Math 3379
Chapter 2 (3 lectures)
Homework 2
2.1
1 (only GSP constructions no more than 2 constructions per page)
2 – 5 (learn the vocabulary as you go)
1 (a, b, c)
4
7
1a
2
(use Google and FULLY explain!)
6
2.2
2.3
Discussion Points:
2.1
Elementary Constructions
Historically, geometers used both a compass and a straightedge neither with
markings for constructions! We’ll just review those constructions…actually there
is some overlap sometimes in an elementary historical construction and how we
build it in GSP.
Page 51
Figure 2.1.3 Constructing an equilateral triangle. AB is a radius!
Page 53 – note the justification steps for Euclid’s proof
Let’s open GSP and build one!
Put a segment on the sketch. Double click the left endpoint – see the
target icon open and close. Single click on the right endpoint. Go to
Transformations and rotate by 60 degrees counterclockwise. Connect
the new side to the original side. Measure an angle or two to make
sure it’s equiangular.
1
Page 53
Figure 2.1.5 Bisecting an angle…it’s interesting, but we can do it
more quickly with GSP and 3 well-placed angle points.
Note that every angle has a unique angle bisector and that unless you
are given that a ray is the bisector you have to PROVE it so before
using bisector facts in a theorem.
Let’s do this in GSP.
Build an angle with a horizontal bottom leg opening to the right –
make sure each leg has a point on it. Click on the top leg point,
vertex, and bottom leg point in THIS order. Go to Construct and click
on the bisector. Check to make sure it is a bisector.
Page 55
Figure 2.1.7 Finding the midpoint of a given segment…again unique
as in “exactly and only one” of them.
Put a 7 cm segment on a new sketch. Select the segment and go to
Construct to find the midpoint. Check to make sure it is the midpoint.
Now let’s build the perpendicular bisector!
What is the definition of a perpendicular bisector in EG?
Construction notes:
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Page 58
Replicating angles. We’ll use the Transformations menu.
Let’s put a 45 degree angle on a new sketch with a horizontal initial
side. Now select the whole angle and Transform it by translating it 5
cm to the right. Next go back to the original angle and Transform it
by rotating it 60 degrees counterclockwise.
Take a moment when you’re reviewing the chapter to look over how
you had to do it back in the technology-free days! You’ll be amazed.
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Page 59
Drawing parallel lines. We’ll use Playfair’s axiom literally. It says:
Through a point not on a given line there is exactly one line parallel to
the given line.
Grab a new sketch and put a segment on it…or a line. Put a point
NOT on the line. Select the line, then the point (in this order). Go to
Construct and construct the parallel line.
Page 60
Let’s focus on building a square. The old fashioned way is in
Example 2.1.8.
Let’s make it 5 cm to the side. Suggestions?
Construction Notes:
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2.2
Exploring Relationships Between Objects
Our focus in the section will be less on construction than on discovering ways to
use construction techniques in proving theorems.
Grab your laptops and build an arbitrary triangle ABC.
Then construct the perpendicular bisectors of each side of the triangle (select a leg,
construct the midpoint, have the leg and the midpoint selected, construct a
perpendicular line).
When you’re done put a point at the intersection of the bisectors and hide the lines.
Take some time to move the triangle’s vertices. Can you move this point (called
the circumcenter) to a side of the triangle? Outside the triangle?
Does it seem like a reasonable assertion to claim that the circumcenter always
exists for every triangle?
Measure the distance from the circumcenter to each vertex…what did you find?
( NB: p. 65 typo, Definition 2.2.1) not “circle”…”center”
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Let’s look at these questions:
Is it possible for the circumcenter to lie on a vertex of the triangle?
If so, what needs to be in place for this to happen?
Is it possible for the circumcenter to lie on a leg of the triangle?
If so, when?
If the triangle is equilateral, where is the circumcenter?
