Q1.
Try to extend this method for constructing line segments of lengths √3 and √5 using a ruler and a compass. Generalise this method to construct a line segment of any length of the form √n, where n is a positive integer.
Answer
Length √3. Start from the √2 already built at P.
erect a perpendicular at P and mark C with PC = 1
OC² = OP² + PC² = (√2)² + 1² = 2 + 1 = 3
so OC = √3
swing OC onto the number line with the compass
OC² = OP² + PC² = (√2)² + 1² = 2 + 1 = 3
so OC = √3
swing OC onto the number line with the compass
Length √5. Repeat twice more, or take a shortcut with legs 1 and 2:
from √3, add a unit perpendicular: 3 + 1 = 4, giving √4 = 2
from 2, add a unit perpendicular: 4 + 1 = 5, giving √5
shortcut: legs 2 and 1 give 2² + 1² = 5 directly
from 2, add a unit perpendicular: 4 + 1 = 5, giving √5
shortcut: legs 2 and 1 give 2² + 1² = 5 directly
The general rule. Having constructed √n, erect a perpendicular of length 1 at its far end. The new hypotenuse satisfies
(new)² = (√n)² + 1² = n + 1
new length = √(n + 1)
new length = √(n + 1)
So from OA = 1 = √1 the construction climbs one step at a time: √1 → √2 → √3 → √4 → … → √n for any positive integer n, in exactly n − 1 steps. Repeating this endlessly draws the square root spiral of Fig. 3.14.
Why it happens: each step adds exactly 1 to the square of the length, never to the length itself. That is why the outer edge of the spiral grows more and more slowly — the gap between √n and √(n + 1) shrinks as n grows, since √(n+1) − √n = 1/(√(n+1) + √n).
Check it yourself: the construction never needs a protractor. A ruler and compass suffice, because a perpendicular at a point can be drawn with a compass alone.