√
1. Another proof that 2 is irrational
√
We already know that 2 is irrational. To prove it, all we have to do is go
through the proof that irrational numbers exist again, but replace the pq with with
1. But the fundamental theorem of arithmetic makes this fact very easy√to prove.
As you may guess, we will prove it by contradiction. So we assume that 2 can be
written as a fully reduced fraction pq . By the fundamental theorem of arithmetic,
both p and q can be written uniquely as products of primes. It is important to
note that no prime can appear in both of the decompositions of p and q. If one
did appear in both, then pq would re reducible, which is a contradiction. Also note
that pq cannot be an integer because no square of an integer will be equal to 2.
√
(That is, 2 is not an integer.)
2
√
2
We now consider 2 = ( 2)2 = ( pq ) = pq2 . Since q ≠ 1, we have that there must
be some prime n that divides q but does not divide p. Notice that if we square
a product of primes, we get a product of the same primes, but with their powers
doubles. For example, 12 = 22 ⋅ 3 and 122 = 24 ⋅ 32 . The important idea is that
2
we have pq2 , we still have that n divides q 2 but not p2 . This implies that q 2 does
2
not divide p. Then pq2 cannot be an integer. But 2 =
√
contradiction. So 2 is irrational.
p2
q2 .
We have reached a
√
This proof works exactly the same to show that a is irration for any positive
integer a that is not the sqaure of an integer. A lot easier than the other proof,
huh?
2. The rebirth of Western mathematics
As the centuries passes the time of the ancient Greeks, Western Europe began
to evolve differently than the Eastern world. With the rise of Christianity and
the Germanic wars, Roman Europe became cut off from the Greek and Islam
worlds. Thus, many of the ancient works of the East were lost to the Latin West.
Fortunately, many of the works of old scholars found their way to Western Europe
by the 13th century via translated Arabic manuscripts.
With the discovery of these manuscripts, along with many other stimuli, Italy,
as well as most of West Europe, entered into the Renaissance. The recovery of
Ptolemy’s maps and Archimedes’ mathematics, the Western world learned the
sciences of the ancient Greeks. From the influx and technology and science that
came with the renaissance came the genius and motivation that would bring the
world calculus.
1
2
Chapter 3 of our text focuses on three important figures described in these Greek
manuscripts : parabolas, ellipses, and hyperbolas. We will focus on parabolas and
ellipses. These shapes are actually conic sections, meaning that they come from
the intersection of a cone with a plane. What the scientists of the Renaissance discovered was that parabolas and ellipses form the basis of the motion of projectiles
and orbits of planets.
3. Area for parallelograms
Before we get into conic sections, we will explore the idea of area. Area is a
fundamental concept for this course that we will revisit repeatedly. It is an ancient
concept that antiquated cultures knew was vital to advancing their societies. Gardeners needed area to figure out the size of their plots of land. Carpenters needed
area to determine the size of their cuts. The applications are quite endless. As
you may expect, we will begin with the simplest of shapes for area - the rectangle.
Suppose we have a rectangle with b as its base and h as its height. The area
will be defined as A = b ⋅ h. The area represents the size of the space enclosed by
the sides of the rectangle. See figure 2 below for reference.
h
b
Figure 2
Now consider a paralleloram with base b and hieght h. Refer to figure 2 below
for a vizualization. Imagine we cut a piece of the parallelogram off as shown
in the figure by the dashed line. Basically, we are cutting off a “triangle” from
the parallelogram. The triangle has height h. Now imagine we flip that triangle
vertically and move it to the right of the parallelogram as shown. This will create
a “rectangle” with height h and base b. So the area of the parallelogram of height
h and base b is the same as the rectangle with similar measurements : A = b ⋅ h.
3
h
h
b
b
Figure 3
Now let’s turn our attention to the Greeks’ favorite shape, the triangle. See
figure 4 below for reference. Our triangle will have height h and base b, and we
would like to find its area. Imagine we add another triangle to our original one.
It will be identical to the first, but we will flip it horizontally and vertically. We
will then attach it to the right side of the first triangle. Refer to the dashed lines
in the right figure for reference. Notice that this addition of a second triangle
creates a parallelogram of height h and base b, which has area b ⋅ h. Since our
second triangle had exactly the same area as the first, we have that the area of the
1
triangle is A = b ⋅ h.
2
h
h
b
b
Figure 4
4. Area for sectors of a circle
Of course, now introduction to area will be complete without considering the
circle. It is a very natural question to ask. If you’d like to create a circular object
made of concrete, how much concrete will you need? Like parallelograms, the
applications for areas of circles is endless.
Imagine we have a circle of radius r. Call the center of it C. We will pick an
arbitrary angle θ. We would like to find the area of the sector enclosed in in the
region determined by r and θ. Call the endpoints of the sector on the circle’s
boundary B and D. See figure 3.14 (how appropriate!) in your text for reference.
This will feel very similar, if not identical, to our proof that the radian is indeed a
good notion for a unit of measurement. Pick a number n. To follow with the text,
we will pick n = 5. We will then split our wedge, or sector, into 5 equal pieces by
4
drawing lines from C to the perimeter of the circle. See figure 3.15 in your text
for reference.
Including B and D, we will have six distinguished points that split the sector
into our five equal pieces. As in our earlier proof, we will draw the chords between
the points, creating five equal isosceles triangles. Notice that for each triangle, two
sides have length r and a base that we will call bn . Each triangle has a height,
determined by the length from C to the center of bn . We will call this height
hn . Therefore the area of each triangle is 21 bn ⋅ hn . Notice that bn and hn vary
depending on our choice for n. We will denote An as the sum of the areas of all
the triangles. So A5 = 5 ⋅ 12 bn ⋅ hn .
It is vital to note that as n gets larger, the triangles better approximate our
sector. This means that as n gets bigger, An gets closer and closer to the actual
area of the sector BCD. Also, the height of the triangles hn gets closer and closer
to the radius of the circle, r. That is, as n gets very large, hn ≈ r. Finally, note
that the sum of all the bn gets closer and closer to the length of the arc BD. That
is, as n gets very large, n ⋅ bn ≈ arc BD. Just as before, we will let n get arbitrarily
large.
1
Area of sector BCD = lim An = lim n ⋅ bn ⋅ hn
n→∞
n→∞
2
1
= r ⋅ (arc BD).
2
arc BD
Recall that θ =
. So arc BD = r ⋅ θ. Therefore, if we let A be the desired
r
area of the sector, we have
1
A = r ⋅ (arc BD)
2
1
= r2 ⋅ θ.
2
1
Note that if θ = 2π, we have that A = r2 ⋅ 2π = π ⋅ r2 , which is our very familiar
2
formula for the area of a circle. Did you ever think you would prove the formula
for the area of a circle using triangles?