Waldronjw

Waldronjw

Answered

2022-07-01

Find the fraction.
Let us consider a fraction whose denominator is smaller than the square of the numerator by unity. If we add 2 to the numerator and the denominator, the fraction will exceed 1/3; now if we subtract 3 from the numerator and the denominator, the fraction remains positive but smaller than 1/10. Find the fraction.
My work:
Let x / y be the fraction. Thus y + 1 = x 2 . Also, ( x + 2 ) / ( y + 2 ) > 1 / 3 and 0 < ( x 3 ) / ( y 3 ) < 1 / 10
Substituting the first equality in the second inequality, 0 < ( x 3 ) / ( x 2 ) ( x + 2 ) < 1 / 10
How do I proceed after this?

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Answer & Explanation

escampetaq5

escampetaq5

Expert

2022-07-02Added 12 answers

HINT From x + 2 y + 2 > 1 3 you get x + 2 x 2 + 1 > 1 3 and from here x 2 3 x 5 < 0 , x N

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antennense

antennense

Expert

2022-07-03Added 7 answers

You have an answer already. To add some detail to it, for x Z
x + 2 x 2 + 1 > 1 3 x 2 3 x 5 < 0 ( x 3 2 ) 2 < 29 4 x 4
Similarly, 0 < x 3 x 2 4 < 1 10
Case x 2 > 4: gives from 0 < x 3 x 2 4 x > 3
Case x 2 < 4: gives from
( × ) x 3 x 2 4 < 1 10 ( x 5 ) 2 < 9
So we are left with x = 4 as the only possibility.

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