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.

1. State whether the following statements are true or false. Justify your answers.

• i) Every irrational number is a real number.
  • True. The set of real numbers includes both rational and irrational numbers.
• ii) Every point on the number line is of the form $\sqrt{m}$, where $m$ is a natural number.
  • False.
    • Justification 1: Negative numbers exist on the number line, but $\sqrt{m}$ (where $m$ is a natural number) is always non-negative. For example, $-2$ is on the number line but cannot be written as $\sqrt{m}$.
    • Justification 2: Numbers like $\frac{1}{2}$ are on the number line, but $\frac{1}{2}$ cannot be expressed as $\sqrt{m}$ for any natural number $m$.
• iii) Every real number is an irrational number.
  • False.
    • Justification: Real numbers include rational numbers as well. For example, $2$ is a real number but it is a rational number, not an irrational number.

2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

No, the square roots of all positive integers are not irrational. Example: Consider the positive integer $4$. Its square root is $\sqrt{4} = 2$. Since $2$ can be written as $\frac{2}{1}$, it is a rational number. Therefore, $\boxed{No, and an example is \sqrt{4} = 2}$.

3. Show how $\sqrt{5}$ can be represented on the number line.

Step 1: Draw a number line and mark a point O as $0$. Step 2: From O, mark a point A at $2$ units to the right (representing the number $2$). So, $OA = 2$ units. Step 3: At point A, draw a line segment AB perpendicular to OA, with length $1$ unit. Step 4: Join O to B. Triangle OAB is a right-angled triangle. Step 5: Using the Pythagorean theorem, the length of the hypotenuse OB is: $OB = \sqrt{OA^2 + AB^2}$ $OB = \sqrt{2^2 + 1^2}$ $OB = \sqrt{4 + 1}$ $OB = \sqrt{5}$ Step 6: With O as the center and OB as the radius, draw an arc that intersects the number line at a point P. Step 7: The point P on the number line represents $\sqrt{5}$.

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:

Step 1: Start with a point O on a large sheet of paper. Draw a line segment OP$_1$ of $1$ unit length. Step 2: At P$_1$, construct a line segment P$_1$P$_2$ of $1$ unit length, perpendicular to OP$_1$. 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. Step 3: At P$_2$, construct a line segment P$_2$P$_3$ of $1$ unit length, perpendicular to OP$_2$. 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. Step 4: At P$_3$, construct a line segment P$_3$P$_4$ of $1$ unit length, perpendicular to OP$_3$. 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. Step 5: Continue this process. 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 $\sqrt{1}, \sqrt{2}, \sqrt{3}, \sqrt{4}, \dots$.