) Show Q⇒P. Therefore, P⇐⇒Q.

proof by contraposition

Since P⇒Q and ~ Q⇒ ~ P are equivalent statements. We can Assume ~Q ... Therefore, ∼P Thus, ~ Q⇒ ~ P Therefore, P⇒Q.

Let x and y be real numbers such that x < 2y. Prove that if 7xy ≤ 3x2 + 2y2, then 3x ≤ y.

Suppose x and y are real numbers and x < 2y < Let P be 7xy ≤ 3x2 + 2y2 and Q be 3x ≤ y. Suppose 3x > y. Then 2y − x > 0 and 3x − y > 0. Therefore( 2y − x)(3x − y) = 7xy − 3x^2 − 2y^2 > 0. Hence, xy > 3x^2 + 2y^2. We have shown that if 3x > y., then 7xy > 3x^2 + 2y^2. Therefore, contraposition if 7xy ≤ 3x2 + 2y2, then 3x ≤ y.

Proof of (∃x)P(x) by Contradiction

Suppose ∼(∃x)P(x) Then (∀x) ∼P(x) ... Therefore, ∼Q ∧ Q a contradiction. Thus ∼(∃x)P(x) is false. Therefore (∃x)P(x) is true.

proof by contradiction

Suppose ∼P ... Therefore, Q. ... Therefore, ∼Q Hence, Q ∧ ∼Q a contradiction. Thus, P. method of proof can be applied to any proposition P. unlike direct and indirect apply to conditionals

For all rational numbers x and y, x + y / 2 is a rational number.

The statement has the form (∀x) (∀y)P(x, y), where the universe is Q and P(x, y) is "x + y / 2 is a rational number. " Let x and y be rational numbers. Then x + y / 2 = 1/ 2 ( p /q + s /t ) = 1 / 2 ( (pt + q s ) / qt )= (pt + qs ) / 2qt The sums and products of integers are integers. The product of three nonzero numbers is not zero. Therefore, x + y / 2 is a rational number.

If x is an even integer, then x^2 is an even integer.

The statement has the form (∀x)(A(x)⇒B(x), where the universe is Z. A(x) is "x is even," and B(x) is "x^2 is even." Let x ∈ Z. < We give a direct proof of A(x)⇒B(x), which we begin by assuming A(x).> Assume x is even .. However, we cannot prove that a universally quantified statement is true by showing that it's true for selected values of the variable.

Between any two rational numbers x and y, where x < y there is always another rational number z.

The statement may be symbolized (∀x Q ) (∀y Q )[x< y⇒ (Ez∈ Q) x < z < y)]. We begin with the two universal quantifers. Suppose x and y are rational numbers. Assume that x< y < Now we must prove the existence of a rational number z with the given property.> We choose z = x + y / 2 x = x + x / 2 < x + y / 2 < y + y / 2 = y. Therefore x < z < y see more examples on page 54

Direct proof of (∀x)P(x)

To prove a proposition of the form (∀x)P(x) we must show that is true for every object x in the universe.So Let x be an arbitrary object in the universe. (The universe should be named or its objects described.) ... Hence P(x) is true. Since x is arbitrary, (∀x)P(x) is true.

indirect proofs

based on tautologies that replace the statement to be proved by an equivalent statement or statements

parity

of an integer is the attribute of being either odd or even. The integer 31 has odd parity while 42 has even parity.

proof of (∀x)P(x) by contradiciton

∼(∀x)P(x) is equivalent to (∃x) ∼P(x) Proof: Suppose ∼(∀x)P(x) Then (∃x) ∼P(x) Let t be an object such that ∼P(t ) Therefore Q ∧ ∼Q. Thus (∃x) ∼P(x) is false, so ( ∀x)P(x) is true. Q ∧ ∼Q.