NCERT Solutions Class 9th Maths Chapter – 1 Number System Exercise – 1.4

Question 1. Classify the following numbers as rational or irrational:

(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) 2π

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, 2π is an irrational number.

Question 2. Simplify each of the following expressions:

(i) (3+√3) (2+√2)

Solution – (3+√3) (2+√2)
Opening the brackets,
we get, (3×2)+(3×√2)+(√3×2)+(√3×√2)
= 6+3√2+2√3+√6

(ii) (3+√3) (3-√3)

Solution – (3+√3) (3-√3)
= 3²-(√3)²
= 9-3
= 6

(iii) (√5+√2)²

Solution – (√5+√2)²
= √5²+(2×√5×√2)+ √2²
= 5+2×√10+2
= 7+2√10

(iv) (√5-√2) (√5+√2)

Solution – (√5-√2) (√5+√2)
= (√5²-√2²)
= 5-2
= 3

Question 3. Recall, π is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That is π = c/d. This seems to contradict the fact that π is irrational. How will you resolve this contradiction?

Solution – Actually c/d = 22/7  is only an approximate value of π and also a non-terminating decimal.

Question 4. Represent √9.3 on the number line.

Solution – Draw a line segment AB = 9.3 units and extend it to C such that BC = 1 unit.
Find mid point of AC and mark it as O.
Draw a semicircle taking O as Centre and AO as radius. Draw BD ⊥ AC.
Draw an arc taking B as Centre and BD as radius meeting AC produced at E such that BE = BD = √9.3 units.

Question 5. Rationalise the denominator of the following:

(i) 1/√7

Solution – Multiply and divide 1/√7 by √7
(1×√7)/(√7×√7)
= √7/7

(ii) 1/(√7-√6)

Solution – Multiply and divide 1/(√7-√6) by (√7+√6)
[1/(√7-√6)]×(√7+√6)/(√7+√6) [denominator is obtained by the property, (a+b)(a-b) = a²-b²]
= (√7+√6)/(√7-√6)(√7+√6)
= (√7+√6)/√7²-√6² 
= (√7+√6)/(7-6)
= (√7+√6)/1
= √7+√6

(iii) 1/(√5+√2)

Solution – Multiply and divide 1/(√5+√2) by (√5-√2)
[1/(√5+√2)]×(√5-√2)/(√5-√2) [denominator is obtained by the property, (a+b)(a-b) = a²-b²]
= (√5-√2)/(√5+√2)(√5-√2)
= (√5-√2)/(√5²-√2²) 
= (√5-√2)/(5-2)
= (√5-√2)/3

(iv) 1/(√7-2)

Solution – Multiply and divide 1/(√7-2) by (√7+2)
1/(√7-2)×(√7+2)/(√7+2) [denominator is obtained by the property, (a+b)(a-b) = a²-b²]
= (√7+2)/(√7-2)(√7+2)
= (√7+2)/(√7²-2²)
= (√7+2)/(7-4)
= (√7+2)/3