Plotting any Irrational Numbers On a Number Line


 
 
Concept Explanation
 

Plotting any Irrational Numbers On a Number Line

Plotting any Irrational Numebrs On a Number Line:

For any positiv real number x, we have

 large sqrt{left ( frac{x+1}{2}right )^{2}-left ( frac{x-1}{2} right )^{2}}=sqrt{frac{x^{2}+2x+1}{4}-frac{x^{2}-2x+1}{4}}=sqrt{frac{4x}{4}}=sqrt{x}

Therefore, to find the positive square root of a positive real number, we may follow the following algorithm.

ALGORITHM:

STEP I  Obtain the positive real number x(say)

STEP II  Draw a line and mark a point A on it.

STEP III Mark a point B on the line such that AB = x units.

STEP IV  From point B mark a distance of 1 unit and mark the new poiint as C.

STEP V  Find the mid-point of AC and mark the point as O.

STEP VI  Draw a circle with centre O and radius OC>

STEP VII  Draw a line perpendicular to AC passing through B and intersecting the semi-circle at D. Length BD is equal to large sqrt{x}.

Justification:  We have,

AB = x units and BC = 1 unit.

large therefore    AC = (x + 1) units

large Rightarrow ;;OA = OC = frac{x+1}{2};units

large Rightarrow ;; OD= frac{x+1}{2};units                      large [because ;OA=OC=OD]

Now,  large OB=AB-OA=x-frac{x+1}{2}=frac{x-1}{2}

Using Pythagoras Theorem in large Delta OBD, we have

  OD^{2}=OB^{2}+BD^{2}

Rightarrow ;;BD^{2}=OD^{2}-OB^{2}

Rightarrow ;;BD^{2}=left ( frac{x+1}{2} right )^{2}-left ( frac{x-1}{2} right )^{2}

Rightarrow ;;BD=sqrt{frac{(x^{2}+2x+1)-(x^{2}-2x+1)}{4}}=sqrt{frac{4x}{4}}=sqrt{x}

This shows that sqrt{x} exists for all real numbers x > 0.

In order to find the position of sqrt{x} on the number line, we consider BC as the number line, with B as the origin to represent zero. Since BC = 1 so, C represents 1. Now, mark points C_{1},C_{2},C_{3},...... such that CC_{1}=BC = 1, C_{1}C_{2}=BC=1 and so on. Clearly, C_{1}C_{2},C_{3},..... represent 2, 3, 4,... respectively.

Now, draw an arc with centre at B and radius equal to BD. Suppose this arc cuts the number line BC with B as the origin at E. Then, BE = sqrt{x}. Consequently, E will represent sqrt{x}.

     

 

 

Sample Questions
(More Questions for each concept available in Login)
Question : 1

Is it possible to say that sqrt {infty }  is defined on number line?

Right Option : B
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Explanation
Question : 2

Can all irrational numbers be plotted on a number line?

Right Option : B
View Explanation
Explanation
 
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