Page 45 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

Solution: Substitute a =8 − b − c − d − e into a + b + c + d + e = 16:
2
2
2
2
2
2 2 2 2 2 2
(8 − b − c − d − e) + b + c + d + e = 16 ⇒ 2b − 2(8 − c − d − e)b + (8 − c −
2
2
2
2
2
2
(8 − b − c − d − e) + b + c + d + e = 16 ⇒ 2b − 2(8 − c − d − e)b + (8 − c −
2
2
2
2
d − e) + c + d + e − 16 = 0
d − e) + c + d + e − 16 = 0. Since b is a real number, then
2
2
2
2
2 2 2 2 2 2
Δ b = 4(8 − c − d − e) − 8[(8 − c − d − e) + c + d + e − 16] ≥ 0 ⇒ 3c − 2(8 −
2
2
2
2
2
2
Δ b = 4(8 − c − d − e) − 8[(8 − c − d − e) + c + d + e − 16] ≥ 0 ⇒ 3c − 2(8 −
2
2
2
d − e)c + (8 − d − e) − 2(16 − d − e ) ≤ 0
d − e)c + (8 − d − e) − 2(16 − d − e ) ≤ 0 . There are real values c satisfying this
2
2
2
inequality if and only if
2 2 2 2 2
Δ c = 4(8 − d − e) − 12[(8 − d − e) − 2(16 − d − e )] ≥ 0 ⇒ 4d − 2(8 − e)d + (8 −
2
2
2
2
2
Δ c =
2 4(8 − d − e) − 12[(8 − d − e) − 2(16 − d − e )] ≥ 0 ⇒ 4d − 2(8 − e)d + (8 −
2
e) − 3(16 − e ) ≤ 0
e) − 3(16 − e ) ≤ 0. There are real values d satisfying this inequality if and only if
2
2
2 2 2 2
Δ d = 4(8 − e) − 16[(8 − e) − 3(16 − e )] ≥ 0 ⇒ 5e − 16e ≤ 0 ⇒ e(5e − 16) ≤
2
2
2
2
Δ d = 4(8 − e) − 16[(8 − e) − 3(16 − e )] ≥ 0 ⇒ 5e − 16e ≤ 0 ⇒ e(5e − 16) ≤
0 ⇒ 0 ≤ e ≤ 16/5
0 ⇒ 0 ≤ e ≤ 16/5 . Hence, the maximum value of e is 16/5.
1.118 Let a positive integer d not equal to 2,5,13, show we can find two
elements a, b from the set {2, 5, 13,d} such that ab − 1 is not a perfect square.
Proof: 2 × 5 − 1 =3 , 2 × 13 − 1 =5 , 5 × 13 − 1= 8 , thus we need to show at least one
2
2
2
of 2d − 1, 5d − 1, 13d − 1 is not a perfect square. We prove this by contradiction. Suppose
these three numbers are perfect squares, that is, 2d − 1= x (i), 5d − 1= y (ii), 13d − 1= z
2
2
2
(iii), where x,y,z are positive integers. (i) implies 2d − 1 ≡ 1 (mod 2), then x is an odd
number, thus 2d − 1 ≡ 1 (mod 4), thus d ≡ 1 (mod 2), that is, d is an odd number.
Similarly (ii)(iii) imply y, z are even numbers. Let y =2y 1 ,z =2z 1 , where y 1 ,z 1 are positive
integers. Substitute them into (ii)(iii) and subtract the two resulting equalities:
z − y =2d ⇒ (z 1 − y 1 )(z 1 + y 1 )= 2d (iv). The right hand side of (iv) is divisible by 2,
2
2
1
1
but the left hand side (z 1 − y 1 ) + (z 1 + y 1 )= 2z 1 is an even number, then z 1 − y 1 and
z 1 + y 1 are multiples of 2, thus the left hand side of (iv) is divisible by 2 . However, d is
2
an odd number, thus the right hand side of (iv) is not divisible by 2 , a contradiction to
2
the assumption.
1.119 Let x,y,z be nonnegative real numbers, and x + y + z =1, find the
maximum value of xy + yz + zx − 2xyz .
Solution:
(1 − 2x)(1 − 2y)(1 − 2z)=(1 − 2y − 2x + 4xy)(1 − 2z)=1 − 2y − 2x +4xy − 2z +
(1 − 2x)(1 − 2y)(1 − 2z)=(1 − 2y − 2x + 4xy)(1 − 2z)=1 − 2y − 2x +4xy − 2z +
4yz +4zx − 8xyz =1 − 2(x + y + z) + 4(xy + yz + zx − 2xyz)
4yz +4zx − 8xyz =1 − 2(x + y + z) + 4(xy + yz + zx − 2xyz)
, thus xy + yz + zx − 2xyz
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