Mathematics_1_oneside.pdf

# If n is an odd number then n 2 is odd proposition 33

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If n is an odd number, then n 2 is odd. Proposition 3.3 P ROOF . If n is odd, then it is not divisible by 2 and thus n can be ex- pressed as n = 2 k + 1 for some k N . Hence n 2 = (2 k + 1) 2 = 4 k 2 + 4 k + 1 which is not divisible by 2, either. Thus n 2 is an odd number as claimed. Contrapositive Method The contrapositive of the statement P Q is ¬ Q ⇒ ¬ P . We have already seen in Problem 2.2 that ( P Q ) ( ¬ Q ⇒ ¬ P ). Thus in order to prove statement P Q we also may prove its contrapositive. If n 2 is an even number, then n is even. Proposition 3.4 P ROOF . This statement is equivalent to the statement: “If n is not even (i.e., odd), then n 2 is not even (i.e., odd).” However, this statements holds by Proposition 3.3 and thus our proposi- tion follows. Obviously we also could have used a direct proof to derive Proposi- tion 3.4 . However, our approach has an additional advantage: Since we already have shown that Proposition 3.3 holds, we can use it for our proof and avoid unnecessary computations. Indirect Proof This technique is a bit similar to the contrapositive method. Yet we as- sume that both P and ¬ Q are true and show that a contradiction results. Thus it is called proof by contradiction (or reductio ad absurdum ). It is based on the equivalence ( P Q ) ⇔ ¬ ( P ∧¬ Q ). The advantage of this method is that we get the statement ¬ Q for free even when Q is difficult to show. The square root of 2 is irrational, i.e., it cannot be written in form m / n Proposition 3.5 where m and n are integers.

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3.4 P ROOFS 13 P ROOF . Suppose the contrary that p 2 = m / n where m and n are inte- gers. Without loss of generality we can assume that this quotient is in its simplest form. (Otherwise cancel common divisors of m and n .) Then we find m n = p 2 m 2 n 2 = 2 m 2 = 2 n 2 Consequently m 2 is even and thus m is even by Proposition 3.4 . So m = 2 k for some integer k . We then find (2 k ) 2 = 2 n 2 2 k 2 = n 2 which implies that n is even and there exists an integer j such that n = 2 j . However, we have assumed that m / n was in its simplest form; but we find p 2 = m n = 2 k 2 j = k j a contradiction. Thus we conclude that p 2 cannot be written as a quo- tient of integers. The phrase “without loss of generality” (often abbreviated as “w.l.o.g.” is used in cases when a general situation can be easily reduced to some special case which simplifies our arguments. In this example we just have to cancel out common divisors. Proof by Induction Induction is a very powerful technique. It is applied when we have an infinite number of statements A ( n ) indexed by the natural numbers. It is based on the following theorem. Principle of mathematical induction. Let A ( n ) be an infinite collec- Theorem 3.6 tion of statements with n N . Suppose that (i) A (1) is true, and (ii) A ( k ) A ( k + 1) for all k N .
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