Page 8 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

√
√ √ √ 3 1
2
2
Solution: Substitute a =2b + 2 into ab+ 3 2 1 =0, then 2b + 2b +( c + )= 0.
c +
2 4 √ 2 4
√ 3 1 √
2
2
2
2
Since b is a real number, Δ b =( 2) − 4 × 2 × ( c + ) = −4 3c ≥ 0 , that is c ≤ 0.
2 4
On the other hand, c is a real number, thus c ≥ 0. As a conclusion, c =0, therefore bc/a =0.
2
1 1 1 1
1.7 Compute + + + ··· + +1 +2+4+ ··· + 1024.
1024 512 256 2
Solution: Since 1 =1 − 1 − 1 − 1 − ··· − 1 , the original sum is equal to 1 − 1 − 1 − 1 − ··· − 1 + 1 + 1 + ··· + 1 + 1+ 2+ 4+ ··· + 1024 =

1024 2 4 8 1024 1 1 1 1 1 1 2 1 1 4 8 1 1 1 1 1024 512 1 1 256 2
1 1 − − − − − − − ··· 1 − − + + + + + ··· + + + 1+ 2+ 4+ ··· + 1024 = =
+ 1+ 2+ 4+ ··· + 1024
+ ···
− ···
2 2 4 4 8 8 1 − 1024 512 256 2 2
1024
512
256
+1+2+4 + ··· + 1024
1 1 1 1 1 1 1 1 1 1024
+1+2+4 + ··· + 1024 .
1
1 − − − − ··· − + + + ··· + + 1+ 2+ 4+ ··· + 1024 = 1 − − +1+2+4 + ··· + 1024
1024
2 4 8 1024 512 256 2 1024
1
1 − +1+2+4 + ··· + 1024
1024 Let S = 1 +2+4+ ··· + 1024 denoted as (i), then 2S = 2 +4+8+ ··· + 2048 denoted
as (ii). (ii)-(i) ⇒ S = 2048 − 1 = 2047. Hence, the original sum is
1 1023
1 − + 2047 = 2047 .
1024 1024
1.8 If the prime numbers x,y,z satisfy xyz = 5(x + y + z), find the values of x,y,z .
Solution: xyz = 5(x + y + z) implies that at least one of the three prime numbers is five.
Without loss of generality, let x =5, then the equation becomes yz = 5+ y + z , that is,
(y − 1)(z − 1) = 6 . Since 6 =2 × 3= 1 × 6 , there are two possibilities (without considering

the order of y and z ): (1) y =3,z =4; (2) y =2,z =7. The case (1) is inappropriate

since z =4 is not a prime number. Therefore, the three prime numbers are 2, 5, 7.

√
2 6 − 1
1.9 Simplify √ √ √ .
2+ 3+ 6
√ √ √
Solution: Let 2+ 3= a , and take square to obtain 2 6= a − 5 , thus
2

√ √ √
2
2 6 − 1 a − 5 − 1 (a + 6)(a − 6) √ √ √ √
√ √ √ = √ = √ = a − 6= 2+ 3 − 6.
2+ 3+ 6 a + 6 a + 6
√
b
b
1.10 If a> 1,b > 0 and a + a −b =2 2, evaluate a − a −b .

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