Question

Explain how to locate the point $\sqrt{2}$on a number line. Can you locate $\sqrt{5}$by a similar method? How can the circle is shown in the figure help us to locate $\pi$on a number line? List some other irrational numbers that you can locate on a number line.

Short answer

We can represent $\pi ,\sqrt{10},\sqrt{17},\sqrt{8},\sqrt{13}$and many more irrational numbers on a number line.

Step-by-step solution

Step 2. Use the concept of the Pythagoras Theorem.

The Pythagoras theorem states that if a triangle is right-angled, then the square of thehypotenuse is equal to thesum of the squares of the other two sides.

Step 2. Locate 2 on number line.

To locate $\sqrt{2}$at the number line, let us consider the integer 2.

We can write 2 as sum of squares of two integers:

$\begin{array}{l}2=1^2+1^2\\ \Rightarrow {\left(\sqrt{2}\right)}^2=1^2+1^2\end{array}$

This equation follows Pythagoras Theorem with$\sqrt{2}$ as hypotenuse, 1 and 1 are the other two sides of the right-angle triangle.

Now we locate$\sqrt{2}$ on the number line:

At origin O, take 1 unit from the origin and represent the point as A.

At A, draw perpendicular of length 1 unit and mark it as AB.

Now, using OB as radius and O as the center, draw an arc on the right side of the triangle, intersecting the number line at P.

Point P represents$\sqrt{2}$

In$\Delta OAB$, using Pythagoras Theorem, we have

$\begin{array}{l}O{B}^2=O{A}^2+A{B}^2\\ \Rightarrow O{B}^2=1+1\\ \Rightarrow O{B}^2=2\\ \Rightarrow OB=\sqrt{2}\end{array}$

OP is also the radius of the circle, so $OP=\sqrt{2}$

Hence, P represents $\sqrt{2}$on a number line.

Step 3. Locate 5 on a number line.

To locate $\sqrt{5}$at number line, let us consider the integer 5.

We can write 5 as the sum of squares of two integers:

$\begin{array}{l}5=2^2+1^2\\ \Rightarrow {\left(\sqrt{5}\right)}^2=2^2+1^2\end{array}$

This equation follows Pythagoras Theorem with $\sqrt{5}$as hypotenuse, 2 and 1 are the other two sides of the right-angle triangle.

Now we locate$\sqrt{5}$on the number line:

At origin O, take 2 units from the origin and represent the point as A.

At A, draw perpendicular of length 1 unit and mark it as AB.

Now, using OB as radius and O as the center, draw an arc on the right side of the triangle, intersecting the number line at P.

Point P represents$\sqrt{5}$

In$\Delta OAB$, using Pythagoras Theorem, we have

role="math" localid="1643282703484" $\begin{array}{l}O{B}^2=O{A}^2+A{B}^2\\ \Rightarrow O{B}^2=2^2+1^2\\ \Rightarrow O{B}^2=5\\ \Rightarrow OB=\sqrt{5}\end{array}$

OP is also the radius of the circle, so$OP=\sqrt{5}$

Hence, P represents $\sqrt{5}$on a number line.

Step 4. Other irrational numbers on a number line.

In a similar fashion as above, we can represent $\pi ,\sqrt{10},\sqrt{17},\sqrt{8},\sqrt{13}$and many more irrational numbers on a number line.