Notes: Chapter 2 Section 2.1: Proof Techniques Direct Indirect • Contradiction - Assume the statement is false. Show that this assumption leads to a contradiction. Remember that (∼ (P ⇒ Q) = P ∧ ∼ Q. • To prove an implication, P → Q, instead prove its contrapositive, ∼ Q ⇒∼ P . P ⇐⇒ Q Prove both P ⇒ Q and P ⇐ Q. The following are equivalent, Tfae The statement that statements A1 , A2 , . . . , An are equivalent means that for each ¡n¢ pair i, j ∈ [n], i < j, Ai ⇐⇒ Aj . So for n statements, there are 2 ⇐⇒ statements, which each give two implications. But due to transitivity, we do not have to prove all 2 · n(n−) implications. We can select an order, for instance A1 , A2 , . . . , An and prove A1 ⇒ A2 , A2 ⇒ A3 , · · · , An−1 ⇒ An , An ⇒ A1 . Existence proof Section 2.1 Topic: Valid reasoning and logic must be used in the proofs of theorems. The Direct Proof. (1) Use complete sentences and paragraphs. (2) Start with a fact, axiom, or definition. (3) Each statement should imply the next using as justification a known fact, axiom, or definition. (4) The justifications should be given, using complete sentences. (5) The final statement in your argument should correspond to the statement of the theorem that you are proving. (The only place the “2-column method” is used is high school geometry. There is nothing wrong with it, logically, but instead we use rules of proper grammar and write with sentences and paragraphs.) Example from Geometry. 1

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We must have a clear understanding of what can be assumed true before starting any proof. Basic facts and definitions: • A triangle A is similar to another triangle B if both A and B have the same angles. • If we join the midpoint of two sides of a triangle, it is parallel to the third side. • Opposite angles of a parallelogram are the same. Used to prove: Theorem 1. If we join the midpoints of the sides of a triangle, we form a triangle inscribed in the original one that is similar to the original one. Proof. Let T be an arbitrary triangle. We draw T in the plane and label the points at the vertices of T , A, B, and C. We label the angles the same as the vertices. Suppose the midpoints of the segments AB, BC, and CA are M1 , M2 , and M3 , respectively. The triangle on the interior has corner points M1 , M2 , and M3 . We label their respective angles the same. We will show that ∠A = ∠M2 . We know that the line segments M1 M2 , M2 M3 , and M3 M1 are parallel to segments CA, AB, and BC, respectively, because if the midpoints of two sides are joined by a line segment, it is parallel to the third side. So that AM1 , M1 M2 , M2 M3 , and M3 A forms a parallelogram. Since opposite angles of a parallelogram are the same, we know that angle A is the same as angle M2 . Similar arguments could be used to show that ∠B = ∠M3 and ∠C = ∠M1 . Therefore, triangle M1 M2 M3 is similar to triangle T . ¤ Topic: How to prove a theorem. • Have an understanding of assumptions. • Working out the details on scrap paper. Playing around with the known facts. • Put the argument into the proper order for a direct proof. • Write out the details as formally or informally as required. • In the most formal form, the written argument should be complete. Any pictures should only be used to supplement the argument so that the reader can understand it better. • When explaining a proof at the board, some words used in the argument may be spoken and not written.

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Example from Calculus. Basic Facts. • Definition of the derivative • sin(u + v) = sin u cos v + cos u sin v • sin x =1 x→0 x lim

