1. Show that
is a rational number. Hint: Find

rational numbers a, b, c, and d such that
and

Fol lowing the hint , we see that if
then

a^{3} + 9ab^{2} = 26 and

3a^{2}b + 3b^{3} = 15:

Checking to see that a = 2, b = 1 is a solution to this
equation, we

then have that
and hence

Similarly, one checks that
so that

Consequently,
which is a rational number.

2. Find a polynomial with rational coefficients that
is a root

of. For this problem, there are a variety of answers. By the previous

problem,
If we write
then we know

that
so that (a - 2)^{2} = 3. Thus

a^{2} - 4a + 4 = 3;

yielding that a^{2} - 4a+1 = 0. Thus a is a root of x ^{2} -
4x+1. To check

this answer, we can use the quadratic formula to see that indeed, a is

one of the roots of this equation.

A second solution comes by writing
so that we obtain

and squaring both sides we obtain that a^{6} - 52a^{3}+676 =

675. Consequently, a^{6} - 52a^{3} + 1 = 0, and a is a root of x^{6} - 52x^{3} + 1.

Is there a relationship between our two answers? Of
course, there

should be since we know from abstract algebra that the minimum polynomial

for a should divide every other polynomial that has a as a root.

Doing long division, we see that

x^{6} - 52x^{3} + 1 = (x^{2} - 4x + 1)(x^{4} + 4x^{3} + 15x^{2} + 4x +
1):

Other students came up with a solution by looking at

and subtracting off 7 to get
Squaring yields that a^{4} -

14a^{2}+49 = 48, so that a^{4} - 14a^{2}+1 = 0, and a is a root of x^{4} -14x^{2}+1.

Again, we can factor this polynomial as

x^{4} - 14x^{2} + 1 = (x^{2} - 4x + 1)(x^{2} + 4x + 1):

3. Solve with radicals by hand (showing your work) the
equation

x^{3} + 6x = 12:

One can do this by simply plugging in for p and q in our
formulas,

but this is not how you do it in practice unless you have the formulas

memorized or easily available. For me, I simply remember the picture

of the cube that we had, so that I know that I want to set up u and v,

where (u - v)^{3} +3uv(u - v) = u^{3} - v^{3}. Thinking of x = u - v, we want

6 = 3uv and 12 = u^{3} - v^{3}. From here, we solve the first equation for v to

get that v = 2/u, so that the second equation becomes 12 = u^{3} - (2/u)^{3}.

Multiplying through by u^{3}, we obtain 12u^{3} = u^{6} - 8, so that we want

u to satisfy u^{6} - 12u^{3} - 8 = 0. This is a quadratic in u^{3}, so that the

quadratic formula gives us

Thus, we can set
We can now solve for v^{3} by noting

that
so that
Since

x = u - v, we have

Checking this on our calculator suggests that it is
correct.

I did the problem this way, to show how one usually solves
cubic equations

in practice. That is, rather than memorize a complicated formula,

it pays to remember the main idea (the picture of the cube), and then

solve the problem algorithmically. You can do something very similar

with the quadratic formula, where you remember only the picture of

the square, and then ll in the numbers as you go along. In particular,

you should note how I didn't exactly follow the steps that we did in

class. Rather than holding onto the terms 6/2 and p /3, I was able to

reduce immediately and use the reduced terms in my formulation.

Compare this to solving a quadratic equation where a = 1,
and b and

c are both even. The quadratic formula requires you to cancel out a 2

along the way, but if you actually solve the equation by completing the

square, or using the geometric algorithm with the square, the 2 that

needs to be cancelled out of the
term and the square root has been

cleaned up.

4. Solve with radicals by hand (showing your work) the
equation

x^{3} + 3x^{2} + 15x + 3 = 0:

We begin by noting that we wish to eliminate the 3x^{2}
term. To do so,

we let x = y - 1. Then plugging in y - 1 for x we obtain

y^{3} - 3y^{2} + 3y - 1 + 3(y^{2} - 2y + 1) + 15(y - 1) + 3 = 0;

which simplifies to

y^{3} + 12y - 10 = 0:

Since we just did the other process by hand, I will work
on this one

by the formula. Putting the above equation for y in our \standard

form," we obtain y^{3}+12y = 10. Thus p = 12 and q = 10, yielding that

p/3 = 4, and q/2 = 5. Using the formula for the standard cubic that

we obtained, we have

As x = y - 1, we have that

5. It is unknown whether eπ and e + π are transcendental
or not. Curiously,

it is known that at least one of them must be transcendental.

