Download Mathematics 30 - Appendix H: Lesson 1

Proof Teacher Notes
1.
2.
3.
4.
5.
6.
7.
8.
The Need For Proof
Inductive Reasoning
Deductive Reasoning
Conditional Statements and Proofs By Counterexample
Integer Property Proofs
Deductive Geometric Proofs and Indirect Proofs
Coordinate Geometry Proofs
Proofs By Mathematical Induction
Lesson 1
The Need For Proof
Problem 1: A motorist drove the 300 km from Saskatoon to Meadow Lake at a speed of 100 km/h.
Poor visibility caused the motorist to make the return trip at 80 km/h. What was the motorist's
average speed for the trip?
(a) What is your intuitive answer to the question? 90 km/h (a frequent guess)
(b) Let's check it out:
How long does it take to drive there? time = distance/rate = 300km/ 100km/h = 3 hrs.
How long does it take to drive back? time = distance/rate = 300km/ 80km/h = 3.75hrs.
What is the total time spent driving? = 3hrs. + 3.75hrs. = 6.75hrs.
What is the total distance travelled? =300km + 300km = 600km
What then is the average speed? Avgspeed = distance/time = 600km/6.75hrs. = 88.8 km/h
If the total trip takes 6.75 hours, and if the average speed had been 90 km/h, you would have
covered a distance of 607.5 km. It is clear that the intuitive answer of 90 km/h is too large .
Problem 2: Two containers, one holding a litre (1000 mL) of cola, the other holding a litre of
coffee, are standing beside one another. A cup (250 mL) of cola is transferred to the coffee
container and thoroughly mixed in with the coffee. A cup of the coffee-cola mix is then transferred
back to the cola container. Is there more coffee in the cola container or is there more cola in the
coffee container?
(a) What is your intuitive answer?
(b) Let's check it out.
Action Taken
Cola Container
Amount Cola Amount Coffee
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Coffee Container
Amount Cola Amount Coffee
original situation
1000
1 cup from cola container to
750
coffee container
1 cup from coffee container
800
to cola container
0
0
1000
0
250
1000
200
200
800
Cola: Coffee= 1:4 (after transfer of 1 cup from cola&endash;coffee)
Therefore 1/5 of the 250ml returned to the cola container will be cola.
As you may have realized from the above examples, intuition needs to be tested by investigating
situations in a precise manner.
Consider the problem below. Your intuition will tell you something is amiss. See if you can
determine what that is.
Problem 3: On a sheet of graph paper draw the figure shown below.
Use large squared
paper to make pieces
easy to handle.
(a) What is the area of the square above? 64u²
(b) From your sheet of graph paper, cut out the four pieces shown in the figure and rearrange them
to form a rectangle. (You do not have to flip over any of the four pieces.)
(c) What is the area of the rectangle formed in part (b)? 65u²
(d) What is the resulting contradiction?
It appears that rearranging the pieces causes the area to increase by
1u².
Explain why this contradiction arises?
The problem lies in that rectangle ABDE formed by the four pieces has "hidden" within it a
parallogram of area 1u². The figure below has been exagerated to show the parallelogram but the
calculations verify the claim above. The eye may not be able to see that points B, F, and E are not
collinear but
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Ө+ CFK + α+ ≠180o (see below).
tan Ө= 8/3
Therefore: Ө= tan-1(8/3) = 69.4440o
LCFK = 90o
tan a = 2/5, therefore: α= tan-12/5 = 21.8014o
Therefore: Ө+ α= 69.4440o + 90o + 21.8014o =
181.2454o
by Pythagoras, BF = sq. root (8² + 3²) = sq. root (73). Similarly, FE = sq. root (5² + 2²) = sq. root
(29). Similarly calculations show EH + sq. root (73) and BH = sq. root (29). Therefore: BFEH is a
llgm (opp. sides are not equal). We know that the area of a triangle is found by 1/2 ab Sin C, so the
area of a llgm (2 Triangle's) is abSin C. (see figure below)
Therefore: llgm BFEH = (sq. root 73)(sq. root 29) sin 178.7546o
= 1.0005 .(The reason this value is not exactly one os because of the rounding in calculating Ө
and α)
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Lesson 2
Inductive Reasoning
Example 2:
(a) Complete each of the following calculations:
1x8+1=9
12 x 8 + 2 = 98
123 x 8 + 3 = 987
1234 x 8 + 4 = 9876
(b) Based on your calculations above, predict the answers to the following calculations.
