Page 19 - Schaum's Outline of Theory and Problems of Advanced Calculus

10 NUMBERS [CHAP. 1

1.9. Prove that between any two rational numbers there is another rational number.
The set of rational numbers is closed under the operations of addition and division (non-zero
a þ 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 þ b a þ b
Then 2a < a þ b,thus a < and a þ b < 2b,therefore < b.
2 2
INEQUALITIES

1.10. For what values of x is x þ 3ð2 xÞ A 4 x?
x þ 3ð2 xÞ A 4 x when x þ 6 3x A 4 x,6 2x A 4 x,6 4 A 2x x,2 A x, i.e. x @ 2.

2
1.11. For what values of x is x 3x 2 < 10 2x?
The required inequality holds when
2
2
x 3x 2 10 þ 2x < 0; x x 12 < 0or ðx 4Þðx þ 3Þ < 0
This last inequality holds only in the following cases.
Case 1: x 4 > 0 and x þ 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 þ 3 > 0, i.e. x < 4 and x > 3. This is possible when 3 < x < 4. Thus the
inequality holds for the set of all x such that 3 < x < 4.

1.12. If a A 0and b A 0, prove that ða þ bÞ A p ffiffiffiffiffi
1
ab.
2
The statement is self-evident in the following cases (1) a ¼ b, and (2) either or both of a and b zero.
For both a and b positive and a 6¼ b,the proof is by contradiction.
1
Assume to the contrary of the supposition that ða þ bÞ < p ffiffiffiffiffi 1 4 2 2
ab then ða þ 2ab þ b Þ < ab.
2
2
2
2
That is, a 2ab þ b ¼ða bÞ < 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 a 1 ; a 2 ; ... ; a n and b 1 ; b 2 ; ... ; b n are any real numbers, prove Schwarz’s inequality
2 2 2 2 2 2 2
ða 1 b 1 þ a 2 b 2 þ þ a n b n Þ @ ða 1 þ a 2 þ þ a n Þðb 1 þ b 2 þ þ b n Þ
For all real numbers ,we have
2
2
2
ða 1 þ b 1 Þ þða 2 þ b 2 Þ þ þða n þ b n Þ A 0
Expanding and collecting terms yields
2 2
2
A þ 2C þ B A 0 ð1Þ
where
2
2
2
2
2
2
2
2
A ¼ a 1 þ a 2 þ þ a n ; B ¼ b 1 þ b 2 þ þ b n ; C ¼ a 1 b 1 þ a 2 b 2 þ þ a n b n ð2Þ
The left member of (1) is a quadratic form in . Since it never is negative, its discriminant,
2
2
2
4C 4A B ,cannot be positive. Thus
2
2
2
2
2
C A B 0 or C A B 2
This is the inequality that was to be proved.
1 1 1 1
1.14. Prove that þ þ þ þ < 1 for all positive integers n > 1.
2 4 8 2 n 1