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Question

Insert a rational number and an irrational number between the following:
$$\dfrac {1}{3}$$ and $$\dfrac {1}{2}$$.


Solution

$$\dfrac {1}{3} = \dfrac {1}{3} \times \dfrac {4}{4} = \dfrac {4}{12}$$ and $$\dfrac {1}{2} = \dfrac {1}{2}\times \dfrac {6}{6} = \dfrac {6}{12}$$
Now, $$\dfrac {5}{12}$$ is a rational number between $$\dfrac {4}{12}$$ and $$\dfrac {6}{12}$$. So, $$\dfrac {5}{12}$$ is a rational number lying between $$\dfrac {1}{3}$$ and $$\dfrac {1}{2}$$.
Again, $$\dfrac {1}{2} = 0.33333...$$ and $$\dfrac {1}{2} = 0.5$$.
Now, $$0.414114111...$$ is a non-terminating and non-recurring decimal.
Hence, $$0.414114111...$$ is an irrational number lying between $$\dfrac {1}{3}$$ and $$\dfrac {1}{2}$$.

Maths

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