homework comments and clarifications

__2.2.9 __
Note this can be done without computing the
volumes of the
polyhedra, since the volumes of the pyramids are easy to express in a
way that cancels out nicely.

__1.3.2__
That is, you are being asked to show that for every
Pythagorean
triple (a,b,c) there are integers p and q such that

a=(p^{2}-q^{2})r, b=2pqr,
c=(p^{2}+q^{2})r.

__1.3.3 __ You're being asked to show that given any right-angled
triangle
with hypotenuse one, there is another right-angled triangle
with hypotenuse one where the other two sidelengths are rational and
arbitrarily close to the sidelengths in the original triangle.
The diagram should help.

__4.2.1 __ For these problems, it's important to use the following
notions
of
set multiplication and addition:

A+B={a+b : a in A, b in B}; A*B={ab : a in A, b in B}

Then he has defined L{r+s : r < x and s < y, r,s rational} = {t : t < x+y, t rational}

and similarly for LHint: the hard part is finding good choices of r and s so that t=r+s for a given t. Show that if r is a rational getting closer and closer to x from below, then s=t-r will eventually work.

__4.2.2 __ Careful: do not use square roots to solve this---that
would
be
circular, since this sequence of exercises is being used to
define the square root of two as a number. Hint: you can solve this
using only squaring, not square-rooting.

__4.2.3 __ This should use really similar logic to what you used to
show
that L_{xy} was a "valid definition" earlier.
The catch is that you need to find some way to construct rationals
getting closer and closer to the square root of two
(or at least argue that they exist) without using a circular argument.

__4.2.4 __ That is, show that the product of the lower set
L_{root-2}
with itself is L_{2}.
Now stop and appreciate that this is cool. You've justified a new
definition of numbers that behaves in the right way.

__4.3.1 __ The figure is meant to suggest that the triangles have
been
approximated with rectangles, in each case, of
height h/4. There's more than one way to pass to the "next step" in
this exhaustion, but a natural one is to divide
up the triangles using rectangles of height h/n. Then you need to show,
using the fact that the area is bh/2, that
these rectangular approximations get arbitrarily close to the true area.

__4.3.4 __ Clarification: you should not assume this is a regular
tetrahedron; that is not needed. The subdivision is defined by
using the midpoints of the edges.
Hint: you do not need to compute the volumes of the figures to argue
about which are bigger than the others.

__4.4.1-3 __ Do NOT use properties of the logarithm function to
answer
these
questions---you are trying to derive properties of the
logarithm function from this unusual definition, so you should only use
facts about area.

__5.4.1 __ In other words: the Greeks had a formula for getting a
new x
and
y from an old one. Recognizing this
as Brahmagupta composition means filling in the blanks making this true:

(x_{n}, y_{n}, (-1)^{n})
* ( ----, ----, ----) =
(x_{n+1}, y_{n+1}, ----)

__5.6.1 __ The problem is correct as stated, but note that if you
apply
the
formulas in the chapter, you may get an answer
with rational sides. There's a different decomposition that works.

__18.1.2 __ All you should need, besides the previous question, is
the
fact
that the total angle of a straight line is pi. You should
not need to use any other theorems of plane geometry.

__21.7.1 __
To do this, you just need to use the fact that i^{2}=-1. Hint:
use the same procedure you'd need to rationalize the denominator
of a fraction of irrationals.

__21.7.3 __
He means a homomorphism from **Z** to some ring R, not
necessarily a map **Z**->**Z**.

__23.2.1 __
The point of this problem is an alternative diagonalization:
you are supposing that the x_m were a sequence containing
all the reals, and then finding smaller and smaller open intervals such
that every x_{m} is left out of infinitely many of the
intervals.
Therefore anything in all of the intervals was NOT on your original
list---and there is something in all the intervals (you're not asked to
prove this, but you can verify it if you like), so your list was
incomplete.

__23.5.1 __
First of all, an integer function is any function
f : **Z**->**Z**. But this problem makes much more sense
as written if you just deal with functions of the natural numbers,
f : **N**->**N**.

There's a typo in the hint: it should say m f_{n}(m)
instead of n f_{n}(m).

