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_2n, so they can t both equal x ○ So, there is at most one such y • Proof (Existence) ○ (To be continued)