Solved Problems

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OPERATIONS WITH NUMBERS 1.1. If x 4, y 15, z 3, p 2, q 1, and r 3, evaluate 3 6 4 (c) p qr , (d) pq r, (e) x p q

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(a) x y z 4 15 3 4 12 16 (b) x y z 4 15 3 19 3 16 The fact that (a) and (b) are equal illustrates the associative law of addition. (c)

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3 2 1 p qr 2 f 1 3 g 2 24 2 1 24 12 3 6 4 3 3 8 2 3 1 (d) pq r f 2 1 g 3 18 3 1 3 36 12 3 6 4 4 9 4 The fact that (c) and (d) are equal illustrates the associative law of multiplication.

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(a) x y z ,

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(b) x y z,

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x p q 4 2 1 4 4 1 4 3 12 2 3 6 6 6 6 6 Another method: x p q xp xq 4 2 4 1 8 4 8 2 6 2 using the distributive 3 6 3 6 3 3 3 law.

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1.2. Explain why we do not consider

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as numbers.

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(a) If we de ne a=b as that number (if it exists) such that bx a, then 0=0 is that number x such that 0x 0. However, this is true for all numbers. Since there is no unique number which 0/0 can represent, we consider it unde ned. (b) As in (a), if we de ne 1/0 as that number x (if it exists) such that 0x 1, we conclude that there is no such number. Because of these facts we must look upon division by zero as meaningless.

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NUMBERS

1.3. Simplify

x2 5x 6 . x2 2x 3

x2 5x 6 x 3 x 2 x 2 provided that the cancelled factor x 3 is not zero, i.e., x 6 3. x2 2x 3 x 3 x 1 x 1 For x 3 the given fraction is unde ned.

RATIONAL AND IRRATIONAL NUMBERS 1.4. Prove that the square of any odd integer is odd.

Any odd integer has the form 2m 1. Since 2m 1 2 4m2 4m 1 is 1 more than the even integer 4m 4m 2 2m2 2m , the result follows.

1.5. Prove that there is no rational number whose square is 2.

Let p=q be a rational number whose square is 2, where we assume that p=q is in lowest terms, i.e., p and q have no common integer factors except 1 (we sometimes call such integers relatively prime). Then p=q 2 2, p2 2q2 and p2 is even. From Problem 1.4, p is even since if p were odd, p2 would be odd. Thus p 2m: Substituting p 2m in p2 2q2 yields q2 2m2 , so that q2 is even and q is even. Thus p and q have the common factor 2, contradicting the original assumption that they had no common factors other than 1. By virtue of this contradiction there can be no rational number whose square is 2.

1.6. Show how to nd rational numbers whose squares can be arbitrarily close to 2.

We restrict ourselves to positive rational numbers. Since 1 2 1 and 2 2 4, we are led to choose rational numbers between 1 and 2, e.g., 1:1; 1:2; 1:3; . . . ; 1:9. Since 1:4 2 1:96 and 1:5 2 2:25, we consider rational numbers between 1.4 and 1.5, e.g., 1:41; 1:42; . . . ; 1:49: Continuing in this manner we can obtain closer and closer rational approximations, e.g. 1:414213562 2 is less than 2 while 1:414213563 2 is greater than 2.

1.7. Given the equation a0 xn a1 xn 1 an 0, where a0 ; a1 ; . . . ; an are integers and a0 and an 6 0. Show that if the equation is to have a rational root p=q, then p must divide an and q must divide a0 exactly.

Since p=q is a root we have, on substituting in the given equation and multiplying by qn , the result a0 pn a1 pn 1 q a2 pn 2 q2 an 1 pqn 1 an qn 0 or dividing by p, a0 pn 1 a1 pn 2 q an 1 qn 1 an qn p 2 1

Since the left side of (2) is an integer, the right side must also be an integer. Then since p and q are relatively prime, p does not divide qn exactly and so must divide an . In a similar manner, by transposing the rst term of (1) and dividing by q, we can show that q must divide a0 .