By Arun-Kumar S.
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Ak−1 , xk ] = For n = k + 1, replace xk by ak + 1 xk+1 ⇒x = [a0 ; a1 , . . , ak−1 , ak + = = and so the result holds for all n. 2 If x > 1 and x + 1/x < i. x < α = ii. 1 x Proof: > −β √ 5 then √ 5+1 2 √ = 5−1 2 √ Note that α and β are roots of equation x + 1/x = 5. √ x + 1/x < 5 ⇒ (x − α)(x − β) < 0 The two possibilities are α < x < −β) or −β < x < α. The first one is ruled out as we are given that x > 1 > −β. So, we√have −β < x < α which proves the first claim. √ 5−1 1 √2 = ✷ Now, x < α ⇒ x < 5+1 ⇒ > which proves the second claim.
M − 1} as the set of positive integers less than m. Consider a relation ≡m ⊂ Z+ × Z+ , where a ≡m b if and only if m | (a − b). ≡m is an equivalence relation • Reflexive: a ≡m a, for all a ∈ Z+ . • Symmetric: If a ≡m b, then a − b = k1 m. So b − a = −k1 m, and b ≡m a. • Transitive: If a ≡m b (implying that a − b = k1 m) and b ≡m c (implying that b − c = k2 m), then a − c = (k1 + k2 )m, and hence a ≡m c. Therefore, we can partition the set of integers into m equivalence classes, corresponding to the remainder the number leaves when divided by m.
1 Every quadratic irrational has SPCF representation. Proof Outline : Let say that x is a quadratic irrational. x= √ b+ d c where b, d, c ∈ Z but d is squarefree integer. let say 33 34 CHAPTER 7. QUADRATIC IRRATIONAL(PERIODIC CONTINUED FRACTION) x= √ m+ d s0 where s0 |(d − m2 ) √ mi + d si mi+1 = ai si − mi d − m2i+1 si+1 = si ai = [xi ] xi = Claim : mi , si are all integers. P roof : By induction on i. Base Case : m0 and s0 are b and c and b, c ∈ Z Let say it is true for i. mi , si are integers and si |(d − m2i+1 ).