• cos x − 1 =0 x→0 x lim

• The limit of a sum is the sum of the limits (if both exist). • The limit of a product is the product of the limits (if both exist). Used to prove: Theorem 2. Let f (x) = sin x. Then, f 0 (x) = cos x. Proof. sin(x + h) − sin x h→0 h sin x cos h + cos x sin h − sin x lim h→0 h sin x(cos h − 1) cos x(sin h) lim + h→0 hµ µh ¶ ¶ cos h − 1 sin h lim sin x + cos x h→0 h h µ ¶ µ ¶ cos h − 1 sin h lim sin x lim + lim cos x lim h→0 h→0 h→0 h→0 h h sin x · 0 + cos x · 1 = cos x

f 0 (x) = lim = = = = =

¤

The quadratic formula Theorem 3. If Ax2 +Bx+C = 0, A 6= 0 then x ∈

n

o √ √ −B+ B 2 −4AC −B− B 2 −4AC , 2A 2A

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Proof. We are given Ax2 + Bx + C Ax2 + Bx B A(x2 + x) A 2 B B A(x2 + x + ) A 4A2 B 2 ) A(x + 2A B 2 (x + ) 2A B x+ 2A B x+ 2A B x+ 2A

= 0 = −C = −C = −C + A ·

B2 4A2

B2 4A −C B2 + Ar 4A2 −C B2 ± + A 4A2 r −4AC B2 ± + 4A2 4A2 √ B 2 − 4AC ± 2A √ B 2 − 4AC B ± − 2A √ 2A 2 −B ± B − 4AC 2A

= −C + = = = =

x = x =

¤ Chapter 2: Section 2.1 We provide a direct proof for each of the statements in Theorem 4 about integers. Use the definition of the odds as integers n that can be written in the form n = 2k + 1 for some integer k and evens as those integers n that can be written in the form 2k for some integer k to prove: Theorem 4. (1) The sum of 2 odds is even. (2) The sum of 2 evens is even. (3) The sum of an odd and an even is odd. (4) The product of 2 odds is odd. (5) The product of an even and an odd is even. (6) The product of 2 evens is even. Proof.

(1) Let m, n be two arbitrary odd numbers. Then there exists two integers, k1 and k2 such that m = 2k1 + 1 and n = 2k2 + 1. Then m + n = 2k1 + 1 + 2k2 + 1. Regrouping, we get

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(2) (3) (4) (5) (6)

m + n = 2(k1 + k2 ) + 2 = 2(k1 + k2 + 1) which is even because it is of the form of 2 times the integer: k1 + k2 + 1. complete this case on your own. complete this case on your own. complete this case on your own. complete this case on your own. complete this case on your own. ¤

We provide two examples of a proof by contradiction. Theorem 5. If a, b, and c are odd integers, then ax2 + bx + c = 0 has no rational solution. Proof. Suppose x is a rational solution, where x = pq is in reduced form. If both p and q were even, x would not be in reduced form, so at least one of p and q is odd. µ ¶2 µ ¶ p p a +b +c = 0 q q µ 2¶ µ ¶ p p a 2 +b +c = 0 q q µ 2¶ µ ¶ p p 2 q (a 2 + b + c) = q 2 · 0 q q ap2 + bpq + cq 2 = 0 Case 1. p even, q odd. p2 is even, so a · p2 is even. b · p is even, so bpq is even. q 2 is odd, and so is cq 2 . Thus, summing the first two terms of ap2 + bpq + cq 2 , we have an even plus an even which is even and adding in the odd term to the even term, we get an odd number. But the right hand side of the equality is zero which is even. We have reached a contradiction, because on the left we have an odd and on the right we have an even. An odd number cannot equal an even number. Case 2. p odd, q even. complete this case on your own. Case 3. p odd, q odd.

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complete this case on your own. ¤ Theorem 6. Among numbers y1 , y2 , . . . , yn , at least one is greater than equal to the average of all n of them. Proof. By contradiction. Suppose not. n . Call this number a. The average can be expressed as y1 +y2 +···+y n By assumption, ∀i ∈ [n], yi < a. That is, y1 < a, y2 < a, . . . , yn < a. Adding up the Left hand sides of each inequality, we get y1 + y2 + · · · + yn . Adding up the right hand sides of the inequalities, we get a, n times, P that is n · a. Thus, ni=1 yi < n · a. We label this inequality (1). But, Pn n X i=1 yi a= and n · a = yi . n i=1 Now, by (1), we have n X i=1

yi