Math 508: Advanced Analysis

Homework 2

Lawrence Tyler Rush

<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 = 2-a2
2a+1

, 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

2 b2 < a2 + t(2a+ 1) = a2 + 2--a-(2a + 1) = a2 + 2- a2 = 2 2a+ 1

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 = 2
c-2c2-

. Since 0 < c < 2 < c² then

2 2 2 2 t > 0 and c - t = c- c---2 = 4c---c-+-2 = 3c-+-2 > 0 2c 2c 2c

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.

2 2 2 2 2 c2---2 2 2 (c- t) = c - 2ct+ t < c - 2ct = c - 2c 2c = c - c + 2 = 2

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.

(c)

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 < x+y-
2

. Similarly we have obtain x + y < y + y, x + y < 2y, and x+y
2

< y. Combining these results we have

x + y x < --2--< y

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 x≠0. The previous problem then implies x² > 0. So if y = 0, then x² + y² = x² > 0. If y≠0, then the previous problem gives y² > 0 which the above lemma then yields x² + y² > 0. Hence, in any case, x² + y²≠0.

(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) > √-
2

. 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 + √-
2

∕n is irrational since √-
2

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.

4

5

(a) Find all sets A ⊂ ℝ such that sup A ≤ inf A

Let A ⊂ ℝ 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 ⊂ ℝ be bound above by α and B ⊂ ℝ be bounded below by β. Then a ≤ α for all a ∈ A and β ≤ b for all b ∈ B, however, since A∩B ⊂ A and A∩B ⊂ 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 ∈ ℂ be complex numbers.

(a) Prove |z - w|≥|z - v|-|v - w|.

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

|x|- |y| ≤ |x- y|

With this, we then have

|z - w| = |z(- v+ v)- w | = |(z - v)- (w - v)| ≥ |z - v|- |w - v| = |z - v|- |v- w |

taking advantage of the fact that |x| = |- x| for all x ∈ ℂ 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 ∈ ℂ.

|| || || w---z|| = -|w---z| 1 - zw |1- zw | = --|w---z|- ||w|- zw| |w - z| = |ww---zw-| = --|w---z|- |w(w - z)| = --|w---z|- |w||w - z| |w---z| = |w-- z| |w - z| = ------ |w - z| = |w---z| |w - z| = 1

7

Let x,y,z ∈ ℝ and define

|x - y| d(x,y) = 1+-|x--y|

(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

|x - z| |x- y| |y - z| |x - y| |y- z| 1+-|x--z| ≤ 1+-|x---z| + 1+-|x--z| ≤ 1+-|x--y| + 1-+-|y---z|

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:

f1(x)∕f2(x)+ 0 = f1(x)∕f2(x ) = 0+ f1(x)∕f2(x)

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

(f (x )∕f (x))(1) = f(x)∕f (x ) = (1)(f (x)∕f(x)) 1 2 1 2 1 2

Then

f (x)∕f(x)+ g (x)∕g(x) = f1(x)g2(x)+-g1(x)f2(x) 1 2 1 2 g2(x)f2(x)

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

(f1(x)∕f2(x))(g1(x )∕g2(x)) = f1(x)g1(x)∕f2(x)g2(x)

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

(f1(x)∕f2(x))(f2(x)∕f1(x )) = 1

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.