MATH 147 – Homework 2 Due in class Monday, September 28 1. Let a1 = 1, and for each n ≥ 1 define an+1 = √ 3 + 2an So, for example, a2 = √ 3 + 2a1 = √ 5. Prove that for every integer n ≥ 1, we have 0 ≤ an ≤ an+1 ≤ 3 2. Let a > 0 be a positive real number.

MATH 147 – Homework 2 Due in class Monday, September 28 1. Let a1 = 1, and for each n ≥ 1 define an+1 = √ 3 + 2an So, for example, a2 = √ 3 + 2a1 = √ 5. Prove that for every integer n ≥ 1, we have 0 ≤ an ≤ an+1 ≤ 3 2. Let a > 0 be a positive real number. Let A = {a n | n ∈ N} = {a, a2 , a3 , . . .} be the set of all positive integer powers of a. (a) If a > 1, show that A is not bounded above. [Hint: Use the Binomial Theorem on (1 + (a − 1))n .] (b) If a < 1, show that inf(A) = 0. [Hint: Use part (a) on 1/a.] 3. Let A be the set of all the real numbers an from problem 1. Prove that sup(A) = 3. [Hint: Define αn = 3 − an, and use the result from problem 2 to show that for every > 0, there is a positive integer n such that αn < .] 4. For every integer n ≥ 1, define an = 1+ 1 1! + 1 2! +. . .+ 1 n! . (If k is a positive integer, we define k! (pronounced “k factorial”) to be k! = 1 · 2 · · ·(k −1)· k.) Let A be the set consisting of all the an. Prove that A is bounded above by 3. (You may assume – without proof – standard facts about geometric series. Like geometric series with common ratio 1/2, for example.) [The fact that A is bounded above means that A has a supremum. This supremum is called e, and is the base of the natural logarithm. It’s also not bigger than 3.] 5. Let A be the set of 10-adic numbers. That is, let A be the set of infinite decimal expansions . . . anan−1 . . . a0.a−1a−2 . . . a−m where the ai are integers between 0 and 9, inclusive, and the expansions are allowed to extend infinitely to the left, but have only finitely many digits after the decimal point. Two 10-adic numbers are added and multiplied in the usual way, from right to left with carry. For example: . . . 444444.2893 + . . . 222362.1832 . . . 666806.4725 because the 3 and the 2 add to 5, the 3 and the 9 add to 12, so write 2 and carry the one to the next column. The 8 and 8 then add to 16, plus the carried 1 gives 7 carry another 1, and so on. Multiplication is similarly analogous to the usual elementary school version: . . .444444.2893 × . . .222362.1832 . . .888.8885786 . . .3333.328679 . . .55555.43144 . . .444444.2893 . . .8888885.786 . . .66666573.58 . . .333332867.9 . . . .2039976 Note that in the above calculation, only the last seven digits in the answer are given, since the other digits would need more lines of long multiplication to calculate. You may assume that this addition and multiplication are associative, commutative, and distributive. You may further assume (although it’s fun to verify this yourself) that . . . 0 is an additive identity element, and . . . 0001 is a multiplicative identity element. Also note that all 10-adic expansions are different: . . . 99999 is (very) different from . . . 00000.1 as a 10-adic number. (a) Prove that . . . 999999 + 1 = 0. (As with ordinary decimal expansions, leading zeroes are left off of 10-adic numbers, so 0 means the same as . . . 0.) (b) Show that 67 = . . . 66667 gives 1 when multiplied by 3. 6. Let A be the 10-adic numbers. Prove that there are two elements a, b ∈ A of A such that a 6= 0 and b 6= 0, but ab = 0. [Notice that this means that A is not a field, because if ab = 0, then a and b can’t have multiplicative inverses: ab = 0 means (1/a)ab = 0, which means b = 0.]

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