Math 521 Archives - Page 7 of 8 - Shawn Zhong

Math 521 – 2/14

$Finite and Countable • Definition ○ Let J_n={1,2,3,…,n} and N={1,2,3,…} ○ For any set A, we say ○ A is finite if A~J_n for some n (∅~J_0 so ∅ is finite) ○ A is infinite if A≁J_n for all n ○ A is countable if A~N ○ A is uncountable if A is neither finite nor countable ○ A is at most coutable if A is finite or countable • Examples ○ N is clearly countable § N={1,2,3,…} ○ Z is countable § Z={0,1,−1,2,−2,3,−3,…} § Define f:N→Z by § f(n)≔{■8(n/2&n is even@(1−n)/2&n is odd)┤ § f is injective □ If f(n)=f(m) □ then n/2=m/2 or (1−n)/2=(1−m)/2 □ Either way, n=m § f is surjective □ Given k∈Z, □ If k0, k=f(2k) □ If k≤0, k=f(−2k+1) § Thus f is bijective ○ Q is countable § There are "less" rational numbers q=m/n (m,n∈Z,n≠0) than § there are ordered pairs of integers (m,n) □ 1/2=15/30 but (1,2)≠(15,30) □ We can also ignore negatives and zeros □ because integers are in 1-1 correspondence with N § Idea: Write ordered pairs of integers in a 2 dimension array § Putting this all together, we have § Q={0,±1,±2,±1/2,±1/3,±3,±4,±3/2,±2/3,±1/4,…} Sequence • Definition ○ A sequence is a function defined on N ○ Notationally, this is often written {x_n } ○ Meaning f(x)=x_n for all n∈N • Example ○ {1/n}={1, 1/2,1/3,…} Theorem 2.8 • Statement ○ Every infinite subset of a countable set is countable • Proof ○ Suppose E⊂A, A is countable and E is infinite ○ Since A is countable, its element will be a sequqnce ○ (order given by the bijective function f:N→A) ○ So, A={x_n } as a set ○ Let n_1 be the smallest n∈N such that x_(n_1 )∈E ○ Let n_2 be the next smallest n∈N such that 〖x_n〗_2∈E ○ So E={x_(n_k ) }={x_(n_1 ),x_(n_2 ),x_(n_3 ),…} (A sequence indexed by k∈N) ○ Now consider g:N→E given by g(k)=x_(n_k ) ○ g is clearly one-to-one and onto by construction ○ So E is countable • Example ○ If A={1, 1/2,1/3,…} and E={1, 1/4,1/9,1/16} ○ Then A={1/n}={1, 1/2,1/3,…} ○ and E={1/n_k }={1, 1/4,1/9,1/16} where n_k=k^2 for k∈N ○ f:A→E by f(k)=1/k^2$

Math 521 – 2/12

$Schwarz Inequality • See Theorem 1.35 in Rudin for a proof of Schwarz Inequality for ℂ ○ For intuition, try proving (x_1 y_2+x_2 y_2 )^2≤(x_1^2+x_2^2 )(y_1^2+y_2^2 ) • Triangle Inequality ○ In a Euclidean Space, |x ⃗⋅y ⃗ |≥|x ⃗ |⋅|y ⃗ | ○ |x ⃗+y ⃗^2 |=|x ⃗ |^2+2x ⃗⋅y ⃗+|y ⃗ |^2≤|x ⃗ |^2+2|x ⃗ ||y ⃗ |+|y ⃗ |^2=(|x ⃗ |+|y ⃗ |)^2 ○ Thus |x ⃗+y ⃗ ||x ⃗ |+|y ⃗ | ○ Let x ⃗≔x ⃗−y ⃗, y ⃗≔y ⃗−z ⃗, we have |x ⃗−z ⃗ ||x ⃗−y ⃗ |+|y ⃗−z ⃗ | Function • Given two sets A and B • A function (or mapping) is a rule that assigns elements in A to elements in B • Notationally, if f is a function from A to B, we write f:A→B • Set A is called the domain of f • Set B is called the codomain of f • For E⊂A, f(E)={b∈B│b=f(e) for some e∈E} is the image of E under f • f(A) is called the range of f • If f(A)=B, then we say that f is onto or surjective • If f(a_1 )=f(a_2 ) implies a_1=a_2, then f is one-to-one or injective • A function that is both one-to-one and onto is said to be bijective • For E⊂B, f^(−1) (E)={a∈A│f(a)∈E} is the inverse image of E under f • Notationally, if y∈B, f^(−1) (y)=f^(−1) ({y}) ○ f^(−1) is at most a single element set for all y∈B if and only if f is injective ○ In this case, f^(−1) can be thought of as a function maps to the single element • Example ○ f:R→R defined by f(x)=x^2 ○ f^(−1) ({1})={1,−1} ○ f^(−1) ({x∈Rx0})=∅ ○ f^(−1) ({0})={0}, we can also write f^(−1) (0)=0 Cardinality • If there exists a one-to-one, onto mapping from set A to set B • We say that A and B can be put in one-to-one correspondence • And that A and B have the same cardinality (or cardinal number) • In this case, we write A~B Equivalence Relation • One-to-one correspondence is an example of an equivalence relation • An equivalence relation satisfies 3 properties ○ Reflexive: A~A ○ Symmetric: If A~B, then B~A ○ Transitivity: If A~B, B~C, then A~C Finite and Countable • Let J_n={1,2,3,…,n} and N={1,2,3,…} • For any set A, we say ○ A is finite if A~J_n for some n (∅~J_0 so ∅ is finite) ○ A is infinite if A≁J_n for all n ○ (To be continued)$

