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# MA104solutions - MA104 SOLUTIONS 1 HW1 1.4 Guess a formula...

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Unformatted text preview: MA104 SOLUTIONS 1. HW1 1.4 Guess a formula for n X i =1 (2 i- 1) and prove it via the PMI. (a) : Get a piece of graph paper and darken 1 unit square, then the 3 squares north, northeast, and east of that square. Now you have a square of area 4. Now darken the 5 unit squares lying north and/ or east of the darkened area thus far. Again the result is a square, of area 9. Now darken the next 7 northeast unit squares to get a square of side length 4, area 16, and so on. (An advantage to this technique is that it even small children and US presidents should be able to understand it, using a stick and dirt.) For those of you who are easily distracted, the guess is n 2 . (b) Proof : We proceed by induction. The formula is true by inspection for n = 1 , 2 , 3 , 4. The inductive assumption is that 1 + 3 + ... + (2 n- 3) + (2 n- 1) = n 2 . Notice that the above formatting implicitly assumes n ≥ 4. Now add 2 n + 1 to both sides to get LHS + (2 n + 1) = n 2 + 2 n + 1 = ( n + 1) 2 . Since LHS + (2 n + 1) is exactly the sum of the first n + 1 odd numbers, that completes the induction. 1.9 Decide for which integers the inequality n 2 < 2 n is true, and prove it using PMI. (a) The inequality is obviously false for n < 0, since then the RHS is smaller than 1, and the LHS is a positive integer. Checking, we see it is true for n = 0 , 1; it is false for n = 2 , 3 , 4 (equality is not allowed); then it is true for n = 5 , 6 , 7 ,... . (b) Proof : We will show the inequality holds for n ≥ 5, so 5 (not 0 or 1) is the base case, and it is true by inspection. The inductive hypothesis assumes n 2 < 2 n . Multiplying both sides by 2 yields 2 n 2 = n 2 + n 2 < 2 n +1 . In light of ( n + 1) 2 = n 2 + 2 n + 1, it suffices to show 2 n + 1 < n 2 for n ≥ 5. For setup, note 0 < n < n + 1 and 0 < 2 < n- 1. Then multiply inequalities: 2 n < ( n + 1)( n- 1) = n 2- 1 . Adding n 2 + 1 to both sides gives 2 n + n 2 + 1 = ( n + 1) 2 < n 2- 1 + n 2 + 1 = 2 n 2 . Combining with 2 n 2 < 2 n +1 gives ( n + 1) 2 < 2 n +1 and completes the induction. 1 2 2.5 Show r := [3 + √ 2] 2 / 3 is irrational. Proof : The real problem here is writing down a polynomial to which we may apply the Rational Zeros Theorem. Some algebra: r 3 = (3 + √ 2) 2 = 9 + 6 √ 2 + 2 = 11 + 6 √ 2 ( r 3- 11) 2 = (6 √ 2) 2 = 36 · 2 = 72 r 6- 22 r 3 + 49 = 0 Assuming that r = p/q is rational and applying the RZT implies q divides 1 (that is, r is an integer), and that p divides 49. One easily checks all possibilites: r = ± 1 , ± 7 , ± 49 , and none satisfy the above. Hence r cannot be rational. 3.1 Which of the field/ order axioms fail for N and Z ? Solution : (a) N satisfies all of the order axioms. Though 0 is not a natural number according to our definition, N will still satisfy some of the properties that are stated in terms of 0. In particular, O5 will hold, since a ≥ 0 automatically for all a ∈ N . N fails to have negatives (additive inverses) and reciprocals (multiplicative inverses). These are A4 and M4. One could also debate whetherinverses)....
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## This note was uploaded on 04/30/2011 for the course MECHANICAL MSC taught by Professor Dr.rahula during the Spring '10 term at University of Moratuwa.

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MA104solutions - MA104 SOLUTIONS 1 HW1 1.4 Guess a formula...

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