Contradiction Inequalities Here are Axioms 3.1.2: 1 Trichotomy: for all real numbers a and b, exactly one of the following holds: a < b, a = b, or a > b. 2 For all real a, b, c: we have a < b () a + c < b + c. 3 For all real a, b, c: if c > 0, then a < b ) ac < bc. If c < 0, then a < b ) ac > bc. 4 Transitivity: a < b and b < c ) a < c. Contradiction Theorem For any real number a, a > 0 if and only if −a < 0. Proof. We need to prove that a > 0 implies −a < 0, and that −a < 0 implies a > 0. First suppose that a > 0. Then we can add −a to both sides and do some easy algebra to get 0 > −a, as desired. For the converse, if −a < 0, adding a to both sides gives 0 < a. Contradiction Theorem 1 > 0. Proof. According to the trichotomy law, since 1 6= 0, we must have either 1 > 0 or 1 < 0 but not both. We need to rule out the possibility that 1 < 0. If 1 < 0, then 1 > 0 (by multiplying by the \negative" number 1). This is a contradiction (because of the trichotomy law), so 1 < 0 is impossible. This finishes the proof. Contradiction Contradiction The last proof was a proof by contradiction: we assumed that the conclusion was false, and we derived a contradiction from it. Formally speaking, to prove a statement Q by contradiction, assume that (not Q) holds, and deduce that (R and (not R)) holds for some statement R. Theorem There is no way to define \<" on the set C of complex numbers so that Axiom 3.1.2 is satisfied. Proof. We use proof by contradiction. Assume that there is a way to define \<" on C so that Axiom 3.1.2 is satisfied. Then we know that 1 > 0 by one of our theorems. We also know that there is a complex number i satisfying i 2 = −1. If i > 0, then i 2 > 0 (multiply i > 0 by i), and if i < 0, then i 2 > 0 (multiply i < 0 by i). Thus whether i is positive or negative, we see that −1 > 0. By an earlier theorem, this means that 1 < 0. Combining the two inequalities 1 > 0 and 1 < 0 contradicts the trichotomy law. Contradiction Theorem p 2 is irrational. A rational number is a real number which may be expressed as a fraction with integer numerator and denominator. An irrational number is a real number which cannot be expressed like this. Proof. p We prove this by contradiction. Suppose thatp 2 is rational. Then there are integers a and b, with b 6= 0, so that 2 = a=b. We choose a and b so that they have no common factors: a=b is a fraction in lowest terms. Then 2 = (a=b)2, so 2b2 = a2. Therefore a2 is even, which means that a must be even: there is an integer k so that a = 2k. Plug this back in: 2b2 = (2k)2 = 4k2, so b2 = 2k2. Therefore b2 is even, which means that b must be even. But then a and b have a common factor, namely 2. This contradicts thep fact that they have no common factor, so our initial assumption, that 2 is rational, must be wrong. This completes the proof. Contradiction Contradiction vs. contrapositive Sometimes when you write a proof by contradiction, it can be simplified to become a proof by contrapositive. Simpler is better, so in this case, the contrapositive proof is better. Theorem If x2 + 3x + 2 6= 0, then x 6= −1. Proof. We use contradiction. Assume that x2 + 3x + 2 6= 0 and that x = −1. Then simple algebra shows us that x2 + 3x + 2 = 0. This is a contradiction. Better proof. We prove the contrapositive. Assume that x = −1. Then simple algebra shows us that x2 + 3x + 2 = 0. Contradiction That is, suppose you are trying to prove P ) Q using contradiction, so you assume (P and not Q). If you then deduce (not P), see if you can use a proof by contrapositive, instead: assume (not Q) and try to prove (not P). This may be simpler than assume (P and not Q) and try to prove (not P)..