Page 19 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

1 1 1 1
1.48 Given + + = =1, show x =1 or y =1 or z =1.
x y z x + y + z
1 1 1
Proof: + + =1 ⇒ xy + xz + yz = xyz (i);
x y z

1 =1 ⇒ x + y + z =1 ⇒ x =1 − y − z (ii).
x+y+z
Substitute (ii) into (i): (1 − y − z)y + (1 − y − z)z + yz = (1 − y − z)yz ⇒ (z − 1)(y + z)(y − 1) = 0

(iii). (ii) is equivalent to y + z =1 − x and substitute it into (iii): (z − 1)(1 − x)(y − 1) = 0,

therefore x =1 or y =1 or z =1.

2
1.49 Show 1+2+2 + ··· +2 5n−1 is divisible by 31.

Proof: 1 + 2 + 2 + ··· +2 5n−1 = 1 − 2 5n =2 5n − 1 = 32 − 1 = (31 + 1) − 1=
2
n
n
1 − 2
0
1
0
1
n
n
C 31 + C 31 n−1 + ··· + C n n−1 31 + C − 1 = 31(C 31 n−1 + C 31 n−2 + ··· + C n n−1 ) which is
n
n
n
n
n
obviously divisible by 31.
a b
1.50 The real numbers a, b, c satisfy = , and x, y are mean values of a, b and b, c
b c
a c
respectively. Show + =2 .
x y

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