Page 42 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

3
2
3
3
2
3
a+b+c+d =0 ⇒ a+b = −(c+d) ⇒ 0 =(a+b) +(c+d) = a +b +3a b+3ab +
3 3 2 2 3 3 3 3 2 2 2 2 3 3 3 3
c +d +3c d+3cd ⇒ a +b +c +d = −3(a b+ab +c d+cd ) ⇒ a +b +c +d −
2
2
2
2
2
3(abc+bcd+cda+dab) = −3(a b+ab +c d+cd )−3(abc+bcd+cda+dab) = −3(a b+
2
2
2
2
2
2
2
ab +c d+cd +abc+bcd+cda+dab) = −3[(a b+ab +abc+abd)+(acd+bcd+c d+cd )] =
Proof: −3[ab(a+b+c+d)+cd(a+b+c+d)] = 0 ⇒ a +b +c +d = 3(abc+bcd+cda+dab) .
3
3
3
3
1.111 Consider a 2n × 2n square grid chessboard, each grid can only have one
piece, and there are 3n grids having pieces, show we can always find n rows and n columns
such that these 3n pieces are within these n rows or these n columns.
Proof: Denote the number of pieces in each row or column as p 1 ,p 2 , ··· ,p n ,p n+1 , ··· ,p 2n with
the order p 1 ≥ p 2 ≥ ··· ≥ p n ≥ p n+1 ≥ ··· ≥ p 2n . The given condition implies that
p 1 + p 2 + ··· + p n + p n+1 + ··· + p 2n =3n (i). If p 1 + p 2 + ··· + p n ≤ 2n − 1 (ii), then at
least one of p 1 ,p 2 , ··· ,p n is not greater than 1. (i)-(ii): p n+1 + ··· + p 2n ≥ n +1, then at least one
of p n+1 , ··· ,p 2n is greater than 1, a contradiction. Hence, we have p 1 + p 2 + ··· + p n ≥ 2n. Hence,
we choose not less than 2n pieces from the n rows and then choose the remaining pieces from the n
columns to include all 3n pieces.
1.112 Find a positive number such that its fractional part, its integer part, and
itself are geometric.

Solution: Denote the number as x> 0, its integer part [x], and its fractional part

x − [x]. The given condition implies that x(x − [x]) =[x] ⇒ x − [x]x − [x] =0, where
2
2
2
√
=
[x] > 0, 0 <x − [x] < 1. The solution is x> 0 1+ 5 [x]. Since 0 <x − [x] < 1, then
2
√ √ √
1+ 5 1+ 5 1+ 5 .
0 < [x] < 1 ⇒ 0 < [x] < < 2 ⇒ [x]=1,x = [x] > 0, 0 <x − [x] < 1
2 2 2
1.113 Consider a sequence a 1 ,a 2 ,a 3 , ··· ,a n satisfying a 1 + a 2 + ··· + a n = n
3
1 1 1
for any positive integer n , evaluate + + ··· + .
a 2 − 1 a 3 − 1 a 100 − 1
Solution: When n ≥ 2, we have a 1 + a 2 + ··· + a n = n (i),
3

a 1 + a 2 + ··· + a n−1 =(n − 1) (ii). (i)-(ii): a n = n − (n − 1) =3n − 3n +1. Thus
3
3
3
2

1 1 1 1 1 1 , n =1, 2, 3, ··· , 100.
= = = −
2
a n − 1 3n − 3n 3n(n − 1) 3 n − 1 n
1 1
1
1 1
1
Hence, 1 1 + 1 + ··· + 1 1 = 1 1 (1 − )+ ( − )+ ··· + ( − 1 )=
1
1
1
1 1
1 1
1
+ a 2 − 1 + ··· + a 100 − 1 − )+ ( − )+ ··· + ( − 3 99 100
)=
=
(1
2
3 2
3
3
a 3 − 1
a 2 − 1 a 3 − 1 1 99 33 3 2 3 2 3 3 99 100
a 100 − 1
1
1
99
1 1 (1 − 1 ) = × 33 . =
(1 − 3 ) = 100 3 = 100 100
×
3 100 3 100 100
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