Proof Teacher Notes

  1. The Need For Proof
  2. Inductive Reasoning
  3. Deductive Reasoning
  4. Conditional Statements and Proofs By Counterexample
  5. Integer Property Proofs
  6. Deductive Geometric Proofs and Indirect Proofs
  7. Coordinate Geometry Proofs
  8. 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.

Cola Container / Coffee Container
Action Taken / Amount Cola / Amount Coffee / Amount Cola / Amount Coffee
original situation / 1000 / 0 / 0 / 1000
1 cup from cola container to coffee container / 750 / 0 / 250 / 1000
1 cup from coffee container to cola container / 800 / 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

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)


Lesson 2

Inductive Reasoning

Example 2:

(a) Complete each of the following calculations:
1 x 8 + 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 / 2 / 3 / 4 / 5 / 6
Number of resulting regions / 2 / 4 / 8 / 16

(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.

/ 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.

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 / 1st # / 2nd # / 3rd #
Choose any number / 11 / 200 / -7
Multiply by 4 / 44 / 800 / -28
Add 10 / 54 / 810 / -18
Divide by 2 / 27 / 405 / -9
Subtract 5 / 22 / 400 / -14
Divide by 2 / 11 / 200 / -7
Add 3 / 14 / 203 / -4
Subtract the original number / 3 / 3 / 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.

(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 .

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.

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.)

(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).

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.

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.