Note about this problem: the point is to show that the set {integer
functions} is uncountable by supposing that there were a list of
them and then constructing a new function not on the list. What you end
up proving is much stronger: you can actually construct
a function that *dominates* everything on the list, in the sense of
growing much faster than any of the others.

__6.6.1 __ For all three of these problems, note that equation (3) is true for *all* n.
(He's just saying you can get a particular consequence from it in the case n=4m+1.)

This one is tricky but very satisfying. Don't look at these hints unless you are stumped for ten minutes!

*Hint*: let theta = pi/2 - alpha. This allows you to rewrite the
left-hand side of the formula as (cis alpha)^{n}, and then apply
(3).

*Another Hint*: the cases n=4m+1 and n=4m+3 determine whether
n(pi)/2 is
the same
angle as pi/2 or 3pi/2. (Do you see why?)

*Final Hint*: sin(pi+A)= - cos(A) and cos(pi+A)= - sin(A). (Do
you see
why?)

__6.6.2-3__
These problems get much, much easier if you use the following fact about complex conjugates:

If (a+bi)^{k}=c+di, then (a-bi)^{k}=c-di.

__9.4.5__ I think this problem calls for induction.
Suppose 1/a_{1} - 1/a_{2} + ... -1/a_{2n} = *what it is*.
Then replace 1/a_{2n} with 1/a_{2n}-1/x to get the next stage.
What is a_{2n}? Use this to simplify.

__10.4.2 __
Clarification: you are being asked to derive the formula
given in the problem for 2/pi, by making the suggested substitution. Using
that, it is an easy matter to turn that into a formula for pi/4, and
check that it matches with Wallis' formula (p153).

__10.6.1__ Hint : if phi is the golden ratio
[1+sqrt(5)]/2, note that -1/phi is [1-sqrt(5)]/2. (Check this.)
That should help the fraction simplify a lot.

__2.3.1__ To see examples of similar triangle constructions, look at
some of the ruler-and-straightedge examples from lecture (like dividing
a segment into n equal
parts, of which one of the methods of bisection was a special case).
Remember that the lengths L_{1}, L_{2}, and 1 are all
given, so you can measure those out whenever you want.

*Hint* : the
idea is that you want to scale up L_{1} by a factor of
L_{2}, or
vice
versa. Similar triangles have sidelengths that are scaled versions of
each other.
To do multiplication, arrange so that the
larger triangle has a side of length
L_{1}L_{2}. What might a second side of that triangle
be, and what would the corresponding
lengths of a smaller triangle be?
You're in business if you can set it up only
using the given lengths
L_{1}, L_{2}, and 1.

__2.3.2__ To find the similar triangles:
two triangles that have the same three angles
are similar. There are three right triangles evident in the picture
(two small ones making up one big one). Suppose the left-most angle
in the biggest triangle is called A, and figure out the angles in all
three triangles in terms of A. Do any of them have the same three
angle measures?

__6.4.2__ Just to clarify: a field is an algebraic structure with addition
and multiplication that are both well-behaved. To check something is a field, you
have to check a bunch of axioms. But it's much easier to check that a subset of
a known field is itself a field, because you know for free that most of the axioms
hold (associativity, distributivity, etc). That's why here you only need to check
closure under the operations: because the complex numbers **C** are algebraically
closed (in particular, the square root of a complex number is complex), each of the
sets F_{k} is in **C**, so you only need to check the four things the problem
tells you to check: you get the rest of the axioms for free.

__21.3.2__ The solutions must be complex numbers since **C**
contains all roots to all polynomials in **C**[x]. Every complex
number can be written in the form x=r e^{iA} where A is an angle
(this is basically polar coordinates). Use that to look for solutions
to x^{3}=1.

__15.2__ What these problems ask you to DO is simple; it's the
interpretation that's subtle. In each case, they ask you to show that a
certain map is
one-to-one and continuous. To show it's one-to-one, you just need to
show that if two inputs are different, the outputs will be different.
To show it's continuous,
you need to show that the map is continuous in each coordinate (x and
y).

