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1.8. Prove that
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p p p p If x 2 3, then x2 5 2 6, x2 5 2 6 and squaring, x4 10x2 1 0. The only possible rational p p roots of this equation are 1 by Problem 1.7, and these do not satisfy the equation. It follows that 2 3, which satis es the equation, cannot be a rational number.
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p p 2 3 cannot be a rational number.
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NUMBERS
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[CHAP. 1
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1.9. Prove that between any two rational numbers there is another rational number.
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The set of rational numbers is closed under the operations of addition and division (non-zero a b denominator). Therefore, is rational. The next step is to guarantee that this value is between a 2 and b. To this purpose, assume a < b. (The proof would proceed similarly under the assumption b < a.) a b a b and a b < 2b, therefore < b. Then 2a < a b, thus a < 2 2
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INEQUALITIES 1.10. For what values of x is x 3 2 x A 4 x
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x 3 2 x A 4 x when x 6 3x A 4 x, 6 2x A 4 x, 6 4 A 2x x, 2 A x, i.e. x @ 2.
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1.11. For what values of x is x2 3x 2 < 10 2x
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The required inequality holds when x2 3x 2 10 2x < 0; x2 x 12 < 0 or x 4 x 3 < 0
This last inequality holds only in the following cases. Case 1: x 4 > 0 and x 3 < 0, i.e., x > 4 and x < 3. This is impossible, since x cannot be both greater than 4 and less than 3. Case 2: x 4 < 0 and x 3 > 0, i.e. x < 4 and x > 3. This is possible when 3 < x < 4. inequality holds for the set of all x such that 3 < x < 4. Thus the
1.12. If a A 0 and b A 0, prove that 1 a b A 2
p ab.
The statement is self-evident in the following cases (1) a b, and (2) either or both of a and b zero. For both a and b positive and a 6 b, the proof is by contradiction. p Assume to the contrary of the supposition that 1 a b < ab then 1 a2 2ab b2 < ab. 2 4 2 2 2 That is, a 2ab b a b < 0. Since the left member of this equation is a square, it cannot be less than zero, as is indicated. Having reached this contradiction, we may conclude that our assumption is incorrect and that the original assertion is true.
1.13. If a1 ; a2 ; . . . ; an and b1 ; b2 ; . . . ; bn are any real numbers, prove Schwarz s inequality a1 b1 a2 b2 an bn 2 @ a2 a2 a2 b2 b2 b2 1 2 n 1 2 n
For all real numbers , we have a1  b1 2 a2  b2 2 an  bn 2 A 0 Expanding and collecting terms yields A2 2 2C B2 A 0 where A2 a2 a2 a2 ; 1 2 n B2 b2 b2 b2 ; 1 2 n C a1 b1 a2 b2 an bn 2 1
The left member of (1) is a quadratic form in . 4C2 4A2 B2 , cannot be positive. Thus C 2 A2 B2 0 or
Since it never is negative, its discriminant, C2 A2 B2
This is the inequality that was to be proved.
1.14. Prove that
1 1 1 1 n 1 < 1 for all positive integers n > 1. 2 4 8 2
CHAP. 1]
NUMBERS
Let Then Subtracting,
1 1 1 1 n 1 2 4 8 2 1 1 1 1 1 S n 1 n 2 n 4 8 2 2 1 1 1 1 Thus Sn 1 n 1 < 1 for all n: S : 2 n 2 2n 2 Sn
EXPONENTS, ROOTS, AND LOGARITHMS 1.15. Evaluate each of the following:
a 34 38 34 8 1 1 14 34 8 14 3 2 2 9 314 3 3 s s 5 10 6 4 102 5 4 10 6 102 p p 2:5 10 9 25 10 10 5 10 5 or 0:00005 5 8 8 10 105 log2=3 27
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