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Notes: Chapter 2

Section 2.1: Proof TechniquesDirect

Indirect

Contradiction - Assume the statement is false. Show that thisassumption leads to a contradiction. Remember that ((PQ) =P Q.

To prove an implication, P Q, instead prove its contraposi-tive,QP.

P Q Prove both P Q and P Q.The following are equivalent, Tfae

The statement that statements A1, A2, . . . , An are equivalent meansthat for each pair i, j [n], i < j, Ai Aj. So for n statements,there are

n

2

statements, which each give two implications. Butdue 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 A1A2, A2A3, , An1An, AnA1.

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 geom-etry. There is nothing wrong with it, logically, but instead we use rulesof 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 Bhave the same angles.

If we join the midpoint of two sides of a triangle, it is parallelto 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 forma triangle inscribed in the original one that is similar to the originalone.

Proof. Let T be an arbitrary triangle. We draw T in the plane andlabel the points at the vertices ofT,A,B , andC. We label the anglesthe same as the vertices. Suppose the midpoints of the segments AB,BC, and CA are M1, M2, and M3, respectively. The triangle on theinterior has corner points M1, M2, and M3. We label their respectiveangles the same.

We will show that A= M2.We know that the line segmentsM1M2,M2M3, andM3M1are paral-

lel to segmentsCA,AB, andBC, respectively, because if the midpointsof two sides are joined by a line segment, it is parallel to the third side.

So that AM1, M1M2, M2M3, and M3A forms a parallelogram. Sinceopposite angles of a parallelogram are the same, we know that angle Ais the same as angle M2.

Similar arguments could be used to show that B= M3and C=M1.

Therefore, triangle M1M2M3 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 com-

plete. Any pictures should only be used to supplement theargument so that the reader can understand it better.

When explaining a proof at the board, some words used in theargument 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

limx0

sin x

x = 1

limx0

cos x 1x

= 0

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. Letf(x) = sin x. Then, f(x) = cos x.

Proof.

f(x) = limh0

sin(x+h) sin xh

= limh0

sin x cos h+ cos x sin h sin xh

= limh0

sin x(cos h 1)h

+cos x(sin h)

h

= limh0

sin x

cos h 1

h

+ cos x

sin h

h

= limh0

sin xlimh0

cos h 1

h

+ lim

h0cos xlim

h0

sin h

h

= sin x 0 + cos x 1 = cos x

The quadratic formula

Theorem 3. IfAx2+Bx+C= 0,A= 0thenxB+

B24AC2A

,BB24AC2A

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Proof. We are given

Ax2 +Bx +C = 0

Ax2 +Bx = CA(x2 +

B

Ax) = C

A(x2 +B

Ax+

B2

4A2) = C+ A B

2

4A2

A(x+ B

2A)2 = C+ B

2

4A

(x+ B

2A)2 =

CA

+ B2

4A2

x+ B2A

= CA

+ B2

4A2

x+ B

2A =

4AC

4A2 +

B2

4A2

x+ B

2A =

B2 4AC

2A

x = B2A

B2 4AC2A

x = B B2 4AC

2A

Chapter 2: Section 2.1

We provide a direct proof for each of the statements in The-

orem 4 about integers. Use the definition of the odds as integersn that can be written in the form n = 2k+ 1 for some integer k andevens as those integers n that can be written in the form 2k for someintegerk 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 thereexists 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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m+n= 2(k1+k2) + 2 = 2(k1+k2+ 1) which is even becauseit is of the form of 2 times the integer: k1+k2+ 1.

(2) complete this case on your own.(3) complete this case on your own.(4) complete this case on your own.(5) complete this case on your own.(6) complete this case on your own.

We provide two examples of a proof by contradiction.

Theorem 5. Ifa, b, andc are odd integers, thenax2 + bx + c= 0 hasno rational solution.

Proof. Supposexis a rational solution, wherex = pq

is in reduced form.

If both pand qwere even,x would not be in reduced form, so at leastone ofp and qis odd.

a

p

q

2+b

p

q

+c = 0

a

p2

q2

+b

p

q

+c = 0

q2(ap2

q

2+bp

q+c) = q

2

0

ap2 +bpq+cq2 = 0

Case 1. p even, q odd.

p2 is even, so a p2 is even.b p is even, so bpqis even.q2 is odd, and so is cq2.Thus, summing the first two terms ofap2 +bpq+ cq2, we have an

even plus an even which is even and adding in the odd term to the eventerm, 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 aneven number.

Case 2. p odd, qeven.

complete this case on your own.

Case 3. p odd, qodd.

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complete this case on your own.

Theorem 6. Among numbersy1, y2, . . . , yn, at least one is greater thanequal to the average of alln of them.

Proof. By contradiction. Suppose not.The average can be expressed as y1+y2++yn

n . Call this number a.

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,ntimes, that is n a.

Thus,n

i=1yi < n a. We label this inequality (1).But,

a=

ni=1yin

and n a=ni=1

yi.

Now, by (1), we have

ni=1

yi