Math 521 – 2/7

$Complex Numbers ℂ • Definition ○ If z∈ℂ, then z=a+bi where a,b∈R and i^2=−1 • Real part and imaginary part ○ For z=a+bi ○ Re(z)=a is the real part of z ○ Im(z)=b is the imaginary part of z • Complex conjugate ○ z ̅=a−bi is the complex conjugate of z ○ zz ̅=(a+bi)(a−bi)=a^2+b^2 • Absolute value ○ |z|=√(zz ̅ )=√(a^2+b^2 ) is the absolute value of z ○ Note § For a real number x § |x|=√(x^2+0^2 )=√(x^2 )≥0 § |x|={■8(x&if x≥0@−x&if x0)┤ • Complex division ○ If z=a+bi, w=c+di∈ℂ, then ○ z/w=(zw ̅)/(ww ̅ )=(a+bi)(c−di)/(c+di)(c−di) =(ac+bd)/(c^2+d^2 )+(bc−ad)/(c^2+d^2 ) i Theorem 1.31 • If z and w are complex numbers, then ○ (z+w) ̅=z ̅+w ̅ ○ (zw) ̅=z ̅⋅w ̅ ○ z+z ̅=2Re(z), z−z ̅=2i Im(z) ○ zz ̅ is real and positive (except when z=0) Theorem 1.33 • If z and w are complex numbers, then (1) |z|0 unless z=0 in which case |z|=0 (2) |z ̅ |=|z| (3) |zw|=|z||w| § Let z=a+bi, w=c+di § Then zw=(ac−bd)+(ad+bc)i § |zw|=√((ac−bd)^2+(ad+bc)^2 ) § =√(a^2 c^2+b^2 d^2+a^2 d^2+b^2 c^2 ) § =√((a^2+b^2 )(c^2+d^2 ) ) § =√(a^2+b^2 ) √(c^2+d^2 ) § =|z||w| (4) |Re(z)|≤|z| (5) |z+w|≤|z|+|w| (Triangle Inequality) § |z+w|^2=(z+w)((z+w) ̅ ) § =(z+w)(z ̅+w ̅ ) § =zz ̅+zw ̅+z ̅w+ww ̅ § =|z|^2+|w|^2+zw ̅+z ̅w § =|z|^2+|w|^2+2Re(zw ̅ ) § ≤|z|^2+|w|^2+2|zw ̅ | by (4) § =|z|^2+|w|^2+2|z||w ̅ | by (3) § =|z|^2+|w|^2+2|z||w| by (2) § =(|z|+|w|)^2 § So |z+w|^2≤(|z|+|w|)^2 § Thus, |z+w|≤|z|+|w|∎ Euclidean Spaces • Inner product ○ If x ⃗,y ⃗∈Rn with § x ⃗=(x_1,x_2,…,x_n ) § y ⃗=(y_1,y_2,…,y_n ) ○ Then the inner product of x ⃗ and y ⃗ is § x ⃗⋅y ⃗=∑_(i=1)^n▒〖x_i y_i 〗 • Norm ○ If x ⃗∈Rn, we define the norm of x ⃗ to be ○ |x ⃗ |=√(x ⃗⋅x ⃗ ) • Euclidean spaces ○ The vector space Rn with inner product and norm ○ is called Euclidean n-space Theorem 1.37 • Suppose x ⃗,y ⃗,z ⃗∈Rn,α∈R, then (1) |x ⃗ |≥0 (2) |x ⃗ |=0 if and only if x ⃗=0 ⃗ (3) |αx ⃗ |=|α|⋅|x ⃗ | (4) |x ⃗⋅y ⃗ |≤|x ⃗ |⋅|y ⃗ | (Schwarz$