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Here is a proof that the circumcenter exists and is equidistance from the vertices:
We consider a nondegenerate triangle
. Since the triangle is nondegenerate,
and
lie on different lines and so their perpendicular bisectors are not
parallel and thus intersect.
Let
be the intersection of these perpendicular bisectors.
Since lies on the perpendicular bisector of
likewise, it is equidistant from and .
, it is equidistant from
and ;
Hence is equidistant from and ; hence also lies on the perpendicular
bisector of
(and is the circumcenter). QED
Would you have thought to toss in the “nondegenerate” remark? It really is
necessary if you’re being careful, you know. And it allows the second sentence in
that paragraph. What is a “degenerate triangle”?
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Points of Concurrency
A point of concurrency is the point where three or more lines intersect.
The point of concurrency of the three perpendicular bisectors of a triangle is called
the circumcenter. [In a triangle, a segment that is perpendicular to a side at its
midpoint is called a perpendicular bisector.] The circumcenter is equidistant
from the vertices of the triangle and is the center of a circle with an inscribed
triangle. The circumcenter is the midpoint of the hypotenuse of a right triangle.
The circumcenter is outside on an obtuse triangle and opposite the obtuse angle.
Let’s take a moment to build the inscribing circle in GSP. This is called the
circumcircle.
Construction notes:
The point of concurrency of the three angle bisectors of a triangle is called the
incenter. We’ve seen this last week. [A special segment of a triangle that divides
an angle of the triangle into two equal angles is called an angle bisector.] The
incenter is equidistant from the sides of the triangle. The incenter is the center of
an inscribed circle. The incenter is always inside the triangle.
The point of concurrency of the three medians of a triangle is called the centroid.
[In a triangle a segment from a vertex to the midpoint of the opposite side is called
a median.] The Centroid Theorem: The medians of a triangle intersect at the
centroid which is 2/3 the distance from each vertex to the opposite sides midpoint.
The centroid is always inside the triangle.
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Let’s illustrate this theorem in GSP.
Construction notes:
The point of concurrency of the three altitudes of a triangle is called the
orthocenter. [In a triangle the perpendicular segment from a vertex of a triangle
to the opposite side or the line that contains the opposite side is called an altitude.]
The orthocenter is the vertex of the right angle in a right triangle. The orthocenter
is outside on an obtuse triangle and behind the obtuse angle.
Let’s construct the orthocenter.
Construction notes:
Explaining and illustrating any one of these 4 would make an excellent
presentation – be sure to provide a mnemonic device to help a student learn which
goes with which and to provide illustrations of the facts. There are really good
questions and construction hints in the book for these to be answered in the
presentation as well. Remember that the presentation deadline is in mid-April.
The presentation is 10% of your grade! See my website for details
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Page 68
The Nine – Point Circle
A famous construction and an amazing theorem. See Figure 2.2.5.
What do
 The midpoints of the legs
 The bases of the altitudes
 And the midpoints of the segments joining the orthocenter and the nearest
vertex
have in common?
They are all points on the 9-point circle (discovered in the early 1800s by Karl
Wilhelm Fuerbach)!
Additionally, the center of this circle is collinear with the orthocenter and the
circumcenter. It bisects the “Euler Line” that joins the orthocenter and the
circumcenter!
Let’s build the circle.
Put a triangle on a new sketch. Find the midpoint of one leg. Find the orthocenter
and the circumcenter (label them). Find the midpoint of the segment joining them.
Click on the center and then the leg midpoint. Construct the circle by center and
point.
Construction notes:
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Now let’s build the Euler Line. Connect the circumcenter and the orthocenter.
Check that the circle’s center bisects the line.
Construction notes:
Questions:
Is it possible for the Euler Line to collapse to one point? If so, under what
circumstances?
Construct the circumcircle of the triangle (use the circumcenter and a vertex to
construct the circle). What is a conjecture about the radius of the circumcircle and
the radius of the 9-point circle? How might you prove your conjecture?
