Number System Class 9 Chapter 1 Ex 1.4 NCERT Solutions

Number System Class 9 Chapter 1 Exercise 1.4 Chapter focuses on the laws of exponents and how they apply to real numbers. Understanding these rules is important because they form the foundation for simplifying expressions in algebra and later topics. 

These NCERT Solutions for Number System Class 9 Exercise 1.4 break down each problem step by step. This makes it easier for students to understand the concept and score well in the exam.

Class 9 Chapter 1 Number System Exercise 1.4 Questions

Class 9 Chapter 1 Number System Exercise 1.4 Questions test your ability to apply the laws of exponents to various real number expressions. These problems help you practise multiplying, dividing, and raising powers of numbers, and also using negative and zero exponents correctly. Below are the NCERT Solutions Class 9 Chapter 1 Ex 1.4: 

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) 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.

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 ) = 3 2 -(√3) 2 = 9-3 = 6

(iii) (√5+√2) 2

Solution: (√5+√2) 2 = √5 2 +(2×√5×√2)+ √2 2 = 5+2×√10+2 = 7+2√10

(iv) (√5-√2)(√5+√2)

Solution: (√5-√2)(√5+√2) = (√5 2 -√2 2 ) = 5-2 = 3

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: There is no contradiction. When we measure a value with a scale, we only obtain an approximate value. We never obtain an exact value. Therefore, we may not realize whether c or d is irrational. The value of π is almost equal to 22/7 or 3.142857…

4. Represent (√9.3) on the number line.

Solution:

Step 1: Draw a 9.3 units long line segment, AB. Extend AB to C such that BC=1 unit.

Step 2: Now, AC = 10.3 units. Let the centre of AC be O.

Step 3: Draw a semi-circle of radius OC with centre O.

Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.

Step 5: OBD, obtained, is a right angled triangle. Here, OD 10.3/2 (radius of semi-circle), OC = 10.3/2 , BC = 1

OB = OC – BC ⟹ (10.3/2)-1 = 8.3/2

Using Pythagoras theorem, We get, OD 2 =BD 2 +OB 2

⟹ (10.3/2) 2 = BD 2 +(8.3/2) 2
⟹ BD 2 = (10.3/2) 2 -(8.3/2) 2
⟹ (BD) 2 = (10.3/2)-(8.3/2)(10.3/2)+(8.3/2)
⟹ BD 2 = 9.3
⟹ BD =  √9.3

Thus, the length of BD is √9.3.

Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of √9.3 from O as shown in the figure.

5. Rationalize the denominators 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)
= (√7+√6)/(√7-√6)(√7+√6)
= (√7+√6)/√7 2 -√6 2 [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ]
= (√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)
= (√5-√2)/(√5+√2)(√5-√2)
= (√5-√2)/(√5 2 -√2 2 ) [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ]
= (√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)
= (√7+2)/(√7-2)(√7+2)
= (√7+2)/(√7 2 -2 2 ) [denominator is obtained by the property, (a+b)(a-b) = a 2 -b 2 ]
= (√7+2)/(7-4) = (√7+2)/3

NCERT Maths Class 9 Number System Exercise 1.4 PDF

NCERT Maths Class 9 Number System Exercise 1.4 PDF provides detailed step-by-step solutions with examples. Students can revise exponent rules quickly and practise as many problems as they want. This way you can have a strong understanding of Number System Class 9 concepts while improving speed and accuracy in solving exponent-based problems.

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