12345 x 8 + 5 = 98765
123456 x 8 + 6 = 987654
123456789 x 8 + 9 = 987654321
(c) Use a calculator to check your predictions in part (b).
(d) Does this pattern continue?
Does 12345678910 * 8 + 9 = 9876543210? NO.
Example 3:
Consider the pattern suggested by the following figures.
Number of points connected
Number of resulting regions
2
2
3
4
4
8
5
16
6
(a) Count the number of regions resulting when 3, 4, and 5 points are connected and complete the
appropriate portion of the table.
(b) Based on the data collected, predict how many regions can be formed when 6 points are
connected.
(c) Place 6 points on the circumference of the circle at the top of the next page and connect them in
every possible way. Count the number of resulting regions.
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only 31 regions (max)
will result
(d) What does this example illustrate about inductive reasoning?
Inductive reasoning can lead to wrong conclusions as well as correct ones.
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Lesson 3
Deductive Reasoning
Example 1:
Try the following number trick. Complete the chart using three different starting numbers. The first
has been done for you.
Directions
Choose any number
Multiply by 4
Add 10
Divide by 2
Subtract 5
Divide by 2
Add 3
Subtract the original number
1st #
11
44
54
27
22
11
14
3
2nd #
200
800
810
405
400
200
203
3
3rd #
-7
-28
-18
-9
-14
-7
-4
3
Inductive reasoning would suggest to us that the result would always be 3. We can use deductive
reasoning to prove what inductive reasoning suggests. If we let our starting number be x, then the
step by step results are shown below. Complete the chart.
If we accept the polynomial manipulations to the right of each statement as being correct, then the
conclusion that we reach, namely that 3 is always the result, is also true. Once again, reasoning in
this manner is termed deductive reasoning.
Directions
1st #
Choose any number
x
Multiply by 4
4x
Add 10
4x + 10
Divide by 2
2x + 5
Subtract 5
2x
Divide by 2
x
Add 3
x+3
Subtract the original number
3
Example 2:
(a) Choose any two positive integers.
(b) Find the square of the sum of the integers chosen in (a).
(c) Find the sum of the squares of the integers chosen in (a).
(d) How does your answer to (b) compare in size to your answer in (d)?
(e) Use deductive reasoning to prove that your conclusion in (d) will hold for any two positive
integers.
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(a) 8,9
(b) (8 + 9)² = 17² = 289
(c) 8² + 9² = 64 + 81 + 145
(d) square of sum > sum of squares
(e) Let a and b be the two positve integers
(a + b)² = a² + 2ab + b²
= a² + b² + 2ab
since a and b are positive so is 2ab
therefore: a² + b² + 2ab will be > a² + b²
therefore: (a + b)² > a² + b²
Example 3:
For any two positive numbers x and y, prove that if x > y, then 1/x < 1/y.
If 1/x - 1/y < 0. then 1/x will be < 1/y
1/x - 1/y = (y - x)/xy. since x and y are positive, xy is positive. Since x > y then y - x is negative.
Therefore: (y - x)/xy = negative# / positive# = a negative#
Since 1/x - 1/y is negative, 1/x < 1/y .
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Lesson 4
Conditional Statements and Proofs By Counterexample
Example 1:
Identify the hypothesis and conclusion in the statement "If the Blades win five in a row, then I'll
buy a season's ticket."
H: The Blades win five in a row.
C: I'll buy a season's ticket.
If-then statements are sometimes called conditional statements or just conditionals. Conditionals
may not always appear in the if p then q form, but they can always be rewritten in that form.
Example 2:
Rewrite each of the following statements in the if p then q form.
(a) All parallelograms have opposite sides that are congruent.
If a figure is a parallelogram, then its opposite sides will be congruent.