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Draw a segment from the orthocenter to a point on the circumcircle…what
happens at the intersection of this segment and the 9-point circle? How might you
prove your conjecture?
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2.3
Formal Geometric Proof
Page 77
Read the material on the Circumcenter. If you chose to do a presention on this
point of concurrency, be sure to include this information in your presentation.
Your job in this section is to read the proofs carefully and see if you can find what
mathematicians call “the crux”…the key point that makes the proof work.
Meanwhile, let’s explore the following information.
A Primer on Proving
In a classroom setting, a problem that starts off with “prove this theorem” is an
exercise in which students are assured that what they are working on can be proven
by the word “theorem” in the problem statement. Neither teachers nor
mathematicians call a statement “theorem” unless it’s been proven. “Prove this
theorem” problems are intended to provide exercise in practicing logic, using
definitions, and mastering facts.
At the professional level of mathematics, “proving” is more often about trying to
find out if a statement or proposition is always true, under what circumstances it is
true, or discovering that it is false. The process hones mathematical intuition and
keeps the mathematician involved on the edge of knowledge. A teacher who is
trying to simulate this situation will use the instruction: “prove the assertion or find
a counterexample”. Exercises with these instructions develop a deeper
understanding of the process of discovering facts rather than being told them.
These exercises also illustrate the discovery process in science and math.
There is, of course, formal logic involved in a formal proof and if the matter to be
proven is done in a formal way, the process of proving can be quite tedious. It is
customary to use an outline or sketch of the formal process – and this convention is
why students have a lot of trouble knowing when or even if the job is done. In this
section we’ll be working on the process of proving in a less formal way.
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Coming up with a proof involves “organized doodling”:
reviewing definitions in the assertion and reviewing related facts,
making sketches,
researching to find related facts, and
performing some calculations.
Someone who is learning to write proofs needs to know a bit of logic and to
practice a lot. One excellent way to learn how to prove is to read other people’s
proofs with the thought of comprehending the logical structure underlying the
words.
We’ll start with some facts that you may use without proving them in this course
and then we’ll look at some types of proofs and some examples of each type.
Real Numbers and Arithmetic Properties
Let a, b, c, and d be real numbers. The following properties are true.
1.
Commutative properties:
a + b = b + a and ab = ba
2.
Associative properties:
a + (b + c) = (a + b) + c and a(bc) = (ab)c
3.
Identity properties:
The additive identity is 0: 0 + a = a + 0.
The multiplicative identity is 1: 1(a) = a(1).
Each of these identities is unique.
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4.
Inverse properties:
For each real number a, there is a real number –a,
called the additive inverse, with a +( –a) = (–a) + a = 0.
For each nonzero real number a, there is a real number 1/a,
called the multiplicative inverse, with a (1/a) = (1/a)a = 1.
Each of these inverses is unique.
5.
Distributive property:
6.
Equality properties:
Addition: a = b  a + c = b + c
Multiplication:
a = b  ac = bc
Substitution a = b  a may be used in place of b as needed
Square root: a > b  1  a ½ > b ½
(note: this is defined to be the positive square root)
7.
The trichotomy property of real numbers: when comparing two real
numbers
exactly one of these statements is true: a = b, a > b, or a < b.
8.
The Sum Inequality: If a, b, and c are positive real numbers and a = b + c,
then
a > b and a > c.
a(b + c) = ab + ac.
You may use these facts freely when you are working on proofs in this course.
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Facts from logic
1.
The logical equivalence of a conditional and its contrapositive:
P  Q  ~ Q ~ P
for example:
If Steve is 18 or older, then Steve is legally an adult.
If Steve is not legally an adult, then Steve is not 18 or older.
or: If Steve is legally a child, then he is under 18 years old.
These are logically equivalent, which means they have the same truth table
outcomes.
2.
Transitivity of implication:
(P  Q  Q  R)  (P  R)
for example:
If A is a square, then A has 4 sides. If a polygon has 4 sides, then it is a
quadrilateral. Thus A is a quadrilateral.
