Page 7 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I reAl numBers

1 REAL NUMBERS

1 1 1
1.1 Compute 2 + 3 + ··· + 2001 .
1
1
1
1
1+ 1 (1 + )(1 + ) (1 + )(1 + ) ···(1 + 1 )
2 2 3 2 3 2001
Solution: This quantity is equal to
1 1 1 1
2 1 + 3 1 1 +···+ 2001 1 1 + 1
1
1
1
+···+
(1 + )(1 + ) ··· (1 +
1+ 12 2 1 + 1 1 3 3 1 3 1 +···+ 1 2 1 1 3 2001 2001 1 + 1 2 1 1 3 1 1 2001 1
(1 + )(1 + )
)
+
) (1 + )(1 + ) ··· (1 +
2001
+
2
1
2
)
) (1 + )(1 + ) ··· (1 +
1
1+ 1 (1 + )(1 + ) (1 + )(1 + ) ··· (1 + 2001 ) (1 + )(1 + ) ··· (1 + 2001 )
1
1
1
1
1
1
1
2 )(1 + )
1 (1 + )(1 + )
(1 +
1 ··· (1 +
1+
2
3
3
3
2
2
− 2 1 1 2 3 =1 − 2 3 1 1 2001 2 3 2001
1
1
(1 + )(1 + ) ···(1 + 1 ) =1 − 1 1 1 )
(1 + )(1 + ) ···(1 +
−
2
3
3
2
1
2001 1
1
1
2001 1
1
− (1 + )(1 + ) ···(1 + 2001 ) =1 − (1 + )(1 + ) ···(1 + 2001 )
)
)
1
1
1
1
1
1
1 (1 + )(1 + ) ···(1 +
1000
2 )(1 +
1 3 ) ···(1 +
(1 +
=1 − 3 2 4 3 1 2001 2002 2001 1 = 2 2 1000 3 3 2001
=1 −
1000
1
1
1001
=
=1 − × × ··· × =1 − 1001 .
=1 2 − 3 3 4 2000 2001 2002 =1 − 1001 = 1001
2001
3
× × ··· × 2001 2002
4
3
2 × × ··· × 2000 2001 1001 1001
2 3 2000 2001
1
1.2 If p, q are prime numbers and satisfy 5p +3q = 19. Compute the value of √ √ .
q − p
Solution: The equation 5p +3q = 19 implies that one of p, q is even. Since p, q are prime
numbers and the only even prime number is 2, we have two possibilities: if q =2, then
p = 13/5, not a prime number, so this case is impossible; if p =2, then q =3, thus
1 1 √ √
√ √ = √ √ = 3+ 2.
q − p 3 − 2
1.3 Solve |x| + x + y = 10 and x + |y|− y = 12 for x, y .
Solution: It is easy to figure out that x ≤ 0 or y ≥ 0 are impossible. Thus x> 0 and y< 0
which lead to x = 32/5,y = −14/5.
√
1001
3
3
(4a − 1004a = 1+ 1001 , compute the value of (4a − 1004a − 1001) 1001 .
1.4 Given − 1001)
2
√ √
1001
3
3
2
Solution: a − 1001) ⇒ 2a − 1=
(4a − 1004 = 1+ 1001 √ 1001 ⇒ 4a − 4a − 1000 = 0 ⇒ (4a − 1004a − 1001) 1001 =
√
1001
3
2
1+ 1001
= 2 3 ⇒ 2 2a − 1= 1001 ⇒ 4a − 4a − 1000 = 0 ⇒ (4a − 1004a − 1001) =
1001
2
1001
2
= −1
[(4a − 4a − 1000a) + (4a − 4a − 1000) − 1]
= (0 − 0 − 1)
3
2
2
[(4a − 4a − 1000a) + (4a − 4a − 1000) − 1] 1001 = (0 − 0 − 1) 1001 = −1.
1 1
3
2
2
1.5 If a, b, x are real numbers and (x + − a) + |x + − b| =0. Show b(b − 3) = a .
x 3 x
a,
3
1
a,
Proof: Since a, b, x are real numbers, the equation implies that b, x + 1 3 = a and b, x + = b .
x x
1
1
1
Hence, a = x + 1 =(x + )(x − 1+ 1 )= (x + )[(x + ) − 3] = b(b − 3).
2
2
2
3
x 3 x x 2 x x
√ √
3 2
+
1.6 If the real numbers a, b, c satisfy a =2b + 2 and a, b, ca, b, c c + 1 =0. Evaluate bc/a .
2 4
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