To compute a n for an arbitrary n note that any

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To compute A N for an arbitrary N , note that any number made out of 1’s and 2’s only the first digit of the number must be either 1 or 2. If we start with 1, then we can complete the number in A N 1 ways, while if we start with 2 we can complete the number in A N 2 ways. Therefore A N = A N 1 + A N 2 . Using this formula we compute A 1 = 1 , A 2 = 2 , A 3 = 3 , A 4 = 5 , A 5 = 8 , A 6 = 13 , A 7 = 21 , A 8 = 34 , A 9 = 55 , A 10 = 89 , A 11 = 144 , A 12 = 233 , A 13 = 377 . The correct answer is (f) . square Solution of problem 1.2: Dividing 2012 by 7 we get a quotient 287 and remainder 3. Thus 2012 equals 3 modulo 7, and so 2012 2012 = 3 2012 modulo 7 . Computing the first few powers of 3 modulo 7 we see that 3 2 = 9 = 2 modulo 7 , 3 3 = 27 = - 1 modulo 7 . In particular since 2012 = 3 · 670 + 2 we get 3 2012 = (3 3 ) 670 · 3 2 = ( - 1) 670 · 2 = 2 modulo 7 . The correct answer is (a) . square 12
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Solution of problem 1.3: (i) Let q = (5 3) / 7. Then the irrational num- ber 3 is equal to (7 q ) / 5. If q is rational, since a product or ratio of rational numbers is rational, we will get that 3 is rational which is a contradicton. Thus (5 3) / 7 is irrational. (ii) The number q = 2( π - 1) can not be rational. Indeed the irrational number π is equal to 1 + q/ 2. But a ratio and sum of rational numbers is rational so if q is rational we will get that π is rational, which is a contradiction. Thus 2( π - 1) is irrational. (iii) The expression simplifies: 8 8 / 72 = 8 1 / 9 = 8 1 / 3 = 3 8 = 2 . Thus 8 8 / 72 is rational. (iv) The expression simplifies: parenleftBig 2 parenrightBig 16 = parenleftBig 2 parenrightBig 4 = 2 2 = 4 . Thus ( 2 ) 16 is rational. The correct answer is (c) . square Solution of problem 1.4: (1) is True . There is a one-to-one correspon- dence between the set of all natural numbers and the set of all natu- ral numbers beginning with 3141592. The correspondence is given by matching a number N eginning with 3141592 with the natural number N obtained from N by deleting the leading digits 3141592. (2) is False . We argued in class that for every two real numbers a < b there exists a rational number c such that a < c < b . 13
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(3) False . The set of all rational numbers with denominators between 1 and 1000 is infinite: it contains the natural numbers since they can be viewed as rational numbers with denominators 1. The set of all natural numbers between 1 and 1000 is finite. Thus the two sets can not have the same cardinality. The correct answer is (b) . square Solution of problem 1.5: The intersection of B with any one of the four dimensional subspaces is given by one constraint so it depends on three parameters. Since we have a two parameter family of four dimensional subspaces, the points in B depend on a tota; pf 3 + 2 = 5 parameters. This shows that B is 5 dimensional. This argument can be made also in terms of formulas. We can solve the constraint for x 3 in terms of the other variables. Then a general point in the slice of B by the subspace satisfying x 1 = a , x 5 = b is given by ( a, x 2 , - x 2 2 - 2 x 3 4 + 5 x 3 6 + ( a + 7 b ) 2 , x 4 , b, x 6 ) .
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