(b) The square of an even integer is even.
If an interger is even, then its square will be even.
(c) Eating candy will cause your teeth to decay.
If you eat candy, then your teeth will decay.
A conditional statement can be proven to be false if an example can be found for which the
hypothesis is true but the conclusion is false. Such an example is called a counter example. It only
takes one counter example to prove that a conditional statement is false.
Example 3:
Provide a counter example to prove that each statement below is false.
(a) If x > 0, then x < X.
1/4 > 0 but 1/4 or 1/2 is notless than 1/4. (The statement is false for any x value such that 0 < x <
1.)
(b) If the diagonals of a quadrilateral are perpendicular, then the quadrilateral is a square.
In a rhombus the diagonals are but a rhombus is not a square.
When you interchange the conclusion and the hypothesis of an if-then statement, you form the
converse of the statement.
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Original statement: if p, then q
Converse statement: if q, then p
Example 4:
Write the converse of each of the following if-then statements. Determine if the converse is true or
false.
(a) If a triangle is equilateral, then it is acute.
If a triangle is acute, then it is equilateral. (False)
(b) If a quadrilateral has four right angles, then it is a rectangle.
If a quadrilateral is a rectangle, then it has four right angles (true)
Example 4 shows us that the converse of a true if-then statement may be true or it may be false. By
using Venn diagrams we can determine whether conclusions reached in an argument are valid. The
statement "if p then q" can be illustrated by the following diagram.
Consider the true statement "If a figure is a square, then it is a polygon". The small circle p in the
diagram below represents all figures that are squares. The larger circle, labelled q, represents all
figures that are polygons. Clearly the relationship of the two circles is correct because all square
figures are also polygons, so the little circle belongs in the big one.
Example 5:
Examine each of the following statements in light of the Venn diagram and determine the
conclusion, if any, that can be reached.
(a) If a figure is a square, then it is a polygon. The figure I am looking at is a square.
Therefore the figure is a polygon. (If you are in circle p, you are in circle q.)
(b) If a figure is a square, then it is a polygon. I am looking at a polygon.
No conclusion possible. (being inside circle q doesn't mean you are inside circle p)
(c) If a figure is a square, then it is a polygon. I am not looking at a square.
No conclusion possible. (being outside circle p doesn't mean you are in circle q. You might be
outside q also.)
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(d) If a figure is a square, then it is a polygon. I am not looking at a polygon.
Therefore I am not looking at a square. (being outside circle q means you are definitely outside
circle p).
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Lesson 5
Integer Property Proofs
Example 1: Prove that the product of any two even integers is even.
Let the two even integers be a and b.
Then a=2k and b=2j where k and j are integers
Therefore ab=(2k)(2j) = 4kj = 2(2kj)
Therefore ab is a multiple of 2 (2*k*j is an integer if k and j are integers - integers are closed under
multiplication)
Therefore ab is even.
Example 2: Prove that the square of any odd integer is odd.
Let a be any odd integer
Therefore a = 2w+1 where w is an integer
Therefore a² = (2w+1)² = 4w²+4w+1
=2(2w²+2w)+1 since w is an integer, so is 2w²+2w.
Therefore a² = 2(integer)+1 which means it's odd.
Example 3: Prove that the sum of a two digit number and the number formed by reversing its digits
will always be divisible by 11.
Let 10t + u be th original two digit number. Then 10u + t is the number with its digits reversed.
Therefore 10t + u + 10u + t = 11t + 11u = 11(t+u) which is a multiple of 11.
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Lesson 6
Deductive Geometric Proofs and Indirect Proofs
The following four examples illustrate two column deductive geometric proofs similar to the kind
of proofs in the Mathematics 20 course.
Example 1: Prove that the pairs of vertical angles formed by two intersecting lines are congruent.
Example 2: Prove that the measure of the exterior angle of a triangle is equal to the sum of the
measures of the two remote interior angles.
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Example 3: Prove that the angle at the circumference of a circle is equal to one-half the angle at the
center of the circle subtended by the same arc.
