Locate $\sqrt 3$ on the number line.
Locate $\sqrt 3$ on the number line.
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.
Similar Questions
Locate $\sqrt 2$ on the number line.
Locate $\sqrt 2$ on the number line.
Write the following in decimal form and say what kind of decimal expansion each has :
$(i)$ $\frac{36}{100}$
$(ii)$ $\frac{1}{11}$
$(iii)$ $4 \frac{1}{8}$
$(iv)$ $\frac{3}{13}$
$(v)$ $\frac{2}{11}$
$(vi)$ $\frac{329}{400}$
Write the following in decimal form and say what kind of decimal expansion each has :
$(i)$ $\frac{36}{100}$
$(ii)$ $\frac{1}{11}$
$(iii)$ $4 \frac{1}{8}$
$(iv)$ $\frac{3}{13}$
$(v)$ $\frac{2}{11}$
$(vi)$ $\frac{329}{400}$
Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$.
Find three different irrational numbers between the rational numbers $\frac{5}{7}$ and $\frac{9}{11}$.
Rationalise the denominator of $\frac{5}{\sqrt{3}-\sqrt{5}}$.
Rationalise the denominator of $\frac{5}{\sqrt{3}-\sqrt{5}}$.
Show that $3.142678$ is a rational number. In other words, express $3.142678$ in the form $\frac {p }{q }$, where $p$ and $q$ are integers and $q \ne 0$.
Show that $3.142678$ is a rational number. In other words, express $3.142678$ in the form $\frac {p }{q }$, where $p$ and $q$ are integers and $q \ne 0$.