Representation of √2 on the number line
Step 1: Draw a line segment of 1cm anywhere on a number line.
Step 2: Mark the line segment A and B

Step 3: Draw a line segment of 1cm perpendicular to AB, and mark the point C.

Step 4: Draw a line connecting A and C.

Step 5: AC is √(12 + 12) = √2

Representation of √3 on the number line
Step 6: Draw a line segment of 1cm perpendicular to AC. Mark the point D

Step 7: Draw line connecting A and D.
Step 8: AD is √ {(√2)2 + (1)2}= √3

Step 9: Take a compass and take the radius to the length of AD.
Step 10: Extend the arc and mark where it touches the number line as E
Step 11: AE is √3

Similarly, locate √n for any positive integer n, after √n-1 has been located. Continue doing this for √4, √5, √6, etc. Square root spiral is obtained.
Square root spiral
Representation of √2 on the number line.
Method 1 is tedious and time-consuming. It is easy to use for smaller numbers. Therefore there is another easy way to represent irrational numbers on the number line which is less time-consuming and can also be used for larger numbers.
Step 1: Draw a line segment of n cm. In this case, it's 2 cm, and mark them A and B.
Step 2: Extend B to C to form a line segment of 1cm.

Step 3: Take a compass, and mark an arch, placing one end of the compass in A. With the same radius, draw another arch placing one end of the compass in B.
Step 4: Mark the point of intersection between the arches as D.

Step 5: Draw a perpendicular line joining D and AC. Mark this point as E.

Step 6: Take AE as radius, and draw a semicircle.
Step 7: Draw a perpendicular line joining B and the semicircle, and mark the point F.

Step 8: BF is √2
Step 9: Take radius AF on the compass on extending it till the number line, and mark the point G.
Step 10: BG is √2 = 1.41421...

Using the same steps you can represent √n on the number line, by taking AB as n cm.
Representation of √3 on the number line

AB = n units, BC = 1 unit.
In this case, n = 3.
BF is = √n
We can prove this by using the Pythagoras Theorem. Note that △EBF is a right-angled triangle and the radius of the circle is (n + 1)/2 units. Therefore, EC = EF = EA = (n + 1)/2 units.
OB = n - (n + 1/2) = (n - 1)/2.
Applying the Pythagoras theorem,
(BF)2 - (EF)2 - (EB)2 = {(n + 1)/2}2 = (4n)/4 = n.
This proves that BF = √n.