Question

find 4 rational no between -1 and 1/3 answer with steps​

Answer 1

ANSWER

method used

• first make denominator same
• divide numerator and denominator of furst numbers by 9 and second by 3
• we get -9/9 and 3 /9
• so numbers are
• -8/9, -7/9, and so on

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Answer 2

let a = -1 ; b = 1/3

we know that

$rational \: number = \frac{a + b}{2}$

substitute all the values in that , we get

$\frac{( - 1) + \frac{1}{3} }{2}$

$\frac{ \frac{ - 3 + 1}{3} }{2}$

$\frac{ - 2}{3} \times \frac{1}{2 } = \frac{ - 2}{6} = \frac{ - 1}{3}$

So,

$- 1 < \frac{ - 1}{3} < \frac{1}{3}$

again a = -1/3 ; b = 1/3

So, again we get

$\frac{ \frac{ - 1}{3} + \frac{1}{3} }{2}$

$\frac{0}{6} \times \frac{1}{2} = \frac{0}{12} = 0$

So,

$- 1 < \frac{ - 1}{3} < 0 < \frac{1}{3}$

again a = 0 ; b = 1/3

So, again we get

$\frac{0 + \frac{1}{3} }{2}$

$\frac{1}{3 } \times \frac{1}{2} = \frac{1}{6}$

So,

$- 1 < \frac{ - 1}{3} < 0 < \frac{1}{6} < \frac{1}{3}$

again a = 1/6 ; b =1/3

So, again we get,

$\frac{ \frac{1}{6} + \frac{1}{3} }{2}$

$\frac{3}{6} \times \frac{1}{2 } = \frac{1}{4}$

So,

$- 1 < \frac{ - 1}{3} < 0 < \frac{1}{6} < \frac{1}{4} < \frac{1}{3}$

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