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Use a proof by contraposition to show that if x + y ≥ 2, where x and y are real numbers, then x ≥ 1 or y ≥ 1. Solution:----- 2. Prove that if m and n are integers and mn is even, then m is even or n is even. Solution:
proving 8x 2D;:q(x) !:p(x). Let’s practice writing the contrapositive (and the negation for when we write proofs by con-tradiction): For each of the following statements, write in predicate logic as an implication, then the contrapositive and then the negation. Assume that the universe is Z+.
Let us use the proof by contradiction. Suppose x 2 Q, y 2 I, and z := x + y 2 Q. But then we have x = z y, and we proved in (1) that sum of rational numbers is rational, so this makes y rational { a contradiction. How can you improve this proof by making it contrapositive instead of contradiction? 3. Assume that. is irrational.
We want to prove the following universally quantified conditional (“for all \(p\)” omitted, domain is positive integers). conditional if (\(p\) is prime and \(p>2\)) then \(p\) is odd.
Proposition. Suppose n 2 Z. If 3 - n2, then 3 - n. Proof. (Contrapositive) Let integer n be given. If 3jn then n = 3a for some. a 2 Z. Squaring, we have. n2 = (3a)2 = 3(3a2) = 3b. where b = 3a2. By the closure property, we know b is an integer, so we see that 3jn2.
Proposition. If P, then Q. Proof. Suppose » Q. ... Therefore » P. ç. ositive proof is very simple. The first line of the proof is the sentence “Suppose Q is not true.” . Or something to that effect.) The last line is the sentence . Therefore P is not true.” Between the first and last line we use logic and definitions to transform the stateme.
A proof by contrapositive is probably going to be a lot easier here. We draw the map for the conjecture, to aid correct identification of the contrapositive. Note that an arrow representing \(T \Rightarrow S\), the contrapositive of the original conjecture, has been added to the map.
