Number Systems

1. Number Systems

Class : IX

Subject: Mathematics

NCERT Text book Solution.

EXERCISE 1.1

1. Is zero a rational number? Can you write it in the form p/q, where p and q are integers and q ≠ 0?

Solution:

Yes zero is a rational number.

It can be written in the form p/q which is 0/1.

2. Find six rational numbers between 3 and 4.

Solution:

Both 3 and 4 are multiplied and divided by (6 + 1 = 7)

3 = 3×7/7 = 21/7

4 = 4×7/7 = 28/7

Required rational numbers are

22/7, 23/7, 24/7, 25/7, 26/7 and 27/7 Ans.

3. Find five rational numbers between ⅗ and ⅘.

Solution:

Both ⅗ and ⅘ multiplied and divided by (5+1=6)

⅗ = 18/30

⅘ = 24/30

Required rational numbers are

19/30, 20/30, 21/30, 22/30 and 23/30 Ans.

4. State whether the following statements are true or false. Give reasons for your answers.

(i) Every natural number is a whole number.

(ii) Every integer is a whole number.

(iii) Every rational number is a whole number.

Solution:

(i)

True, Natural numbers are a part of whole numbers.

(ii)

False, Negative integers are not whole numbers.

(iii)

False, ½ is a rational number but not a whole number.

प्रश्नावली 1.1

1. क्या शून्य एक परिमेय संख्या है? क्या इसे आप p/q के रूप में लिख सकते हैं, जहाँ p और q पूर्णांक हैं और q ≠ 0 है?

हल:

हां, शून्य एक परिमेय संख्या है। इसे p/q के रूप में लिखा जा सकता है जो 0/1 है।

2. 3 और 4 के बीच में छः परिमेय संख्याएँ ज्ञात कीजिए।

हल:

3 और 4 दोनों के अंश एवं हर को 7 से गुना करने पर।

3 = 21/7

4 = 28/7

अतः 6 परिमेय संख्या है

22/7, 23/7, 24/7, 25/7, 26/7 एवं 27/7।

3. ⅗ और ⅘ के बीच पाँच परिमेय संख्याएँ ज्ञात कीजिये ।

हल:

⅗ एवं ⅘ दोनों के अंश एवं हर को 6 से गुना करने पर।

⅗ = 18/30

⅘ = 24/30

अतः 5 परिमेय संख्या है

19/30, 20/30, 21/30, 22/30 एवं 23/30

4. नीचे दिए गए कथन सत्य हैं या असत्य? कारण के साथ अपने उत्तर दीजिए।

(i) प्रत्येक प्राकृत संख्या एक पूर्ण संख्या होती है।

(ii) प्रत्येक पूर्णांक एक पूर्ण संख्या होती है।

(iii) प्रत्येक परिमेय संख्या एक पूर्ण संख्या होती है।

हल:

(i)

सत्य है, क्योंकि पूर्ण संख्या के समूह में प्राकृत संख्या शामिल है।

(ii)

असत्य है, क्योंकि ऋणात्मक पूर्णांक पूर्ण संख्या नही होता है।

(iii)

असत्य है, जैसे ½ परिमेय संख्या है लेकिन पूर्ण संख्या नही है।

EXERCISE 1.2

1. State whether the following statements are true or false. Justify your answers.

(i) Every irrational number is a real number.

Solution:

True, every irrational numbers are represented on a number line.

(ii) Every point on the number line is of the form, √m where m is a natural number.

Solution:

False, every point on the number is of the form of real number.

(iii) Every real number is an irrational number.

Solution:

False, every real number is either rational or irrational.

2. Are the square roots of all positive integers irrational? If not, give an example of the square root of a number that is a rational number.

Solution:

The square root of all positive integers are not irrational.

Example

√4 is a rational number.

3. Show how √5 can be represented on the number line.

Solution:

Following steps are used to show √5 on number line.

Step 1.

Apply Pythagoras theorem to get base for unit perpendicular and for hypotenuse of √5.

b² = (√5)² - 1² = 4 = 2² 

Therefore, 

b = 2.

Step 1.

Draw perpendicular on 2 of unit that is P.

Step 2.

Join O to P.

Step 3.

Draw an arch of radius OP and centre O. The arch intersects the number line at Q. This represent √5 on number line.

Figure is given below.

1. Real Numbers

Class X

NCERT Text Book Solution

Exercise 1.1

1. Use Euclid's division algorithm to find the HCF of 

(i)135 and 225 (ii) 196 and 38220 (iii) 867 and 255

Solution:

(i)

From question it is clear that

225 > 135.

Apply Euclid's division lemma, to 135 and 225.

225 = 135x1 + 90

135 = 90x1 + 45

90 = 45x2 + 0

The remainder has now become zero.

The HCF of 225 and 135 is 45 Ans.

(ii)

From question it is clear that 

38220>196

Apply Euclid's division lemma, to 38220 and 196.

38220 = 196x195 + 0

The remainder has now become zero.

The HCF of 38220 and 196 is 195 Ans.

(iii)

From question it is clear that

867>255

Apply Euclid's division lemma, to 867 and 255.

867 = 255x3 + 102

255 = 102x2 + 51

102 = 51x2 + 0 

The remainder has now become zero.

The HCF of 867 and 255 is 51 Ans

2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.

Solution:

Let a be any positive integer and b = 6.

Now, by Euclid's division algorithm,

a = 6q + r for some integer q ≥ 0,

r = 0, 1, 2, 3, 4, 5

Because 0 ≤ r < b

So, a can be

6q,or 6q + 1, or 6q + 2, or 6q + 3, or 6q + 4, or 6q + 5.

Since a is odd, a cannot be 6q, or 6q + 2, or 6q + 4 ( since they are all divisible by 2)

Therefore, any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5. Proved.

3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march ?

Solution:

For this, it will be find out the HCF of 616 and 32.

Apply Euclid's algorithm, to 616 and 32.

616 = 32x19 + 8

32 = 8x4 + 0

So, the HCF of 616 and 32 is 8.

Therefore, the maximum number of columns in which they can march = 8 Ans.

4. Use Euclid's division lemma to show that the square of any positive integer is either of the form 3m, or 3m + 1 for some integer m.

Solution:

Let a be any positive integer and b = 3.

Now, Euclid's division lemma,

a = 3q + r for some integer q ≥ 0,

Since, 0 ≤ r < 3

a can be 3q, or 3q + 1, or 3q + 2

Therefore,

a² = 9q², or 9q² + 6q + 1, or 9q² + 12q + 1.

or, a² = 3x3q², or 3( 3q² + 2q) + 1, or 3(3q² + 4q) + 1

So, the square of any positive integer is either of the form 3m, or 3m + 1. Proved

5. Use Euclid's division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution:

Let a be any positive integer and b = 9.

Now, Euclid's division lemma,

a = 9q + r for some integer q ≥ 0,

Since, 0 ≤ r < 9

a can be 9q, or 9q + 1, or 9q + 2, 9q + 3, or ...... 9q + 8.

Therefore,

a³ = 9x81q³, or 9(81q³ + 27q² + 3q)+ 1, or 9(81q³ + 54q² + 6q) + 8 ..................

So, the cube of any positive integer is of the form 9m,or 9m + 1 or 9m + 8. Proved.