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Representation of Irrational Numbers on Number Line: Definition and Examples | EDU.COMEDU.COMResourcesBlogGuidePodcastPlanBackHomesvg]:size-3.5">Math Glossarysvg]:size-3.5">Representation of Irrational Numbers on Number LineRepresentation of Irrational Numbers on Number Line: Definition and ExamplesTable of ContentsRepresentation of Irrational Numbers on Number Line Definition of Irrational Numbers and Number Lines

Irrational numbers are real numbers that cannot be written as a ratio (fraction) pq\frac{p}{q}qp​, where p and q are integers and q≠0q \neq 0q=0. In simple words, all real numbers that are not rational are irrational. These numbers have non-terminating, non-repeating decimal expansions. Examples of irrational numbers include 2\sqrt{2}2​, 3\sqrt{3}3​, π\piπ, and many others.

A number line is a straight line on which numbers are marked at equal intervals. It extends infinitely in both directions and is usually drawn horizontally. Every point on the number line matches with exactly one real number. To show irrational numbers on a number line, we use the Pythagorean theorem. We break down the number inside the square root into parts, form a right triangle, and then use the length of the hypotenuse to mark the location of the irrational number on the number line.

Examples of Representing Irrational Numbers on Number Line Example 1: Showing 2\sqrt{2}2​ on a Number Line Problem:

Represent 2\sqrt{2}2​ on a number line.

Step-by-step solution: Step 1, Draw a number line and mark points 000 and 111. This gives us a starting point for our work.

Number Line

Step 2, Break down 2\sqrt{2}2​ as 1+1\sqrt{1 + 1}1+1​. Draw a line straight up from point 111 that is 111 unit long. Let's call this point A.

Step 3, Connect point O (at zero) to point A, making a right-angled triangle. In this triangle, the base OB is 111 unit, the height AB is 111 unit, and OA is the slant side (hypotenuse).

Number Line

Step 4, Find the length of OA by using the Pythagorean Theorem: OA2=AB2+OB2OA^2 = AB^2 + OB^2OA2=AB2+OB2 OA2=12+12OA^2 = 1^2 + 1^2OA2=12+12 OA2=1+1=2OA^2 = 1 + 1 = 2OA2=1+1=2 OA=2OA = \sqrt{2}OA=2​

Number Line

Step 5, Now, keeping OA as radius and O as the center, draw an arc on the number line and name the point as P. Now, the distance OP is equal to 2\sqrt{2}2​. The point P represents 2\sqrt{2}2​ on the number line.

Number Line

Example 2: Showing 5\sqrt{5}5​ on a Number Line Problem:

Represent 5\sqrt{5}5​ on the number line.

Step-by-step solution: Step 1, Draw a number line and mark points 000, 111, and −1-1−1. This gives us reference points to work with.

Number Line

Step 2, At point 111, draw a line going straight up that is 222 units long. This creates a perpendicular line segment.

Step 3, Make a right-angled triangle ABC where C is at 000, B is at 111, and A is at the top of our perpendicular line. In this triangle, AB is 222 units high, BC is 111 unit long, and AC is the slant side (hypotenuse).

Number Line

Step 4, Find the length of AC using the Pythagorean Theorem: AC2=AB2+BC2AC^2 = AB^2 + BC^2AC2=AB2+BC2 AC2=22+12AC^2 = 2^2 + 1^2AC2=22+12 AC2=4+1=5AC^2 = 4 + 1 = 5AC2=4+1=5 AC=5AC = \sqrt{5}AC=5​

Number Line

Step 5, Using C (at 000) as the center point and AC as the distance (radius), draw a curved line (arc) that crosses the number line. Let's call this point D. The point D shows 5\sqrt{5}5​ on the number line.

Number Line

Example 3: Showing 3\sqrt{3}3​ on a Number Line Problem:

Represent 3\sqrt{3}3​ on the number line.

Step-by-step solution: Step 1, First, we need to show 2\sqrt{2}2​ on the number line (as explained in Example 111). Let's say we have already marked point P representing 2\sqrt{2}2​ on the number line.

Number Line

Step 2, Now we can use point P to find 3\sqrt{3}3​. From point P, draw a line going straight up that is 111 unit long.

Step 3, Connect this new point to the origin (000), forming a right triangle. In this triangle, the base is OP which equals 2\sqrt{2}2​ units, and the height is 111 unit.

Step 4, Find the length of the hypotenuse using the Pythagorean Theorem:

Hypotenuse2=(2)2+(1)2\text{Hypotenuse}^2 = (\sqrt{2})^2 + (1)^2Hypotenuse2=(2​)2+(1)2

Hypotenuse2=2+1=3\text{Hypotenuse}^2 = 2 + 1 = 3Hypotenuse2=2+1=3

Hypotenuse=3\text{Hypotenuse} = \sqrt{3}Hypotenuse=3​

Step 5, Using the origin (0) as the center point and the hypotenuse length as the distance (radius), draw a curved line (arc) that crosses the number line. The point where this arc meets the number line, let's call it M, represents 3\sqrt{3}3​.

Number Line

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