State whether the following statements are true or false. Justify your answers: (i) Every irrational number is a real number. (ii) Every point on the number line is of the form sqrt(m), where m is a natural number. (iii) Every real number is an irrational number.

So, $OA = 2$ units.

Triangle OAB is a right-angled triangle.

4. Classroom activity (Constructing the 'square root spiral'): Take a large sheet of paper and construct the 'square root spiral' in the following fashion. Start with a point O and draw a line segment OP$_1$ of unit length. Draw a line segment P$_1$P$_2$ perpendicular to OP$_1$ of unit length (see Fig. 1.9). Now draw a line segment P$_2$P$_3$ perpendicular to OP$_2$. Then draw a line segment P$_3$P$_4$ perpendicular to OP$_3$. Continuing in this manner, you can get the line segment P$_{n-1}$P$_n$ by drawing a line segment of unit length perpendicular to OP$_{n-1}$. In this manner, you will have created the points P$_2$, P$_3$,..., P$_n$,..., and joined them to create a beautiful spiral depicting $\sqrt{2}$, $\sqrt{3}$, $\sqrt{4}$, ...

To construct the square root spiral:

Draw a line segment OP$_1$ of 1 unit length.

Join O to P$_2$. By the Pythagorean theorem, $OP_2 = \sqrt{OP_1^2 + P_1P_2^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$ units.

Join O to P$_3$. By the Pythagorean theorem, $OP_3 = \sqrt{OP_2^2 + P_2P_3^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2+1} = \sqrt{3}$ units.

Join O to P$_4$. By the Pythagorean theorem, $OP_4 = \sqrt{OP_3^2 + P_3P_4^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3+1} = \sqrt{4} = 2$ units.

For each subsequent step, draw a line segment P$_{n-1}$P$_n$ of 1 unit length, perpendicular to the previous hypotenuse OP$_{n-1}$. Then join O to P$_n$. The length of the new hypotenuse $OP_n$ will be $\sqrt{n}$ units.
This sequence of hypotenuses (OP$_1$, OP$_2$, OP$_3$, ...) forms a spiral, with lengths $\sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}, \dots$.