Math 508: Advanced Analysis

Homework 1

Lawrence Tyler Rush

<me@tylerlogic.com>

September 10, 2014

http://coursework.tylerlogic.com/courses/upenn/math508/homework01

1 For rational r≠0 and irrational x, prove that r + x and rx are irrational.

Since r is rational, both 1∕r and -r are rational. Thus, if r + x were rational then

x = (r+ x)+ (- r)

would imply x is rational by the closure of addition in the field of rationals; but since x is assumed to be irrational, so must be r + x.

Similarly, rx must be irrational since

x = (rx)(1∕r)

and the field of rationals is closed under multiplication.

2 Prove that there’s no rational number with a square of 12

Lemma 2.1. The square root of 3 is irrational.

Proof. Assume for later contradiction that 3 is rational. Then we can find integers m,n which are not both even such that m∕n = √ -
3 . Therefore

m2 = 3n2

This equation then demands that if n² is even, then so must be m² (and therefore m), and similarly if n² is odd, then so must be m² (and therefore m). Because m and n were assumed to not both be even, then m and n must both be odd. Thus we can set m = 2p + 1 and n = 2q + 1. Furthermore the above equation implies that

(2p+ 1)2 = 3(2q+ 1)2 2 2 4p +4p + 1 = 12q + 12q+ 3 4p2 + 4p = 12q2 + 12q+ 2 2p2 + 2p = 6q2 + 6q+ 1 2(p2 + p) = 2(3q2 + 3q)+ 1

however, the left-hand side of the last line above is even and the right-hand side is odd; a contradiction. So 3 must be irrational. __

Since √ --
12 = 2 √ -
3 the previous problem and the above lemma together imply that is irrational.

3 Prove the following from the field axioms for multiplication.

(a) If x≠0 and xy = xz, then y = z

Let x≠0 and xy = xz, then

( ) ( ) y = (1)y = -1x y = 1-(xy) = 1(xz) = 1x z = 1(z) = z x x x x

(b) If x≠0 and xy = x, then y = 1

This follows from the previous part for z = 1.

(c) If x≠0 and xy = 1, then y = 1∕x

Again, this follows from the first part of this problem for z = 1∕x.

(d) If x≠0 then 1∕(1∕x) = x

Let x≠0, then

( 1 ) 1 1 1 x = (1)x = ---(1∕x) x = --- ((1∕x)x) = ---(1) =--- 1∕x 1∕x 1∕x 1∕x

4 Rudin p.23 #17

For x,y ∈ ℝⁿ with the Euclidean distance, we have

|x +y |2 + |x - y |2 = (x ⋅x+ 2x ⋅y+ y ⋅y)+ (x⋅x - 2x⋅y + y⋅y) = x ⋅x+ y ⋅y+ x ⋅x+ y ⋅y = 2x ⋅x+ 2y ⋅y = 2|x|2 + 2|y|2

Geometrically, if |x| and |y| are the two lengths of two adjacent sides of a parallelogram, then |x + y| and |x-y| would be the lengths of the two diagonals. So this equation reveals that the sum of the squares of the diagonals of a parallelogram is equal to twice the sum of the squares of its sides.

5 Fix a,b ∈ ℝ^k. Show there exist c ∈ ℝ^k and r ∈ ℝ such that ∀x ∈ ℝ^k, |x - a| = λ|x - b| iff |x - c| = r, for λ > 1

Letting (λ² - 1)c = λ²b - a and r = λ(λ² - 1)^-1|b - a| we obtain

(λ2 - 1)2|c|2 = |λ2b- a|2 = λ4|b|2 - 2λ2 ⟨b,a⟩+ |a|2

(5.1)

Now assuming that |x - a| = λ|x - b| for some point x ∈ ℝ^k we get the following sequence of equations.