Example 4: Prove that the opposite angles of a quadrilateral inscribed in a circle (a cyclic
quadrilateral) are supplementary.
Example 5:
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Example 6:
Example 7: Prove that √2 is an irrational number.
1) Assume that √2 is rational
2) therefore: √2 = a/b where a and b are integers with no common factor.
3) therefore: a = 2b
4) therefore: a² = 2b²
5) therefore: a² is even [4) shows a² is a multiple of 2]
6) therfore: a is even (if a wereoddd, a² would be odd)
7) therefore: a = 2w (definition of even)
8) therefore: a² = 4w² [squaring 7)]
9) therefore: 4w² = 2b² [substitute 8) into 4)]
10) therefore: 2w² = b² [divide 9) by 2]
11) therfore: b² is even (def'n even: it's a multiple of 2)
12) therefore: b is even ( if b were odd its square would be odd)
13) therfore; both a and b are even [steps 6) and 12)]
14) therefore: a and b have a common factor of 2. this contradicts the fact that a and b are integers
with no common foctor [step 2)]
15) therefore: our assumption is false so √2 is irrational.
Example 8: To prevent Al, Betty, and Cory from dominating the High Rollers Social Club, the
membership of the executive committee of that club was restricted by the conditions below. Prove
that Betty is not a member of the executive.
(a) If Al is a member, then Betty is not a member.
(b) If Al is not a member, then Cory is a member.
(c) If Betty is a member, then Cory is not a member.
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Assume that Betty is a member
Therefore Al is not a member (If p, then q, and not q, implies not p). statement (a)
Therefore Cory is a member (statement (b))
Therefore Betty is not a member (statement (c) If p, then q, and not q, implies not p).
Example 9: Prove that there are infinitely many prime numbers.
1) Suppose the number of prime number is finite.
2) Then there must be some greatest prime number, say p.
3) Let q be the integer which is 1 greater than the product of all the prime numbers.
Therefore q = (2)(3)(5)(7)(11)(13)...(p)+1
4) Therefore q has no prime factors. (If we divide q by any prime we will always have a remainder
of 1)
5) Therefore q is prime.
6) Since q = (2)(3)(5)...(p)+1 it is greater than p.
7) This contradicts #2. Therefore our assumption is false. Therefore the number of primes is
infinite.
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Lesson 7
Coordinate Geometry Proofs
Example 1:
Use coordinate geometry to prove that the midpoint of the hypotenuse of a right triangle is
equidistant from the vertices of the triangle.
Let A and B have coordinates of (0,a) and (b,0)
therefore: M = [(0 + b)/2 + (a + 0)/2] = (b/2,a/2)
AM = √[(0 - b/2)² + (a - a/2)²] = √ (b²/4 + a²/4) = √ (a² + b²)/2
MB = √ [(b/2 - b)² + (a/2 + 0)² = √ (b²/4 + a²/4) = √ (a² + b²)/2
MC = √ [(b/2 - b)² + (a/2 + 0)² = √ (b²/4 + a²/4) = √ (a² + b²)/2
scince MA = MB = MC, M is equidistant from A, B, and C.
Remarks: The most important step in a coordinate geometry proof
is placing the figure onto the coordinate plane in such way as to
make the mathematical manipulations as simple as possible. Clearly
had we chosen the coordinates of the vertices of the right triangle as
shown at the left, our calculations would have been much more
involved. We would have had to use the fact that AC and BC have
slopes that are negative reciprocals of each other somewhere in our
proof because of the right angle at C. Not having any of the vertices
at the origin or on an axis complicates our calculations as well.
Example 2:
Prove that the median drawn to the base of an isosceles triangle is perpendicular to the base.
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Remark: It may have been tempting to choose the coordinates of
the isoceles triangle as shown at left, but this immediately would
assume that the median is perpendicular to the base since the axes
are perpendicular. With this orientation, one would not use the
given fact that the triangle is isoceles. If we were given that the
median drawn to one side of a triangle was perpendicular to that
side and were then asked to prove that the triangle was isoceles, the
layout would be ideal.
