Visualise $4. \overline{26}$ . on the number line, up to $4$ decimal places.

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We can magnify an interval endlessly using successive magnification.

To visualize $4 . \overline{26}$ or $4.2626 \ldots$ on the number line up to $4$ decimal places, we use the following steps.

$I$. The number $4.2626 \ldots$ lies between $4$ and $5 .$ Divide the interval $[4,\,5]$ in $10$ smaller parts :

$II.$ Obviously, the number $4.2626 \ldots$ lies between $4.2$ and $4.3 .$ We magnify the interval $[4.2,\,4.3]$.

$III.$ Next, we magnify the interval $[4.26, \,4.27]$ :

$IV.$ Finally magnify the interval $[4.262,\,4.263]$ :

In Fig. $(iv)$, we can easily observe the number $4.2626 \ldots$ or $4 . \overline{26}$.

1098-s30

Similar Questions

Simplify the following expressions :

$(i)$ $(5+\sqrt{7})(2+\sqrt{5})$

$(ii)$ $(5+\sqrt{5})(5-\sqrt{5})$

$(iii)$ $(\sqrt{3}+\sqrt{7})^{2}$

$(iv)$ $(\sqrt{11}-\sqrt{7})(\sqrt{11}+\sqrt{7})$

Simplify

$(i)$ $2^{\frac{2}{3}} \cdot 2^{\frac{1}{3}}$

$(ii)$ $\left(3^{\frac{1}{5}}\right)^{4}$

$(iii)$ $\frac{7^{\frac{1}{5}}}{7^{\frac{1}{3}}}$

$(iv)$ $13^{\frac{1}{5}} \cdot 17^{\frac{1}{5}}$

Rationalise the denominators of the following :

$(i)$ $\frac{1}{\sqrt{7}}$

$(ii)$ $\frac{1}{\sqrt{7}-\sqrt{6}}$

$(iii)$ $\frac{1}{\sqrt{5}+\sqrt{2}}$

$(iv)$ $\frac{1}{\sqrt{7}-2}$

Simplify each of the following expressions :

$(i)$ $(3+\sqrt{3})(2+\sqrt{2})$

$(ii)$ $(3+\sqrt{3})(3-\sqrt{3})$

$(iii)$ $(\sqrt{5}+\sqrt{2})^{2}$

$(iv)$ $(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})$

Find : 

$(i)$ $64^{\frac{1}{2}}$

$(ii)$ $32^{\frac{1}{5}}$

$(iii) $ $125^{\frac{1}{3}}$