λ2|x- b|2 = |x- a|2 2( 2 2) 2 2 λ |x| - 2⟨x,b⟩ + |b| = |x| - 2⟨x,a⟩+ |a| (λ2 - 1)|x|2 - 2λ2⟨x,b⟩ +2 ⟨x,a⟩ = |a|2 - λ2|b|2 (λ2 - 1)|x|2 - 2⟨x,λ2b+ a⟩ = |a|2 - λ2|b|2 2 2 ⟨ 2 ⟩ 2 2 2 (λ⟨ - 1)|x| -⟩2 x,(λ - 1)c = |a| - λ |b| (λ2 - 1)|x|2 - 2 x,(λ2 - 1)c + (λ2 - 1)|c|2 = (λ2 - 1)|c|2 + |a|2 - λ2|b|2 (λ2 - 1)(|x|2 - 2⟨x,c⟩ + |c|2) = (λ2 - 1)|c|2 + |a|2 - λ2|b|2 2 2 2 2 2 2 2 (λ - 1)|x- c| = (λ - 1)|c|+ |a| - λ |b|

At this point, Equation 5.1 yields the fact that (λ² - 1)|c|² = (λ² - 1)^-1 ( 4 2 2 2)
λ |b| - 2λ ⟨b,a⟩+ |a|

, so making that substitution into the last line from the sequence of equations above, we continue:

2 2 2 -1( 4 2 2 2) 2 2 2 (λ - 1)|x - c| = (λ - 1) (λ |b|- 2λ ⟨b,a⟩+ |a| + |a| - λ |b| ) (λ2 - 1)|x - c|2 = (λ2 - 1)-1 (λ2 - 1)2|c|2 + (λ2 - 1)|a|2 - (λ2 - 1)λ2|b|2 |x - c|2 = (λ2 - 1)-2((λ2 - 1)2|c|2 + (λ2 - 1)|a|2 - (λ2 - 1)λ2|b|2) 2 2 -2(( 4 2 2 2) 2 2 2 2 2) |x - c| = (λ - 1) ( λ |b|- 2λ ⟨b,a⟩+ |a| +) (λ - 1)|a| - (λ - 1)λ |b| |x - c|2 = (λ2 - 1)-2 λ2|b|2 - 2λ2 ⟨b,a⟩+ λ2|a|2 |x - c|2 = λ2(λ2 - 1)-2(|b|2 - 2 ⟨b,a⟩+ |a|2) 2 2 2 -2 2 |x - c| = λ( (λ - 1) |b- a|) |x - c|2 = λ(λ2 - 1)-1|b- a|2 |x - c|2 = r2

With the final line above we must have |x - c| = r since both |x - c| and r are positive values (r being positive because λ > 1). So we have that |x - a| = λ|x - b| implies that |x - c| = r; however, reversing the order of equations above also reveals that |x - c| = r implies |x - a| = λ|x - b|, giving us the “iff” we desire.

What happens when λ = 1? When λ = 1, we’d be interested in all points x ∈ ℝ^k where |x-a| = |x-b|. However the points that satisfy this are the points in the hyper-plane that perpendicularly bisects the line segment, which of course has no notion of a radius since it is a hyper-plane and not a hyper-sphere. Hence it makes no sense to inquire about a sphere of radius r = λ(λ² - 1)^-1|b - a|; this is dually supported by the denominater, (λ² - 1), having a value of zero when λ = 1. In a silly way of thinking, it’s as if the sphere gets so big (as λ approaches 1) that it “becomes a plane”.

6 Prove that a union of a countable number of finite sets is countable

Lemma 6.1. If S₁,S₂,S₃,… are finite sets that are mutually disjoint, then ∪_j=1^∞S_i is countable.

Proof. First note that if any S_i is the empty set it can simply be neglected. So fix an ordering of each of the elements of each set S_i and label them as follows

S1 = {x11,x12,...,x1n1} S2 = {x21,x22,...,x2n2} S = {x ,x ,...,x } 3 31 32 3n3 ...

where n_i = |S_i|. With this labeling, the following correspondence shows that ∪_j=1^∞S_j is countable

x11 ↦→ 1 x12 ↦→ 2 . .. x1n1 ↦→ n1 x ↦→ n + 1 21 1 x22 ↦→ n1 + 2 ... x2n2 ↦→ n1 + n2 x31 ↦→ n1 + n2 + 1 .. .

