Math - Observe That 4n − 1 = 2n + (2n − 1)

Exercise 1.1. Ross 1.10 Proof. Observe that 4n − 1 = 2n + (2n − 1) so the sum can be rewritten as (2n + 1) + (2n + 3) + · · · + (2n + (2n − 1)) 2(1) − 1 = 1 so for n = 1 (2(1) + 1) = 3 = 3(12) Assume P (n) is true. Consider the sum (2(n + 1) + 1) + (2(n + 3) + 1) + · · · + (2(n + 1) + (2(n + 1) − 1) = (2n + 3) + (2n + 5) + · · · + (2n − 1) + (2n + (2n + 1)) + (2n + (2n + 3)). Using the IH, this can be rewritten as 3n2 − (2n + 1) + (4n + 1) + (4n + 3) = 3n2 + 6n + 3 = 3(n + 1)2 , as desired. Exercise 1.2. Ross 1.12

Proof. a. (a + b)1 = a + b = 10 a + 11 b


(a + b)2 = a2 + ab + ba + b2 = a2 + 2ab + b2 = 20 a2 = 21 ab + 22 b2 2 2 2 3 3 2 2 3 (a
+ b)3 =
a3 + a2 b
+ aba + ab
+ ba + bab + b a + b = a + 3a b + 3b a + b = 3 3 3 3 3 3 2 2 0 a
+ 1 ab
+ 2 ab + 3 b n n! n! n! n! b. nk + k−1 = k!(n−k)! + (k−1)!(n−k+1)! = k(k−1)!(n−k)! + (k−1)!(n−k)!(n−k+1) =
(k)n!+(n−k+1)n! (n+1)! n+1 = k!(n+1−k)! = k (k)!(n−k+1)! c. the case where n = 1 follows from part a n+1 Assume P (n) is true. Consider to
(a +nb)
n−1. By the IH,n this
is equivalent
n n n b + · · · + n−1 abn−1 + nn bn ) = (a + b)(a + b) = (a + b)( 0 a + 1 a







n n+1 n + ( n0 + n1 )an b + ( n1 + n2 )an−1 b2 + · · · + ( n−1 + nn )abn + nn bn . 0 a



Since n0 = 1 = n+1 and nn = 1 = n+1 is equivalent to 0 n+1 , this




n+1 n+1 n+1 n n+1 n−2 n+1 n+1 n+1 n a + a b + a b + · · · + 0 1 2 n ab + n+1 b Exercise 1.3. Ross 2.1 √ Proof. Observe that for any positive integer c, c solves the equation x2 − c = 0 so we can apply the rational roots test to this polynomial to determine if c is rational. √ √ 2 √3 solves √ x − 3 = 0 so the possible rational √ roots are ±1, ±3. Since 1 = 1 < 3 < 4 = 2, none of these roots work so 3 is irrational. √ √ Similarly since 5, 7, and 31 are all prime, using the bounds 2 = 4 < 5 < 1 √ √ √ √ √ √ √ 3 =√9, 2√= 4 <√ 7 < 3 = 9, and 5 = 25 < 31 < 36 = 6 we can see that√ 5, 7, and 31 are √ irrational. √ √ For 24, observe that 24 = 2 6 so it suffices to show 6 is irrational. The 2 polynomial √equation √ x − 6 = 0 has possible roots ±1, ±2, ±3, ±6. Observing √ 2 = 4 < 6 < 9 = 3, gives the desired result. Exercise 1.4. Ross 2.2 √ √ Proof. Since 2,√5, and 13 are all prime, using the √ inequalities 1= 31< 32< √ √ √ √ √ 3 8 = 2, 1 = 7 1 < 7 5 < 7 128 = 2, 1 = 4 1 < 4 13 < 4 16 = 2 and similar reasoning as 2.1 we can conclude all are irrational. Exercise 1.5. Ross 2.7 p √ √ Proof. a. Suppose 4 + 2 3 − 3 = x. Rearranging yields this yields x2 + √ √ 2 3x − 1 − 2 3 = 0 which implies the sum is rational. So if we assume x is rational, x2 is rational so we can subtract out x2 − 1 and still have a rational √ number. Thus 2 3(x √ − 1) is rational. Since we assumed x to be rational, x − 1 is rational. So since 2 3 is irrational in order for the product to be rational we √ must have 2 3(x − 1) = 0 so x = 1. Plugging this in, we see it satisfies the original relation x = 1. p so √ √ √ 2 b.√Suppose 6 + 4 2 − 2 = x. Rearranging yields this √ yields√x + 2 √2x − 4 − 4 2 = 0. Applying similar reasoning as part a we see 2 2x − 4 2 = 2 2(x − 2) is rational so x = 2. Exercise 1.6. Ross Theorem 3.1 Proof. (i). If a + c = b + c, then adding (−c) to both sides yields a + c + (−c) = b + c + (−c) so a + 0 = b + 0 so a = b (ii). Observe a · 0 = a · (0 + 0) = a · 0 + a · 0 so by part (i) a · 0 = 0 (iii). Observe that (−a)b + ab = (−a + a)b = (0)b = 0 so (−a)b is an additive inverse of ab. Since inverses are unique it follows that (−ab) = −ab. (iv). By part (iii), (−a)(−b) = −(−ab) which is that additive inverse of −ab. Since ab is also an additive inverse of −ab, it follows −(−ab) = ab. (v). If ac = bc with c ̸= 0 then c has a multiplicative inverse c−1 . Multiplying both sides by c−1 yields acc−1 = bcc−1 so a(1) = b(1) so a = b. (vi). Suppose ab = 0 and assume WLOG a ̸= 0. Observe that ab = 0 = a(0) so by part (v), it follows that b = 0. Exercise 1.7. Ross Theorem 3.2 Proof. (i). If a ≤ b observe that a + (−a − b) ≤ b + (−a − b) so −b ≤ −a (ii). If a ≤ b and c ≤ 0, then by part (i), −c ≥ 0 so −ac ≤ −bc so by part (i) bc ≤ ac (iii). Follows since 0b = 0 ≤ ab by axiom. (iv). Either 0 ≤ a or a ≤ 0. If 0 ≤ a, then by part (iii) 0 ≤ a2 . If a ≤ 0 then 0 ≤ −a so 0 ≤ (−a)(−a) = a2 . (v). Observe that 1 = 12 so by part (iv), 0 ≤ 12 . Also by definition of a field, 2 since 0 and 1 are distinct we have 0 ̸= 1 so 0 < 1. (vi). If 0 < a observe that 0 < a−1 a−1 since a−1 ̸= 0 for any a. So multiplying both sides by a−2 yields 0a−2 = 0 < aa−2 = a−1 . (vii). If 0 < a < b observe that a, b ̸= 0 so they have defined inverses with 0 < a−1 , b−1 by part (vi) so 0 < a−1 b−1 . Multiplying by a−1 b−1 yields 0a−1 b−1 < aa−1 b−1 < ba−1 b−1 so 0 < b−1 < a−1 . 3