Let me place √5, √6, √7, -√6, -√8, -√11 on same Number Line.

Let the point O represent 0 on the number line.

We place a point A on the number line such that OA = 2 units.

Now, we draw AB perpendicular to OA at point A such that AB = 1unit.

By Pythagoras’ Theorem, we know that,

OB² = OA² + AB²

⇒ OB² = 4 + 1 = 5

⇒ OB = √5

Now, we draw BC perpendicular to OB at point B such that BC = 1unit.

Again, by Pythagoras’ Theorem, we know that,

OC² = OB² + BC²

⇒ OC² = 5 + 1 = 6

⇒ OC = √6

We draw CD perpendicular to OC at point C such that CD = 1unit.

Again, by Pythagoras’ Theorem, OD = √7

Again, we draw ED perpendicular to OD at point D such that ED = 1unit.

Again, by Pythagoras’ Theorem, OE = √8

If we extend ED till F such that EF = 2 units and applying Pythagoras’ theorem on ΔODF, we get OF = √11

Now, taking O as a centre and OB, OC, OD, OE and OF as the radius, the arcs are drawn which intersects the number line at G, H, I, K and L.

∴ OG = √5 units, OH = √6 units, OI = √7 units, OJ = -√6 units, OK = -√8 units and OL = -√11 units