Let S₁,S₂,S₃,… each be a finite set. The problem here is that these sets are not mutually disjoint, but nevertheless, we can define the sets T₁,T₂,T₃,… as T₁ = S₁ and T_n = S_n -∩_j=1^n-1S_j and furthermore, ∩_j=1^n-1T_j = ∩_j=1^n-1S_j. So due to the above lemma, ∩_j=1^n-1T_j is countable and therefore so is A = ∩_j=1^n-1S_j.

7 Prove that the algebraic numbers are countable

Define the sets S₁,S₂,S₃,… by

Sn = {z ∈ ℂ | ∃ a0,...,an ∈ ℤ ∩ [- n,n] ∋ anzn + an-1zn-1 + ⋅⋅⋅+a1z + a0 = 0}

Now with this definition, the set A = ∪_i=1^∞S_i will contain only algebraic numbers, but it will furthermore contain all of the algebraic numbers since if z is the root of a_nxⁿ + ⋅⋅⋅ + a₁x + a₀ and some a_i has |a_i| > n, then z will be in the S_N where N = max(|a₀|,…,|a_n|) since z will also be a root of 0x^N + 0x^N-1 + + 0xⁿ⁺¹ + a_nxⁿ + + a₁x + a₀. Also, because a polynomial of degree n has at most n roots and the set ℤ ∩ [-n,n] is finite for any n, then each of the sets S_i are finite. Therefore, the previous problem gives us that A (the set of algebraic numbers) is a countable set.

8 Defining real exponents

Let b > 1 be real and m,n,p,q ∈ ℤ, r ∈ ℚ such that r = m∕n = p∕q

(a) Prove ^1∕n = ^1∕q

Since m∕n = p∕q, then mq = np and through use of the laws of integer exponents we obtain

m 1∕n ( (( 1∕q)q)m )1∕n (( 1∕q)mq)1∕n (( 1∕q)np)1∕n (( p1∕q)n)1∕n p1∕q (b ) = b = b = b = (b ) = (b )

So it makes sense to define b^r = (bm) ^1∕n

(b) Prove b^r+s = b^rb^s for rational r,s

Let s = a∕b for a,b ∈ ℤ. Then we have

mb + an r+ s = m ∕n +a∕b =---nb--

which in turn allows

( )1∕nb br+s = bmb+an = (bmbban)1∕nb ( ) = bmb 1∕nb(ban)1∕nb ( ( )b)1∕b(( )n)1∕n = (bm)1∕n (ba)1∕b = (bm)1∕n(ba)1∕b = brbs

through the use of properties of integer exponents and the previous part of this problem.

(c) Prove that b^r = sup B(r)

Define B(r) = {b^t | t ∈ ℚ and t ≤ r}. Because b^r ∈ B(r), f(t) = b^t is monitonically increasing since b > 0, and any b^t ∈ B(r) for rational t would have t ≤ r, then b^r must be supB(r).

(d) Definining b^x = sup B(x) ∀x ∈ ℝ, show that b^x+y = b^xb^y for x,y ∈ ℝ

Define B(x,y) = {b^sb^t | s,t ∈ ℚ and s ≤ x,t ≤ y}. By an argument identical to the previous part of this problem, we have supB(x,y) = b^xb^y. However B(x,y) is also {b^s+t | s,t ∈ ℚ and s ≤ x,t ≤ y}, which is the same as {b^t | t ∈ ℚ and t ≤ x + y}. Of course this last set is just the definition of B(x + y), so b^x+y = supB(x + y) = supB(x)supB(y) = b^xb^y.

9 Prove that there exist real numbers that are not algebraic.

The set of real numbers is uncountable, whereas the set of algebraic numbers is countable, as per problem seven. So there are “too many” real numbers for all of them to be algebraic.

10 Is the set of all irrational numbers countable? Why?

No, the irrational numbers are not countable.

Each rational number r = p∕q with p,q ∈ ℤ, q≠0, is the root of f(z) = qz -p, implying that the rational numbers are a subset of the countable algebraic numbers, i.e. they’re countable. So because the rational numbers are countable, the union of the rational and irrational numbers are the reals, and the reals are uncountable, then the irrational numbers are necessarily uncountable.