Homework 2 <me@tylerlogic.com>
September 12, 2014
http://coursework.tylerlogic.com/courses/upenn/math508/homework02

### 1 Density of rationals in reals

#### (a) For a ∈ ℚ with a > 0 and a2< 2 find rational b > a with a2< b2< 2

First, if a < 1, then we can simply let b = 1 since a2 < 12 < 2 in this case. So assume that a 1. This implies 2a + 1 3. Furthermore, this implies that 2 - a2 < 2 - a < 2, from which we conclude 2 - a2 < 2a + 1. Thus, defining t = , means that 0 < t < 1.

With this definition of t, setting b = a + t we see b2 = (a + t)2 = a2 + 2at + t2 < a2 + 2at + t where the last inequality comes from the fact that 0 < t < 1. Continuing, we have

Thus since b = a + t and both a and t are positive, we have a2 < b2 < 2. Finally, t is rational since it is constructed by a combination of multiplication and addition of rational numbers; this implies that rationality of b since it is the sum of two rationals, a and t.

#### (b) For c ∈ ℚ with c > 0 and 2 < c2, find rational d > 0 with d < c and 2 < d2< c2

Define t = . Since 0 < c < 2 < c2 then

Now let d = c - t. The above equations imply that 0 < d < c and also allow for the use of the inequality in the following equation.

Given this equation, we now have 2 < d2 < c2, and furthermore, since d is equal to some combination of rational numbers which are added/multiplied together, then it too is rational.

### 2 Some properties of an ordered field.

Let F be an ordered field containing elements x and y.

#### (a) Show that x < y implies x << y

Let x < y. Then the definition of an ordered field and Rudin’s Proposition 1.18 yield x + x < x + y, implying 2x < x + y and thus x < . Similarly we have obtain x + y < y + y, x + y < 2y, and < y. Combining these results we have

as desired.

#### (b) Prove x≠0 implies x2> 0

First suppose x > 0. Then Rudin’s Proposition 1.18 (b) implies x(x) > x(0) which is equivalent to x2 > 0.

Now suppose x < 0. Rudin’s Proposition 1.18 (c) implies x(x) > x(0), i.e. x2 > 0.

#### (c) Prove x2+ y2= 0 implies x = y = 0

Lemma 2.1. If F is some ordered field and a,b F are such that a > 0 and b > 0, then a + b > 0.

Proof. Let F be an ordered field with a,b F such that a > 0 and b > 0. Adding b to the both sides of a > 0 yields a + b > b, but b > 0, so a + b > 0. __

Let x,y not be both identically zero. Without loss of generality, assume x0. The previous problem then implies x2 > 0. So if y = 0, then x2 + y2 = x2 > 0. If y0, then the previous problem gives y2 > 0 which the above lemma then yields x2 + y2 > 0. Hence, in any case, x2 + y20.

#### (d) Show that 2xy ≤ x2+ y2. When does equality occur?

Part (b) of this problem implies that for any x,y we have (x - y)2 0. Thus x2 - 2xy + y2 0, and therefore x2 + y2 2xy, as desired.

When does equality occur? Stepping backwards through this proof, we see that x2 + y2 = 2xy when (x - y)2 = 0. Then part (b) of this problem implies that x - y = 0. Thus we have equality when x = y.

### 3 More rational density

#### (a) Show there is an irrational between rationals x < y

If x < y, then 0 < y -x. Hence the archimedean property of the reals yields an integer n > 0 such that n(y - x) > . Since > 0 we have
 0 < < n(y - x) 0 < ∕n < y - x x < x + ∕n < y
Now we saw in the last homework that both the multiplication and sum of a rational with an irrational is irrational, so x + ∕n is irrational since is irrational.

#### (b) Show there is a rational between any real numbers x < y

If x < y, then 0 < y -x. So again, the archimedean property gives us an integer n > 0 such that ny -nx > 1. This implies, since consecutive integers have a difference of one, that there must some integer m with nx < m < ny. Thus x < m∕n < y, and m∕n is rational.

### 5

#### (a) Find all sets A ⊂ ℝ such that supA ≤infA

Let A with a = inf A and bsupA. If x,y A and x < y then we would have a x < y b, and so it’s not possible for a set with two or more elements to have a supremum that’s less than or equal to the infimum. Hence only singleton sets have the desired property.

#### (b) If A ⊂ ℝ is bounded above and B ⊂ ℝ is bounded below, prove A ∩ B is bounded.

Let A be bound above by α and B be bounded below by β. Then a α for all a A and β b for all b B, however, since AB A and AB B, then we must have that β x α for all x A B. In other words, A B is bounded.

### 6

Let z,w,v be complex numbers.

#### (a) Prove |z - w|≥|z - v|-|v - w|.

We know for a,b that |a + b|≤|a| + |b|, which implies |a + b|-|b|≤|a|. Since a and b are arbitrary, then we can find x,y when a = x - b and b = y, then |(x - y) + y|-|y|≤|x - y| which implies

With this, we then have

taking advantage of the fact that |x| = |- x| for all x in the rightmost equality.

#### (b) Graph the points z ∈ ℂ such that 1 < |z - i| < 2

The following region is the set of points, note that the edges of the region are not included.

#### (c) For z,w ∈ ℂ with |z| < 1 and |w| = 1, prove |(w - z)∕(1 -zw)| = 1

We have the following sequence of equations due to the fact that |a| = |a|, |ab| = |a||b|, |w| = 1, and a - b = a -b for any a,b .

### 7

Let x,y,z and define

#### (a) Prove the above d(⋅,⋅) satisfies the triangle inequality

If |x - z|≤|y - z| or |x - z|≤|x - y|, then the fact that f(t) = t∕(1 + t) is an increasing function implies the triangle inequality for d.

So assume that |x - z| is greater than both |x - y| and |y - z|. Since |x - z|≤|x - y| + |y - z| we have

with the rightmost inequality comming from our initial assumption. Thus the triangle inequality holds for this d(,).

#### (b)

The proof for this is identical to the proof of in the previous part of this problem after substituting in the function |⋅-⋅| for g(,). This is because the only property of |⋅-⋅| that was used in the previous proof was that it upholds the triangle inequality, which g(,) also does.

### 8

#### (a)

Let f1(x)∕f2(x),g1(x)∕g2(x) .

By the following, the zero constant is the additive identity:

By the following, the one constant is the multiplicative identity:

Then

so is closed under addition. Since polynomial addition is commutative and associative, then addition is commutative and associative in . Finally, since f(x) + (-f(x)) = 0, additive inverses exist in .

Since

then is closed under multiplication. Since polynomial multiplication is commutative and associative, then multiplication is commutative and associative in . Finally, if f1(x)∕f2(x) is not zero, then

and so multiplicative inverses exist in .

Finally, since polynomial addition and multiplication obides by the distributive law, then so does addition and multiplication in .

So is indeed a field.