__18.6__
You can draw the hyperbolic plane either as the upper half of the
complex plane (in this case, the boundary is the x-axis)
or as the unit disk in the complex plane (in this case, of course, the
boundary is the unit circle). In either picture, the
geodesics (the "lines") are the *orthocircles*---lines/semi-circles that are perpendicular to
the boundary. That's what you want to build your triangles and polygons
out of...

Here are some helpful ingredients: (a) the total area of a triangle in the hyperbolic plane is Pi - (A+B+C) if the interior angles are A, B, and C; (b) it follows from that that very small triangles (with small total area) have angle sum nearly Pi; (c) curves that are tangent make angle zero with each other.

__19.5.1__ A rigid motion of the plane is any combination of translations,
reflections, and rotations about the origin. All of these "preserve length,"
in the sense that the length of a line segment or curve before you apply a rigid motion
is the same as the length afterwards! They also preserve angle. (This is easy to prove
for each of the three types on its own, so it must be true for a composition of the generators.)

A similarity of the plane should be something that "preserves shape," but that's not a very good definition. So in this context, it's any combination of rigid motions and rescalings (maps that send the point (x,y) to the point (kx,ky), where k is any positive real number). For instance, the "doubling" map sends a circle to one of twice the radius; it sends an ellipse to one of double the size; it sends a triangle to one with twice the side-lengths. In each case, the new figure has the "same shape" as the old figure.

What's the orbit of a triangle under the group of similarities? In other words, if you start with one triangle, is there a succinct way of describing all the ones you'll get? To support your point, explain whether or not similarities preserve angle and/or length.

__19.5.2__ The "relation" is being in the same orbit-- in other
words, X is equivalent to Y if there is some group element g that
carries X to Y:

X ~ Y if and only if there exists g in G such that gX=Y

So for reflexivity, you just need to show that there's something in the group that carries X to itself. And so on...
__19.6.4__ What does it mean for W to be in the group? It means
that if the generators are a,b, and c, for example, then W can be
written as a "word" in
the "alphabet" of those letters and their inverses;
for instance, W=ab^{-1}aaacc. Now suppose that there is a
relation in the group that says ac=bb. That means that anytime you see
the string of letters "ac", you can replace it with "bb". So that word
from before can be re-written as the equivalent word
ab^{-1}aa(bb)c. This new word
is also in the equivalence class [W], since it is equivalent to W.

This problem asks you to show that if you take anything equivalent to
W^{-1}, it serves as an inverse to anything equivalent to W.
(That is, their
product is equivalent to the identity.)

Note that the notation C^{r}(x) means the composition of the function C, r times, applied to x.
So for instance C^{3}(56)=C(C(C(56)))=7.

__(3)__ Just to check if you're understanding the definitions right,
the set A_{3}(16) should have eight elements. You can use this
to check your answers for the first part of the question.

__(6)__ Here's the graph for the hailstone transformation reduced
modulo 4. The arrows indicate what might happen after you apply the
transformation. So, for instance, suppose you start with 0. A number
of the form 4k must be even, so the rule says to cut it in half;
when you take half of it, it is of the form 2k, which may be congruent
to either 2 or 0 (mod 4). That's why there are arrows from 0 to both 2
and 0.

__(2)__
The problem uses the "floor function," which takes the integer part of a real number.
On this site, I'll use [x] for the floor, which is slightly different notation than on the problem set.
(For instance, the [pi]=3.) Floor just truncates the decimal part, that is.
To approach the problem, think about how big the k-th perfect square is, and how big the (k+1)-st perfect square is.
For instance, 10^{2}=100 and 11^{2}=121. So if you're in between those two, how many squares precede you? For instance, what is sq(105)?

Also, a hint: We can introduce the notation {x} for the fractional part of x, so that x=[x]+{x}.
Use this to help proving the limit part of the question if you're stuck; recall how to prove
that the limit of (3x^{2}+5x)/4x^{2} as x goes to infinity is 3/4 and use the same idea.

__(3)__There was a typo in 3(d) that has now been fixed (it used to say N_{s} less than
or equal to 2^{k}N instead of square root of N).

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