Exercise-1.2, Class 9th, Maths, Chapter 1, NCERT

1. 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 $m$, where $m$ is a natural number.
(iii) Every real number is an irrational number.

Answers & Justifications

(i) True.
Definition: real numbers $=$ rational numbers $\cup$ irrational numbers. Every irrational number belongs to the set of real numbers by definition.
So every irrational number is a real number.

(ii) False.
A point on the number line corresponds to a real number. Natural numbers are $1,2,3,\dots$— only those points whose coordinates are natural numbers are of the form $m$. But there are many points whose coordinates are integers, rational numbers (e.g. $\frac{1}{2}$), or irrationals (e.g. $\sqrt{2}$, $\pi$).
Hence not every point on the number line is of the form $m$ (natural number).

(iii) False.
Real numbers include both rational and irrational numbers. Rational numbers (for example $\frac{1}{2},3,0$) are real but not irrational. So not every real number is irrational.

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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.

Answer & Explanation

No — the square roots of all positive integers are not irrational.
If the integer is a perfect square, its square root is an integer (hence rational). For example:

• $\sqrt{9}=3$ (rational),

• $\sqrt{16}=4$ (rational),

• $\sqrt{36}=6$ (rational).

If the integer is not a perfect square (e.g. 2, 3, 5, 6, 7, 10, …), then $\sqrt{n}$ is irrational. So the property depends on whether the integer is a perfect square.

Example requested: $\sqrt{9}=3$ is rational.

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

Construction (step-by-step, easy to copy/paste):

One convenient geometric method uses Pythagoras, because $1^2+2^2=5$. So the hypotenuse of a right triangle with legs $1$ and $2$ has length $\sqrt{5}$. Steps:

1. Draw a number line and mark the origin $O$ (0) and point $A$ at distance $2$ units to the right of $O$. (So $OA=2$)

2. At point $A$, draw a perpendicular segment $AB$ of length $1$ unit (uptowards). So $AB=1$.

3. Join $O$ to $B$. By the Pythagorean theorem, the length $OB$ equals $\sqrt{O{A}^2+A{B}^2}=\sqrt{2^2+1^2}=\sqrt{5}$.

4. With center $O$ and radius $OB$ use compass to draw an arc that intersects the number line (to the right of $O$). The intersection point $P$ on the number line corresponds to the positive number $\sqrt{5}$.

5. Label that point $P$. Thus $OP=\sqrt{5}$

(You can reverse the orientation: some texts put the leg of length 2 along the number line and erect the perpendicular of length 1 — the same idea. The key is a right triangle with legs 1 and 2.)

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4. Classroom activity (Constructing the ‘square root spiral’):
(Instructions / brief write-up describing how to do it — ready for classroom display.)

Goal: construct a visual spiral in which the line segments from the origin represent $\sqrt{2},\sqrt{3},\sqrt{4},\dots$

Materials: large sheet of paper, ruler, compass, pencil.

Procedure:

1. Mark a point $O$ (the origin). From $O$ draw a horizontal segment $O{P}_1$ of unit length. (So $O{P}_1=1$)

2. At point $P_1$ draw a segment $P_1{P}_2$ of unit length perpendicular to $O{P}_1$. Now $O{P}_2$ is the hypotenuse of a right triangle with legs 1 and 1, so $O{P}_2=\sqrt{2}$.

3. At $P_2$, draw $P_2{P}_3$ of unit length perpendicular to $O{P}_2$. The new segment $O{P}_3$ will have length $\sqrt{3}$. (Reason: using successive right triangles and Pythagoras.)

4. Continue this process: at each step $P_{n-1}$ draw a unit segment $P_{n-1}{P}_n$ perpendicular to $O{P}_{n-1}$. Then $O{P}_n=\sqrt{n}$. (Each new triangle adds one more unit square in the Pythagorean sum.)

5. Join the points $P_1,P_2,P_3,\dots$ smoothly (or simply mark them). The locus of these tips makes a spiral-like curve commonly called the “square-root spiral” (or the “odious spiral” / “Theodorus spiral” in classical references).

Discussion / learning points:

• You physically see $\sqrt{2},\sqrt{3},\sqrt{4},\dots$ laid out on the paper.

• $\sqrt{n}$ increases but not linearly; the spiral visually demonstrates growth of square roots.

• Useful classroom extension: measure $O{P}_n$ and compare with a calculator value of $\sqrt{n}$. Also discuss which $\sqrt{n}$ are rational (only perfect squares) and which are irrational.