**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)

[a^{2} – b^{2} = (a + b) (a – b)]

= 3^{2}-(√3)^{2}

= 9-3

= 6

**(iii) (√5+√2) ^{2}**

**Solution:**

(a + b)^{2} = a^{2} + b^{2} +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) = (a^{2} – b^{2})

= (√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:**

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,

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} = 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)/√7^{2}-√6^{2}

[Denominator is obtained by the property, a^{2 }– b^{2} = (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)/(√5^{2}-√2^{2})

= (√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)/(√7^{2}-2^{2})

= (√7+2)/(7-4)

= (√7+2)/3

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