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# Classify the following numbers as rational or irrational: (i) 2 –√5, (ii) (3 +√23), (iii) 2√7/7√7,(iv) 1/√2. Q.1 (1),(2), (3),(4)

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Today i am solving the important question of ncert class 9th chapter number systems . Its very tricky question of this exercise find the best solution of exercise 1.5 question number 1 .Please help me to find the easiest solution of this question .Classify the following numbers as rational or irrational: (i) 2 –√5, (ii) (3 +√23), (iii) 2√7/7√7,(iv) 1/√2

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1. (i) 2 –√5

Solution:

We know that, √5 = 2.2360679…

Here, 2.2360679…is non-terminating and non-recurring.

Now, substituting the value of √5 in 2 –√5, we get,

2-√5 = 2-2.2360679… = -0.2360679

Since the number, – 0.2360679…, is non-terminating non-recurring, 2 –√5 is an irrational number.

(ii) (3 +√23)- √23

Solution:

(3 +23) –√23 = 3+23–√23

= 3

= 3/1

Since the number 3/1 is in p/q form, (3 +√23)- √23 is rational.

(iii) 2√7/7√7

Solution:

2√7/7√7 = ( 2/7)× (√7/√7)

We know that (√7/√7) = 1

Hence, ( 2/7)× (√7/√7) = (2/7)×1 = 2/7

Since the number, 2/7 is in p/q form, 2√7/7√7 is rational.

(iv) 1/√2

Solution:

Multiplying and dividing numerator and denominator by √2 we get,

(1/√2) ×(√2/√2)= √2/2 ( since √2×√2 = 2)

We know that, √2 = 1.4142…

Then, √2/2 = 1.4142/2 = 0.7071..

Since the number , 0.7071..is non-terminating non-recurring, 1/√2 is an irrational number.

(v) 2

Solution:

We know that, the value of = 3.1415

Hence, 2 = 2×3.1415.. = 6.2830…

Since the number, 6.2830…, is non-terminating non-recurring, 2 is an irrational number.

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