Math 521 – 2/5

$Theorem 1.21 • Notation ○ For a positive integer n § x^n≔⏟(x⋅x⋅x⋯x)┬(n times) ○ For a negative integer n § x^n≔⏟((1/x)⋅(1/x)⋅(1/x)⋯(1/x) )┬(−n times) • Statement ○ For every real x 0, and positive integer n ○ There is one and only one positive real number y s.t. y^n=x ○ In this case, we write y=x^(1/n) • Intuition ○ Try this for n=2 and x=2, so y=√2 • Proof (Uniqueness) ○ If there were y_1 and y_2 s.t. ○ y_1^n=x,y_2^n=x, but y_1≠y_2 ○ Without loss of generality, assume y_1 y_2 ○ Then y_1^n y_2^n, so they can t both equal x ○ So, there is at most one such y • Lemma ○ If n is a positive integer, then § b^n−a^n=(b−a)(b^(n−1)+ab^(n−2)+…+a^(n−2) b+a^(n−1) ) ○ Moreover, if b a 0, then § b^n−a^n (b−a) ⏟((b^(n−1)+b^(n−1)+…+b^(n−1)+b^(n−1) ) )┬(n terms) § b^n−a^n (b−a)nb^(n−1) § Also, (b^n−a^n)/(nb^(n−1) ) b−a b • Proof (Existence) ○ Let E≔{t∈Rt 0 and t^n x} ○ E is not empty § Let t≔x/(x+1), then 0 t 1 and t x § So, 0 t^n t x § Thus, t∈E § Therefore E is not empty ○ E is bounded above § Let t∈R s.t. t 1+x § Therefore t^n t 1+x x § So t∉E and E is bounded above by 1+x § By least upper bound property, sup⁡E exists § Let y≔sup⁡E ○ We now show that y^n≮x and y^n≯x ○ Assume y^n x § Choose h∈R s.t. § 0 h 1 and h (x−y^n)/(n(y+1)^(n−1) ) § Then hn(y+1)^(n−1) y^n § Use the lemma b^n−a^n (b−a)nb^(n−1) § Set a≔y,b≔y+h § (y+h^n−y^n (y+hy)n(y+h^(n−1) § (y+h^n−y^n hn(y+1)^(n−1) § (y+h^n−y^n y^n § (y+h^n x § Since y+h h and y+h∈E § y is not an upper bound of E § This contradicts y=sup⁡E § Thus, y^n≮x ○ Assume y^n x § Let k≔(y^n−x)/(ny^(n−1) ) 0 § Use the lemma (b^n−a^n)/(nb^(n−1) ) b § Set b≔y,a^n≔a, then k=(y^n−x)/(ny^(n−1) ) y § Thus,0 k y § Let t∈R s.t. t≥y−k, then § y^n−t^n≤y^n−(y−k)^n § Use the lemma b^n−a^n (b−a)nb^(n−1) § Set a≔y,b≔y−k, then § y^n−t^n≤y^n−(y−k)^n kny^(n−1)=y^n−x § Therefore, t^n x § By definition of E={t∈Rt 0 and t^n x} § t∉E and t is greater than everything in E § Also t≥y−k, so y−k is an upper bound for E § But y−k y, which contradicts y=sup⁡E § Thus, y^n≯x ○ Therefore y^n=x • Corollary: If a,b∈R+, and n∈Z+, then a^(1/n)⋅b^(1/n)=(ab)^(1/n) ○ Let α=a^(1/n), β=b^(1/n), then ○ α^n β^n=ab ○ (αβ)^n=ab ○ So αβ=(ab)^(1/n) Complex Numbers ℂ • Definition ○ The complex numbers ℂ can be thought of ○ as numbers of the form a+bi, where a,b∈R, and i^2=−1 • Addition, multiplication, subtraction, and division ○ If a+bi, c+di∈ℂ, then ○ (a+bi)+(c+di)=(a+c)+(b+d)i ○ (a+bi)−(c+di)=(a−c)+(b−d)i ○ (a+bi)⋅(c+di)=(ac−bd)+(ad+bc)i ○ (a+bi)/(c+di)=((a+bi)/(c+di))((c−di)/(c−di))=(a+bi)(c−di)/(c^2+d^2 )$