3.
The Law of the Excluded Middle:
not both nor neither.
Either P or ~P is true.
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Style Sheet
1.
Be sure you understand what is to be proved. Rewrite the proposition or
theorem in manglish* at the very least. If you can, rewrite the proposition or
theorem in a conditional format so that you can focus on the hypotheses and
the conclusion separately. See the text of the counterexample that follows
for an illustration of how important this can be.
2.
Assemble the facts that you think you’ll need, especially the definitions or
theorems proved previously that are on the same subject. You need to
consider your audience with respect to inserting definitions and the text of a
theorem into the proof you are writing. An experienced audience needs less
information than a novice reader. A homework proof needs plenty of
explaining and few assumptions about what “everybody knows”.
3.
Choose the type of proof you want to try first. Don’t get too fond of any
style or type; if a direct proof gets very, very messy sometimes a change to
contradiction or contrapositive can help.
4.
Proofs are usually written as prose with complete sentences, correct spelling,
and standard paragraph development. High schools are now getting away
from “two column” proofs and going to paragraphs. A “guided proof” is
usually in two columns – these are handy for tests.
5.
If you need to use a pronoun in order to write a complete sentence, use “we”.
This choice assumes that you, as the author, and your reader are following
the logic of the argument together.
*a contraction of Math and English – like Spanglish or Franglais.
Invented by Mary Droz in Math 2303 many years ago.
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6.
If you need to use equations or inequalities in your sentences, fit them in as
clauses. They should be grammatically correct as math expressions and as
clauses. Often, symbols or variables are used as the subject of a sentence.
These symbols or variables are to be regarded as pronouns and should have a
sentence that introduces each of them prior to being used.
For example: “M is the set of all natural numbers greater than or equal to 2.
M can be divided into two subsets: the prime numbers and the composite
numbers.” Using the second sentence without having the first sentence in
place will inevitably cause misunderstandings for your readers.
7.
Be sure to choose an appropriate object as an illustration or subject. If you
are proving the Pythagorean Theorem, you will need a right triangle, not just
any triangle. If you are proving a statement about all triangles, then you
must choose a triangle that represents all triangles and not some special
subset of them. This is the idea behind the phrase: “Let A be an arbitrary
triangle” or “Let A be a right triangle”. Look for ideas in the hypothesis of
your conditional statement to guide your choices.
8.
At or near the beginning of a proof, select the appropriate objects with which
to work. If your hypothesis uses natural numbers and you need three of
them, choose three arbitrary natural numbers and call them something like x,
y, and z. If you need a right triangle chose an arbitrary one – which means
the only facts you have is that the object is a triangle, one angle measures
ninety degrees, and the side across from this angle is called the hypotenuse.
9.
Indicate the end of your argument with QED or  or a phrase like “This
completes the proof”.
Note that your proofs will be graded using this style sheet as a standard.
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Proof Formats
Direct Reasoning:
Start with the hypothesis, use definitions and previously established facts in a
straightforward logical string right to the conclusion.
a formal example:
[(P  Q)(Q  R) (R W)]  (PW)
another example:
If Les is taller than Pat and Pat is taller than Lee and Lee is taller than
Lucky, then Les is the tallest of these four people
another example:
If x, m, n, r. and y are real numbers and x – 1 = m and m = n + y and
y = r – 1, then x = n + r.
There’s a difference between the last two examples, though. I’ve IMPLICITLY
used some properties of equality that are separate from the logic in the last
example.
What are these properties?
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An example of a direct proof
The Pythagorean Theorem
Given a right triangle with sides of length A and B and hypotenuse of length C,
A2 + B2 = C2.
You are to “see” that each shape is a square.
The smaller square cuts each side of the
larger square into pieces a + b. We’ll
call the side of the smaller square c.
The area formula for a square is
½ side 1 x side 2.
You may add and subtract areas at will.
If you subtract the smaller area from the
larger area you get the area of the 4 right
triangles.
