Mathematics – Class 9 – Chapter 1- Number Systems – Exercise 1.5 – NCERT Exercise Solution

1.Classify the following numbers as rational or irrational:

(i) 2 –√5

Solution:
√5 = 2.2360679…
Here, 2.2360679…is non-terminating and non-recurring. So, it is an irrational number.
Now,
2-√5 = 2-2.2360679… = -0.2360679
Since the number, – 0.2360679…, is non-terminating non-recurring.
Hence, 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, which is a rational number.
Hence, (3 +√23)- √23 is rational number.

(iii) 2√7/7√7

Solution:
According to question,
2√7/7√7 = (2/7) × (√7/√7)
So, (√7/√7) = 1
Now,
(2/7) × (√7/√7) = (2/7)×1 = 2/7

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

(iv) 1/√2

Solution:

We have to multiply and divide the numerator and denominator by √2,
(1/√2) × (√2/√2) = √2/2 (since √2×√2 = 2)
√2 = 1.4142…
Then, √2/2 = 1.4142/2 = 0.7071..

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

(v) 2π

Solution:

The value of π = 3.1415

Hence, 2 = 2 × 3.1415.. = 6.2830…

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

2. Simplify each of the following expressions:

(i) (3+√3) (2+√2)

Solution:

(3+√3) (2+√2)

{(3×2) + (3×√2)} + {(√3×2) + (√3×√2)}

= 6+3√2+2√3+√6

(ii) (3+√3) (3-√3)

Solution:

= (3+√3) (3-√3)

[a² – b² = (a + b) (a – b)]

= 3²-(√3)²

= 9-3

= 6

(iii) (√5+√2)²

Solution:

(a +  b)² = a² + b² +2ab)

= (√5+√2)²

= (√5)²+ (2×√5×√2) + (√2)²

= 5+2×√10+2

= 7+2√10

(iv) (√5-√2) (√5+√2)

Solution:

(a + b) (a – b) = (a² – b²)

= (√5-√2) (√5+√2)

= (√5²-√2²)

= 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:

When we measure a value with a scale or any device, 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… .

Let c and d both are irrational.

c/d is irrational and hence π is irrational.

Hence, there is no contradiction in saying that it is irrational.

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

Solution:

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

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

Step 3: With center o draw a semi-circle of radius OC.

Step 4: At B point draw a BD perpendicular to AC which intersecting the semicircle at D. Now, Join OD.

Step 5: We obtained a right angled triangle OBD.

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

Now, using Pythagoras theorem,

OD² = BD² + OB²

⟹ (10.3/2)² = BD²+(8.3/2)²

⟹ BD² = (10.3/2)² – (8.3/2)²

⟹ BD² = 9.3

⟹ BD = √9.3

Hence, the length of BD is √9.3.

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

5. Rationalize the denominators of the following:

(i) 1/√7

Solution:

We have to multiply and divide 1/√7 by √7

= (1×√7)/(√7×√7)

= √7/7

(ii) 1/(√7-√6)

Solution:

We have to multiply and divide 1/(√7-√6) by (√7+√6)

= [1/(√7-√6)]×(√7+√6)/(√7+√6)

= (√7+√6)/(√7-√6)(√7+√6)

= (√7+√6)/√7²-√6²

[Denominator is obtained by the property, a² – b² = (a + b) (a – b)]

= (√7+√6)/(7-6)

= (√7+√6)/1

= √7+√6

(iii) 1/(√5+√2)

Solution:

We have to multiply and divide 1/(√5+√2) by (√5-√2)

[1/(√5+√2)]×(√5-√2)/(√5-√2)

= (√5-√2)/(√5+√2)(√5-√2)

= (√5-√2)/(√5²-√2²)

= (√5-√2)/(5-2)

= (√5-√2)/3

(iv) 1/(√7-2)

Solution:

We have to multiply and divide 1/(√7-2) by (√7+2)

= 1/(√7-2)×(√7+2)/(√7+2)

= (√7+2)/(√7-2)(√7+2)

= (√7+2)/(√7²-2²)

= (√7+2)/(7-4)

= (√7+2)/3