Page 118 - Elementary Algebra Exercise Book I

ELEMENTARY ALGEBRA EXERCISE BOOK I inequAlities

3.86 Let a, b, c are integers and at least one of them is nonzero, and their
√ √
6
absolute values are not greater than 10 , show |a + b 2+ c 3| > 10 −21 .
Proof: When b =0,c =0, the conclusion is obviously valid. When one of b, c is nonzero, we consider
√ √ √ √ √ √ √ √
the following four numbers: t 1 = a + b 2+ c 3,t 2 = a + b 2 − c 3,t 3 = a − b 2+ c 3,t 4 = a − b 2 − c 3

√ √ √ √ √ √ √ √ √ √ √
√ 2 2 √ 2 2 2 √ 2 2 2
2
2
2
2
2
2
t 1 = a + b 2+ c 3,t 2 = a + b 2 − c 3,t 3 = a − b 2+ c 3,t 4 = a − b 2 − c 3. They are all irrational numbers, and t = t 1 t 2 t 3 t 4 = [(a + b 2) − 3c ][(a − b 2) − 3c ]= (a +2 2ab +2b − 3c )(a −
2
2√
√
√
√
√
√ t = t 1 t 2 t 3 t 4 = [(a + b 2) − 3c ][(a − b 2) − 3c ]= (a +2 2ab +2b − 3c )(a −
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
√
√ √ √ 2 2ab +2b − 3c ) = [(a +2b − 3c ) +2 2ab][(a +2b − 3c ) − 2 2ab] = [(a +
√
2 = [(a + b 2) − 3c ][(a − b 2) − 3c ]= (a +2 2ab +2b − 3c )(a −
√ t = t 1 t 2 t 3 t 4
2
2
2
2
2
2
2
2 √
2 √
√ 2 2 √ 2 2 2 √ 2 2 2 √ 2 2ab +2b − 3c ) = [(a +2b − 3c ) +2 2ab][(a +2b − 3c ) − 2 2ab] = [(a +

t = t 1 t 2 t 3 t 4 = [(a + b 2) − 3c ][(a − b 2) − 3c ]= (a +2 2ab +2b − 3c )(a −
2
2
2
2
2
2
2
2
2
√t = t 1 t 2 t 3 t 4 = [(a + b 2) − 3c ][(a − b 2) − 3c ]= (a +2 2ab +2b − 3c )(a − 2 2ab 2 2 2 2 2 2 2 2 2 2 2 2
2 +2b − 3c ) = [(a +2b − 3c ) +2 2ab][(a +2b − 3c ) − 2 2ab] = [(a +
√
√
2b − 3c ) − 8a b ] ∈ Z
2
2
2
2 2
2
2
2
2
2
2 2
2
2
2 2ab +2b − 3c ) = [(a +2b − 3c ) +2 2ab][(a +2b − 3c ) − 2 2ab] = [(a +
√
√
√
2 2ab +2b − 3c ) = [(a +2b − 3c ) +2 2ab][(a +2b − 3c ) − 2 2ab] = [(a + 2b − 3c ) − 8a b ] ∈ Z . Thus |t|≥ 1,
2b − 3c ) − 8a b ] ∈ Z
2 2
2
2
2 2
2
2
2
2
2
2
2
2
2 2
2 2
2
2b − 3c ) − 8a b ] ∈ Z √ √
2 2
2
2 2
1
6
2b − 3c ) − 8a b ] ∈ Z which implies that |t 1 |≥ |t 2 |·|t 3|·|t 4| . In addition, since 1+ 2+ 3 < 10 and |a|, |b|, |c|≤ 10 ,
√ √
7
6
we have |t i |≤ (1 + 2+ 3) · 10 < 10 , thus |t 1 | > 1 7 = 10 −21 .
7
7
10 ·10 ·10
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