1
(a  b) 2  c 2  4[( )ab]
2
a 2  2ab  b 2  c 2  2ab
a 2  b2  c2  0
a 2  b2  c2
** This particular version is in the style usually used on the Indian subcontinent
1500 years ago. Along this line of thought, Joseph Rotman notes that Elisha Scott
Loomis published a book of 370 proofs of the Pythagorean Theorem in 1940
(see p. 51 in Journey into Mathematics ).
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Theorem
The least distance from a point P to a line L, not containing P, is the perpendicular
distance PQ.
Proof
Let P and L be as stated above and suppose Q is a point on L with PQ
perpendicular to L.
Let R be another point on L, distinct from Q. Connect R and P to make the line
segment PR and note that triangle PQR is a right triangle. By the Pythagorean
Theorem, we have the following equation:
QR2 + PQ2 = PR2.
This means that PQ2 < PR2 which means that PQ < PR. Since PR is arbitrary, the
theorem is proven.
Do you see where I’ve used the arithmetic properties in this proof?
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Proof by Contradiction and Contrapositive:
In order to prove some conditional statement X  P : assume not-P is true and then
make logical and correct deductions from that assumption. When you end up with
a statement that is logically derived and totally contradicts a known fact (X is
always a true statement) then you may announce that not-P cannot be true. (As it
is a precept of logic that a statement is either true or false, you know that P is true.)
On a more formal note, proofs by the contrapositive can work well because a
conditional statement and it’s contrapositive are logically equivalent. This means
that proving not-Q implies not-P is the same as proving directly that P implies Q.
Both of these proceed by negating the second clause and working from that point.
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An example of a proof by contradiction
Theorem
If x is a real number and x2 = 0, then x is zero.
Proof
Let x be any real number and x2 = 0.
Suppose x  0.
Since x is not zero*, we may use x as a divisor and divide both sides of the
equation x2 = 0 by x.
On the left hand side x2  x = x and on the right hand side 0  x = 0.
In other words, by dividing by our non-zero x, we find that x = 0.
This contradicts our supposition that x  0, and we must conclude that x is zero. 
* Division by zero is not allowed while division by a non-zero number is.
Depending on the level of your audience, you may or may not need to review this
point thoroughly. For an audience of math sophisticates, you might even omit any
mention of it and assume that OF COURSE “everybody knows that”. (see Style
sheet #2)
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Another example of a proof by contrapositive with an additional twist
Theorem:
A counting number is odd if and only if its square is odd.
Strategy:
If and only if statements are two conditional statements compressed
into one sentence. You have to unpack them before you can prove them.
Unpacked theorem:
O  OS
If a counting number is odd, then its square is odd. (O  OS) – and –
If the square of a counting number is odd, then the number itself is odd. (OS  O)
We have to prove both of these statements. Note that you’re expected to know that
an even number is 2x where x is a natural number and an odd number is 2x  1 or
2p + 1 where p is a natural number.
Proof of statement 1:
[ O  OS ]
Let n be an odd counting number.
Thus n = 2p + 1 for some whole number p.
The square of n is, then, (2p  1)2  4p2  4p  1  2(2p2  2p)  1 . The last
representation shows that the square is odd. 
This is a direct proof.
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Proof of statement 2:
OS  O
[ notO  notOS ]
[Contradict the conclusion. ]
Suppose the number, n, is even (which is to say “not odd”). Then n = 2x for some
natural number x. Therefore, n 2  4x 2  2(2x 2 ) …which says that the square of the
number is even. We have shown that the square is not odd (i.e. even). 
This is a proof by contrapositive.
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Elimination Proofs
If, in looking at an assertion, you find that the list of all outcomes is finite, you
may eliminate all the false ones in your proof. If there’s only one outcome left, it
is, then, the correct or true conclusion.