Math 521 – 2/2

$Proposition 1.18 • Let F be an ordered filed, for x,y,z∈F (1) If x 0 then −x0, and vice versa § x 0 § x+(−x) 0+(−x) § 0 −x∎ (2) If x 0 and yz then xyxz § x 0, z−y 0 § x(z−y) 0 § xz−xy 0 § xyxz∎ (3) If x0 and yz then xy xz § x0 § By (1), −x 0 § By (2), (−x)y(−x)z § 0(−x)(z−y) § By (1), x(z−y)0 § xzxy∎ (4) If x≠0 then x^2 0. In particular 1 0 § If x 0, by (2), x^2 0⋅x=0 § If x0, by (3), x^2 0⋅x=0 § 1=1^2=1×1 0 § So 1 0∎ (5) If 0xy, then 01/y1/x § If y 0, then 1/y⋅y=1 0=0⋅1/y by (4) § So, 1/y must have been positive by (2) § Similarly, 1/x 0 § Therefore (1/x)(1/y) 0 § Multiply both sides of xy by (1/x)(1/y) § We get 1/y1/x § Therefore 01/y1/x∎ Theorem 1.19 • There exists an ordered filed with the least upper bond property called R • Moreover R has Q as a subfield • Proof: See appendix Theorem 1.20 • The Archimedean property of R ○ Given x,y∈R, and x 0 ○ There is a positive integer n such that nx y • Proof: The Archimedean property of R ○ Let A={nx│n is a positive integer} ○ Assume the Archimedean property is false ○ Then A has an upper bound ○ i.e. sup⁡A exists ○ Let α=sup⁡A ○ x 0, so α−xα ○ And α−x is not an upper bound for A ○ By definition of A={nx│n is a positive integer} ○ α−xmx for some positive integer m ○ So, αmx+x=(m+1)x∈A ○ This contradicts α=sup⁡A ○ Therefore the Archimedean property is true • Corollary ○ Given x 0 ○ Let y=1, then ○ ∃n∈Z+ s.t. nx 1 ○ Therefore given x 0, ∃n∈Z+ s.t. 1/nx • Q is dense in R ○ If x,y∈R, and xy, then there exists a p∈Q s.t. xpy ○ We can always find a rational number between two real numbers • Proof: Q is dense in R ○ Let x,y∈R, and xy ○ So y−x 0 ○ By the Archimedean property of R § There exists a positive integer n s.t. § n(y−x) 1 § ⇒ny−nx 1 § ⇒ny nx+1 ○ By the Archimedean property of R again § There are positive integers m_1,m_2 s.t. § m_1 nx, m_2 −nx § i.e. −m_2nxm_1 § So there is an integer m s.t. § −m_2≤m≤m_1 § And more importantly, m−1≤nxm ○ Combining two parts together, we have § nxm≤1+nxny § In particular, nxmny § Since n 0, we can multiply by 1/n and get § 1/n (nx)1/n (m)1/n (ny) § Therefore xqy, where q=m/n∈Q Theorem 1.21 • Notation ○ For a positive integer n § x^n≔⏟(x⋅x⋅x⋯x)┬(n times) ○ For a negative integer n § x^n≔⏟((1/x)⋅(1/x)⋅(1/x)⋯(1/x) )┬(−n times) • Statement ○ For every real x 0, and positive integer n ○ There is one and only one positive real number y s.t. y^n=x ○ In this case, we write y=x^(1/n) • Intuition ○ Try this for n=2 and x=2, so y=√2 • Proof (Uniqueness) ○ If there were y_1 and y_2 s.t. ○ y_1^n=x, y_2^n=x, but y_1≠y_2 ○ Without loss of generality, assume y_1y_2 ○ Then y_1^ny_2^n, so they can t both equal x ○ So, there is at most one such y • Proof (Existence) ○ (To be continued)$

Shawn Zhong

Shawn Zhong

Math 521

Math 521 – 2/14

Math 521 – 2/12

Math 521 – 2/7

Math 521 – 2/5

Math 521 – 2/2

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