History of Mathematics

#### HW1

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.

#### HW2

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=(p2-q2)r, b=2pqr, c=(p2+q2)r.

Do the p and q from the previous question work? (What should r be?)

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.

#### HW3

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 Lx+y to be Lx + Ly and he is asking you to verify that it is equal to Lz where z=x+y. In other words, you need to show

{r+s : r < x and s < y, r,s rational} = {t : t < x+y, t rational}

and similarly for Lxy.

Hint: 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 Lxy 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 Lroot-2 with itself is L2. 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:

(xn, yn, (-1)n) * ( ----, ----, ----) = (xn+1, yn+1, ----)

5.4.3 The proof that square root of two is irrational is classical: see page 11. It's a little trickier for the square root of N, but similar. The difficulty is that some of the prime factors can appear with powers higher than one.

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.

#### HW4

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 i2=-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 xm 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 fn(m) instead of n fn(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.

#### HW5

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.

That is, taking the conjugate commutes with exponentiation. In particular, this means that the n-th root of (cos nA - isin nA) is the conjugate of the n-th root of (cos nA + isin nA).

9.4.5 I think this problem calls for induction. Suppose 1/a1 - 1/a2 + ... -1/a2n = what it is. Then replace 1/a2n with 1/a2n-1/x to get the next stage. What is a2n? 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.

#### HW6

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 L1, L2, and 1 are all given, so you can measure those out whenever you want.

Hint : the idea is that you want to scale up L1 by a factor of L2, 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 L1L2. 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 L1, L2, 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 Fk 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 eiA where A is an angle (this is basically polar coordinates). Use that to look for solutions to x3=1.

#### HW7

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-1aaacc. 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-1aa(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.)

#### Hailstone Numbers

Note that the notation Cr(x) means the composition of the function C, r times, applied to x. So for instance C3(56)=C(C(C(56)))=7.

(3) Just to check if you're understanding the definitions right, the set A3(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.

(7) For instance, the 4x+1 version isn't "interesting" because there are (lots of) divergent trajectories. Explain.

#### Primes

(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, 102=100 and 112=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 (3x2+5x)/4x2 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 Ns less than or equal to 2kN instead of square root of N).