Locate $\sqrt 3$ on the number line.

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Construct $BD$ of unit length perpendicular to $OB$ (as in Fig.). Then using the Pythagoras theorem, we see that $OD =\sqrt{(\sqrt{2})^{2}+1^{2}}=\sqrt{3}$. Using a compass, with centre $O$ and radius $OD ,$ draw an arc which intersects the number line at the point $Q$. Then $Q$ corresponds to $\sqrt{3}$.

In the same way, you can locate $\sqrt n$ for any positive integer $n$, after $\sqrt {n - 1}$ has been located.

1098-s8

Similar Questions

Simplify each of the following expressions :

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

$(ii)$ $(3+\sqrt{3})(3-\sqrt{3})$

$(iii)$ $(\sqrt{5}+\sqrt{2})^{2}$

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

Are the following statements true or false ? Give reasons for your answers.

$(i)$ Every whole number is a natural number.

$(ii)$ Every integer is a rational number.

$(iii)$ Every rational number is an integer.

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