Question

If a and b are integers and a<b, b not congruent to 1 then compare a-1/b-1 , a+b/b+1.

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Answer 1

$\bf \frac{a - 1}{b - 1} \: - \frac{a + 1}{b + 1} \:$

$\bf = \frac{(a - 1)(b + 1) - (a + 1)b - 1)}{(b - 1)(b \times + 1)}$

$\bf= \frac{(ab - b + a - 1) - (ab + b - a - 1)} { {b}^{2} - 1}$

$\bf= \frac{ab - b + a - 1 - ab - b + a + 1}{ {b}^{2} - 1}$

$\bf = \frac{2a - 2b} { { {b}^{2} - 1 } } \:$

$\bf = \frac{2(a - b)}{ {b}^{2} - 1}...(1)$

Now a < b

$\therefore$a - b <0

also b² - 1 > 0 b is not congruent 1

$\bf\frac{2(a - b)}{ {b}^{2} - 1} < 0...(2)$

$\bf\frac{a - 1}{ab - 1} - \frac{a + 1}{b + 1} < 0...from(1)(2)$

$\bf = \frac{a - 1}{b - 1} < \frac{a + 1}{b + 1}$

Answer 2

$\bf \frac{a - 1}{b - 1} \: - \frac{a + 1}{b + 1} \:$

$\bf = \frac{(a - 1)(b + 1) - (a + 1)b - 1)}{(b - 1)(b \times + 1)}$

$\bf= \frac{(ab - b + a - 1) - (ab + b - a - 1)} { {b}^{2} - 1}$

$\bf= \frac{ab - b + a - 1 - ab - b + a + 1}{ {b}^{2} - 1}$

$\bf = \frac{2a - 2b} { { {b}^{2} - 1 } } \:$

$\bf = \frac{2(a - b)}{ {b}^{2} - 1}...(1)$

Now a < b

$\therefore$a - b <0

also b² - 1 > 0 b is not congruent 1

$\bf\frac{2(a - b)}{ {b}^{2} - 1} < 0...(2)$

$\bf\frac{a - 1}{ab - 1} - \frac{a + 1}{b + 1} < 0...from(1)(2)$

$\bf = \frac{a - 1}{b - 1} < \frac{a + 1}{b + 1}$