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\sqrt{2}

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1.Mark OA unit length line segment on number line at 0&1

2.construct AB of unit lentgh perpendicular to OA.

3.Join O and B then you can see by the pythagorus theorem OB=square root of 2

4

Case-1

5. construct BD of unit length perpendicular to OB.

\sqrt{3}

6. Join O and D then you can see by pythagoras therem OD=squre root 3

7.

Case-2

To  \; know  \;the  \;location  \;of \; \sqrt{2} \; on \; number \; line  \;check \; this  \;box  \;and  \;sequentually  \;check  \;the  \;boxes  \;that  \;appears.

To  \; know  \;the  \;location  \;of \; \sqrt{3} \; on \; number \; line  \;check \; this  \;box  \;and  \;sequentually  \;check  \;the  \;boxes  \;that  \;appears.

Using a compass with centre O and radius OB  draw an arc interesting
 the number line.intersectin point is the location of squre root of 2 on number line 

Using compass with O as centre and radius OD draw an arc that intersect 
number line the pointof intersection is the root three value

Author- Sucheta SS, GHS, Thyamagondlu, Bengaluru Rural
License- CC BY SA 4.0

O = Point(xAxis)
A = Point(xAxis)
f = OrthogonalLine(A, xAxis)
c = Circle(A, Distance[O, A])
B = Intersect(c, f, 2)
g = Segment(O, B)
C = Point(c)
D_1 = Point(c)
d = CircleArc(A, C, D_1)
e = Circle(O, Distance[O, B])
E = Point(e)
F = Point(e)
h = CircleArc(O, E, F)
P = Intersect(h, xAxis, 1)
i = OrthogonalLine(B, g)
k = Circle(B, Distance[B, D_1])
D_2 = Intersect(k, i, 2)
G = Point(k)
H = Point(k)
j = Segment(O, D_2)
q = Circle(O, Distance[O, D_2])
I = Point(q)
J = Point(q)
r = CircleArc(O, I, J)
K = Intersect(r, xAxis, 1)
l = Segment(O, A)
a = false
b = false
m = false
n = false
s = false
o = false
t = false
u = false
v = false
text1: "Pythagoras   \;   theorem  \;  
OB=\sqrt{OA^2+AB^2}
=\sqrt{" + (LaTeX[text2]) + "+" + (LaTeX[text3]) + "}=" + (LaTeX[text4]) + ""
text7: "Pythagoras \;  theorem \;  OD=\sqrt{OB^2+BD^2}= \sqrt{(\sqrt{2})^2+1^2}=" + (LaTeX[text6]) + ""
L = (0.7030482964257587, 1.8632195811522074)
M = (0.15502402447674715, 1.2877940956057456)
c_1 = CircleArc(B, L, M)
D_3 = Intersect(c_1, i, 1)
D = Intersect(c_1, i, 1)