Use the following out line to prove this:

(a) Show that either e+π or e - π is transcendental. (You
are allowed

to use that e and π are transcendental, and that the algebraic

numbers form a eld.)

(b) Calculate (e +π )^{2} - 4eπ and factor.

(c) Using that the algebraic numbers are closed under
square roots,

show that if both e+π and eπ are algebraic then e - π would also

be algebraic.

(d) Using the first part, show that either e+π or eπ is
transcendental.

I will write this proof in one short proof, rather than do
the step-by-step

outline.

Let us begin by assuming that e+π is algebraic (otherwise
there would

be nothing to prove). If e - π were also algebraic, then e + π +(e - π ) =

2π would also be algebraic since the algebraic numbers are closed under

addition. As the algebraic numbers are also closed under multiplication,

this would imply that
was algebraic, which is a contradiction.

Consequently, e - π must be transcendental in this case.

Since (e+pi)^{2} - 4eπ = e^{2} - 2eπ + π^{2} = (e - π)^{2}, if
both e+pi and 4eπ

were algebraic, then (e - π)^{2} would also have to be algebraic. However,

the algebraic numbers are closed under square roots since if α is a root

of p(x), then
is a root of p(x^{2}) which is a polynomial if p(x) is.

Consequently, if both e+π and eπ are algebraic, then e - π would have

to be algebraic. This, however, is a contradiction to the first paragraph,

so if e+π is algebraic, then it must be the case that eπ is not algebraic

and is thus transcendental.

6. Discuss the difference between transcendental and
algebraic numbers

and how this affects a student's ability to understand each. How might

this influence your teaching about π and e?

An algebraic number is a number that is the root of a
polynomial with

integer coefficient. A real number is transcendental if it is not algebraic.

Thus, a transcendental number is not the root of any polynomial with

integer coefficients. In addition to making these concepts opposites,

this also means that a transcendental number is defined negatively, in

that it is only be defined by properties that it doesn't satisfy rather

than by properties that it does satisfy. When students encounter algebraic

numbers, they have a pretty solid foundation for working with

it. Graphing the polynomial by hand or on a calculator, they can zero

in on where the roots of the polynomial lie, and pretty quickly, they

can assign the number to a place on the number line. They also have

one of the main properties that defines this number at their fingertips,

namely that it is a root of the polynomial. For example,
despite

being a complex number , is relatively easy for students to deal with

algebraically, even if they have some concerns over what it represents.

In contrast to algebraic numbers, transcendental numbers must be de-

fined by properties that the students are less likely to be familiar with.

For example, transcendental number that students might see (without

ever hearing that it is transcendental) is
This
number is much

harder to zero in on. Certainly, the student's calculator can calculate

an approximation for it, but for the most part, the decimal approximation

of
is how the students will define the number internally,

as opposed to the number
which the students will internally define

as a number that squares to 2.

Thus, when introducing students to transcendental numbers,
the teacher

has two basic alternatives. The first is to introduce the number and

try and have the students think about it by its decimal approximation,

and the second is to give the students some other defining property. As

I believe that students are far too likely to treat decimal expressions as

the definition of a number (leading to difficulties with ,),
it seems

to me that the teacher should choose the second alternative whenever

possible.

For π this isn't too bad. Most students will be introduced
to π early on

as the special number that helps us find the area of a circle . They also

will hear of it as being defined as the ratio of the circumference to the

diameter of a circle. Consequently, when teaching π, you can emphasize

the property, and then come back to the idea of approximations.

The number e on the other hand is a problem this way.
While many

classical texts introduce e via instantaneously compounding interest ,

the very next statement is that e is the number that is approximately

2:71828182845945…This, of course, leads the students to treat e

as its decimal approximation. The main problem with this is that e

becomes mostly unmotivated to the students, and thus it becomes a

number that they treat as one of those silly things that math teachers

force us to learn. So, what properties could we use to introduce e?

We want them to somehow be natural properties for students to think

about. One high school math textbook uses two such properties, that

e is the number so that the function e^{x} has slope 1 at x = 0, and that

e is the number so that the area under the graph of 1/x between 1 and e

is equal to 1. From the advanced viewpoint, we recognize that both of

these definitions require some fundamental understanding of calculus.

However, neither actually requires calculus. The first requires that

the students understand the idea of slopes of tangent lines, which is

really the idea of rate of change. The second requires that the students

understand the idea of area. The trick is that when we divide algebra

and geometry, defining a number by area makes some students feel

uneasy. The number π is typically introduced when students are used

to thinking about many areas of mathematics at once, so it doesn't run

into this problem, while the number e does.