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Lesson 8
Proofs By Mathematical Induction
Example 3:
Prove that for all natual numbers n,
1 + 1 + 1 + 1 +...+
1
= n
1*2 2*3 3*4 4*5
n(n + 1) n + 1
Basis Step: Is it true when n = 1?
Does 1/(1*2) = 1/2 ? yes
What if n = 3 ?
Does 1/(1*2) + 1/(2*3) + 1/(3*4) =? 3/4
1/2 + 1/6 + 1/12 =? 3/4
6/12 + 2/12 + 1/12 =? 3/4
9/12 =? 3/4 (yes)
Induction hypothesis:
Assume that the statement is true for any natural number k. that is assume that 1/(1*2) +
1/(2*3) + 1/(3*4) + ....+ 1/[k(k + 1)] = k/(k + 1)
Prove now that the statement is true when n = k + 1
i.e. Prove that
1/(1*2) + 1/(2*3) + 1/(3*4) + ....+ 1/[k(k + 1)] + 1/[(k + 1)(k + 1 + 1) = (k + 1)/(k + 1 +
1)
k/(k + 1) + 1/[(k + 1)(k + 2)
[k(k + 2) + 1]/[(k + 1)(k + 2)
[(k + 1)(k + 1)]/[(k + 1)(k + 2)]
(k + 1)/(k + 2)
Since L.H.S. = R.H.S. the statement is true when = k + 1. Therefore the statement os true for
all natural numbers
Example 4:
Prove that the number of interior diagonals in a convex polygon of n sides is [n(n - 3)]/2, where n ³
3
Basis Step.
Is the statement true when n = 3? A convex polygon of three sides is a . A has 0 interior
diagonals. if n= 3 [3(3- 3)]/2 = 0. Therfore it is true for n = 3. If n = 4, the polygon is a
quad, which has two interior diagonals. [4(4 - 3)]/2 = 2.
Induction Hypothesis: Assume that a convex polygon of k sides has [k(k - 3)]/2 interior
diagonals. We must now prove that the statement is true if n = k + 1. That is we must show
that a convex polygon of k + 1 sides has [(k + 1)(k + 1 - 3)]/2 or [(k + 1)(k - 2)]/2
diagonals.
The extra vertex obtained when we change from a k gon to a k+1 gon can be connected to k
+ 1 - 3 vertices to form diagonals. (To form diagonals you can't connect the k + 1st vertex
to itself nor to the two adjacent vertices hence the k + 1 - 3). by introducing the k + 1st
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vertex you however will allow for 1 more diagonal formed by connecting the two vertices
adjacent to (one on either side of) the k + 1st vertex. Thus the total number of new diagonals
formed by adding the k + 1st vertex is k + 1 - 3 + 1 or k - 1 diagonals. Since a convex
polygon of k sides has [k(k - 3)]/2 interior diagonals (induction hypothesis) and since
creating a k + 1 gon adds k - 1 diagonals to the figure, the total number of diagonals is [k(k
- 3)]/2 + k -1 = [k(k - 3) + 2(k - 1)]/2 = (k² - 3k + 2k - 2)/2 = (k² - k -2)/2 = [(k - 2)(k +
1)]/2. This is what we had to show to prove the statement true for n is less than k + 1.
Therefore true for all natural numbers > three.
Example 5:
Prove that for every natural number n, n < 2n?
Basis Step:
Is the statement true for n = 1, 2, 3 ?
1 < 2¹, 2 < 2², 3 < 2³ -yes all are true
Induction Hypothesis:
Assume the statement is true for n = k. That is assume that k < 2k. We must now prove that
the statement is true when n = k + 1. We must show that k + 1 < 2k + 1.
By our assumption k < 2k.
Multiplying both sides by 2 yields 2k < 2*2k Þ
Therefore k + k < 2k + 1 Þ
but k > 1 or 1 < k since we know our statement is true for n = 1 our assumptions that the
statement is true for any number k implies k > 1. So if 1 < k then 1 + k < k + k or 1 + k < 2k
Therefore 1 + k < 2k + 1 (see Þ).
Therefore our statement is true when n = k + 1. Therefore it is true for all natural numbers.
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