Q3:

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

Initial Setup & Theoretical Foundation

To represent an irrational number of the form $\sqrt{n}$ on a real number line, we utilize the geometric interpretation of the Pythagorean Theorem. The theorem states that in a right-angled triangle, the square of the hypotenuse ($c$) is equal to the sum of the squares of the base ($a$) and the perpendicular ($b$).

Mathematically, this is expressed as:
$c^2 = a^2 + b^2 \implies c = \sqrt{a^2 + b^2}$

Our objective is to represent $\sqrt{5}$. We must find two integers, $a$ and $b$, such that the sum of their squares equals $5$.

Step 1: Determining the Base and Perpendicular

We decompose the radicand ($5$) into the sum of two perfect squares:
$5 = 4 + 1$
$5 = (2)^2 + (1)^2$

[By the principles of Euclidean Geometry], if we construct a right-angled triangle with a base of $2$ units and a perpendicular height of $1$ unit, the length of the hypotenuse will be exactly $\sqrt{5}$ units.

Step 2: Geometric Construction on the Number Line

The construction proceeds through the following precise steps:

• Define the Origin: Draw a horizontal number line and mark a point $O$ representing the zero ($0$) coordinate.
• Establish the Base: Mark a point $A$ on the number line at the coordinate $2$. The line segment $OA$ now represents the base of our triangle, with a magnitude of $2$ units ($OA = 2$).
• Construct the Perpendicular: At point $A$, construct a perpendicular line segment extending upwards. Mark a point $B$ on this perpendicular such that the length of $AB$ is exactly $1$ unit ($AB = 1$).
• Form the Hypotenuse: Draw a straight line segment connecting the origin $O$ to point $B$. This forms the right-angled triangle $\triangle OAB$.

Step 3: Mathematical Proof of the Construction

Applying the Pythagorean Theorem to the constructed right-angled triangle $\triangle OAB$ (where $\angle OAB = 90^\circ$):

$OB^2 = OA^2 + AB^2$
$OB^2 = (2)^2 + (1)^2$
$OB^2 = 4 + 1$
$OB^2 = 5$
$OB = \sqrt{5}$ units

[Per the geometric construction], the length of the segment $OB$ is precisely $\sqrt{5}$.

Step 4: Transferring the Magnitude to the Number Line

To represent this magnitude on the one-dimensional number line, we use a compass. Placing the compass point at the origin $O$ and setting the pencil point at $B$ (setting the radius to $OB = \sqrt{5}$), we draw an arc downwards until it intersects the positive side of the number line.

Let the point of intersection be $P$. Because $OP$ and $OB$ are radii of the same circular arc, their lengths are equal. Therefore, the distance from $O$ to $P$ is exactly $\sqrt{5}$ units.

High-Precision Geometric Visualization

Below is the exact geometric representation of the construction. The base $OA$ is $2$ units, the perpendicular $AB$ is $1$ unit, and the arc transfers the hypotenuse $OB$ to point $P$ on the number line.

Final Solution: By constructing a right-angled triangle with a base of $2$ units and a perpendicular of $1$ unit at the origin, the resulting hypotenuse measures exactly $\sqrt{5}$ units. Projecting this hypotenuse onto the number line using a circular arc accurately locates the point $P$, which represents the irrational number $\sqrt{5}$.