Suppose you have two segments and you are comparing their lengths A and B. As
length is a real number you know, then, that the lengths are equal or one is shorter
than the other. If you can show that A is not equal to B and A is not less than B,
you may then conclude without further discussion that A is greater than B.
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Induction Proofs:
The procedure is to
demonstrate the assertion is true for the smallest natural number that is specified,
assume that the assertion is true for some natural number, and
demonstrate that the assertion is true for the next natural number.
The demonstration part of the proof is usually algebraic and involves resisting the
temptation to work on both sides of the comparative symbol. Note that the work in
the examples is exclusively on a single side of the inequality or equality.
Induction as a method of finding out truths or illustrating patterns is an excellent
teaching tools to have in your repertoire.
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An example of a proof by induction
Theorem
The number 3 is a factor of 22n – 1 for all natural numbers n.
Proof
Let Pn = 22n – 1.
P1 = 4 – 1 = 3 and 3 is a factor of itself.
We will assume that Pk is true, which is to say that 3 is a factor of 22k – 1.
(This means that 22k – 1= 3x for some number x.)
We must now show that 3 is a factor of Pk+1 = 22(k+1) – 1.**
22(k+1) – 1 =
22k+2 – 1 =
22k22 – 1 =
4(22k) – 4 + 3 =
4(22k – 1) + 3 =
4(3x) + 3 =
3(4x + 1). 
** the strategy at this point is algebraic: show that Pk+1 is true by using algebra in
such a way that our hypothesis that Pk is true can be used. Note that the spacing of
the body of this proof is somewhat non-standard. In reality most of the new lines
would be run continuously as “text” in situation other than “teaching mode”.
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Another example of a proof by induction
Prove that n!  nn
**
Proof:
The statement is true for n = 1 because 1! = 1 and 11 = 1 so that 1!  11.
Assume that the statement is true for n = k : k!  kk
Show that it is true for n = k + 1
(k + 1)! = (k + 1)k!
Since k!  kk is true by assumption,
we have that (k + 1)! = (k + 1)k!  (k + 1)kk  (k + 1)(k + 1)k = (k + 1)k + 1 
Be sure to understand the reasoning and the choices that make every step of this
work out. Often it is a good idea to write out what the goal is – in this case we
wanted to show that (k + 1)!  (k + 1)k + 1 . The procedure is to work with the
expression (k + 1)! exclusively until the goal is reached.
** you’ll need to do some doodling to figure out what this is saying…
2!  2 2
3!  3 2
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Finding a counterexample
To demonstrate or prove that a claim is false, we need to find one object that meets
all the hypotheses and doesn’t demonstrate the conditions in the conclusion. This
process is called finding a counterexample and finding it uses “creative doodling”.
In this class we’ll need some facts from College Algebra, some facts about real
number arithmetic, and some of the information from your previous math courses,
including high school courses. You’ll need to use these facts to figure out whether
or not a conclusion is true.
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An instance of finding a counterexample
Prove or find a counterexample:
P(n) = n2 + n + 11 is prime for every natural number n.
The first thing to do is spend a little time thinking about it and working with the
conjecture creatively. This includes

Review the definition of prime numbers:
A natural number greater than or equal to two is prime if and only if it has no
factors other than one or itself. It cannot be factored into a product of numbers
bigger than one and smaller than itself.

Rewriting the conjecture in implication form:
If n is a natural number, then P(n) is a prime number.

Trying out a few natural numbers:
P(1) = 13
P(2) = 17
ok
ok
[Some conjectures might be best understood by making some sketches. The form
of this conjecture suggests sample calculations]

A bit of wondering or internal chatting:
A natural thought might be to ask if the formula is incorrect because it’s not giving
us 2 or 3 or 5, the smaller prime numbers. Look at the statement again. It does
NOT say that the formula generates ALL prime numbers, it just says that
if you use a natural number the answer is a prime number. (#1 stylesheet)
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
Keep playing with it. Do you think you could factor the formula into linear
factors? If you can, then it’s not true. Try that and keep on doing some
calculations.
