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How to Rationalize Irrational Numbers?

The number system is nothing but just the representation of different kinds of numbers on the number line (for example, integers, whole numbers, rational numbers, irrational numbers, natural numbers). Let’s understand more briefly suppose one has a number line from -∞ to ∞ and has to collect 0, 5, -6, 582, -633, etc. From the number line so first, check what kind of numbers these are (as it is known, one has collected integers from the number line), and like this more different kinds of numbers can be assumed from the number line for different purposes.

Rational and Irrational numbers

Rational numbers are those numbers that can be written in the form of p/q, that is numerator divided by denominator, and can say that the number can be represented infraction. here p and q are integers and q can’t be zero (q ≠ 0). Example: 2/4, 10/30, -25/1, etc. Every integer is a rational number but not every rational number is an integer.



Irrational numbers are those numbers that can not be represented in the form of p/q, where p and q are integers and q can’t be zero (q ≠ 0). Example: √2, √3, (3 + √7) .

How to Rationalize Irrational Numbers?

What’s the need to rationalize irrational numbers? To represent rational numbers on the number line is easy but if one wants to represent irrational numbers on the number line it becomes so much difficult to represent. So to get an easy solution for this problem there is a need to rationalize the irrational number, and even in further study of advanced topics like integration,  differentiation, this kind of sub-topics helps a lot to solve them.



To Rationalize irrational numbers, suppose there is an irrational number 1/√2,

√2/2 = 1.41/2 now one can easily represent this point on a number line.

Let’s take another complex problem 3/(√5 + √3). The given problem can be solved or can be rationalized using some similar operation,

So in this manner, an irrational number can be rationalized by multiplying and dividing the denominator in the given question.

Sample Problems

Question 1: Solve, (2√3 + √2)/(1 + √2).

Solution:  

Multiply and divide by denominator with a question (make sign opposite from denominator) means (2√3 + √2)/(1 + 2) × (1 – √2 )/(1 – √2)

= ((2√3 + √2) – 2√6 – 4)/1 – 2

= -((2√3 + √2) – 2√6 – 4)

= (2√3 – √2) + 2√6 + 4.

Question 2: Solve, √5/(√6 – √3).

Solution:

√5/(√6 – √3) × (√6 + √3)/(√6 + √3)

(√5 × (√6 + √3))/6 – 3

(√30 + √15)/ 3

Question 3: Solve, (1 + √17)/√39.

Solution:

= ((1 + √17)/√39) × √39/√39

= (√39 × (1 + √17))/√39 × √39

= (√39 + √663)/39.

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