Try P(10)
P(10) = 121 = 112 .
So n = 10 is a counterexample because 121 is NOT a prime number. And the
assertion is NOT a theorem
Another example:
Prove or find a counterexample.
If an integer is a multiple of both 10 and 15, then it is a multiple of 150.
First, recall that a multiple is defined to be a natural number times a natural
number…no fractions or negatives allowed.
Next recall that there’s a number called the least common multiple. There’s lots of
ways to find the lcm. You may need to review them. The least common multiple
of x and y is denoted (x, y).
(10, 15, 150) = 30 and note that 30 is NOT a multiple of 150 even though it is a
multiple of both 10 and 15. We found a counterexample because 30 meets the
hypothesis and
And is not a multiple of 150.
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Inequality Theorems for Triangles
33
Let’s jump right into a proof using these facts.
Theorem:
The measure of an exterior angle is greater than that of either
remote interior angle.
C
Illustration:
The angle labeled “Exterior Angle” is
bigger
than A or C .
Exteri or
Angl e
A
B
Proof
Let’s start by stating that m  A + m  C + m  CBA = 180° by an earlier theorem.
Now since  CBA and the Exterior Angle are a linear pair, they’re supplementary
by axiom:
m  CBA + m  Exterior = 180°.
Now subtract the second equation from the first. You will find that:
m  A + m  C − m  Exterior = 0
so m  Exterior = m  A + m  C. And since angles measures are just numbers
between 0 and 180, we know that the sum is greater than either summand (#2
above, Arithmetic Properties). Thus the Exterior Angle is larger than either remote
interior angle.
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Moving along:
The transitive nature of inequalities is used in the next proof and the Tricotomy
Law gets used in the following proof.
Our strategy with the Tricotomy Law is to show that a given side is not shorter nor
equal to another side. Thus it must be longer. This is not a comfortable strategy
for beginners…it’s called an elimination proof…you eliminate two possibilities so
only the third remains.
Unequal Sides and Angles Theorem and proof:
If the measures of two sides of a triangle are not equal, then the measures of
the angles opposite those sides are
A
unequal in the same order.
Given AB > AC.
D
B
Show mACB > m B
C
We have that side AB > side AC. Locate point D on side AB so that you can
construct a segment on the interior of the triangle, AD, so that AD = AC, thus
creating an isosceles ADC. [Note that point D is in the interior of  ACB.]
Now the base angles of ΔADC are congruent.
By Axiom 13, we have that
mACB = mACD + mDCB.
From arithmetic then mACB is bigger than either summand, in particular
mACB > mACD = mADC > mB
as desired, because  ADC an exterior angle to ΔBDC and is thus larger than the
remote interior angle B.
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Unequal Sides Exercise:
List the angles of this triangle in order, largest to smallest:
C
AC = 6.12 cm
CB = 2.75 cm
AB = 7.28 cm
A
B
Unequal Angles and Sides Theorem and proof:
If the measures of two angles of a triangle are unequal, then the lengths of the
sides opposite these angles are unequal in the same order.
Given:
m B > mC
Show: AC > AB
A
C
B
Let’s suppose AB > AC.
Then, by the preceeding theorem, mC > mB.
This is impossible because of what we are given.
Ok, then, let’s suppose AB = AC,
then mB = mC because they’d be the base angles of an isosceles triangle.
Which isn’t possible because we’re given that angle B is larger.
Ok, now let’s review: AB is NOT larger than AC and it’s NOT equal to AC. And
these measurements are numbers so the only remaining possibility is:
AB < AC by AP4, the Tricotomy Law
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Two reasonable books for your math library are:
The Nuts and Bolts of Proofs
Antonella Cupillari
ISBN 012 199 4511
Journey into Mathematics: An Introduction to Proofs
Joseph Rotman
ISBN 013 842 3601
Neither is new. You can find them used on Amazon or ExLibris or some other
book selling site.
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