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NUMBER SYSTEM IN MATHS FOR CLASS 9 NOTES

NUMBER SYSTEM IN MATH FOR CLASS 9 NOTES

 EXERCISE: 1.3

 

 Question: 1. Write the following in decimal form and say what kind of decimal expression each has:

 i) 36/100

 ii) 1/11

 iii) 4?1/8

 iv) 3/13

 v) 2/11

 vi) 329/400

 Solution: On dividing 36 by 100 , we get


 Therefore, 36/100=0.36 which is a terminating decimal.

 ii) 1/11

 On dividing 1 by 11, we get


We observe that while dividing 1/11, the quotient=0.09 is repeated.Therefore,1/11=0.090909….. or 1/11=0.?0?9 which is non-terminating decimal.

 iii) 4×1/8=4+1/8=32+1/8=33/8

On dividing 33 by 8, we get



While dividing 33 by 8, the remainders is “0”.Therefore,4×1/8= 33/8 =4.125 which is a terminating decimal.

 iv) 3/13

On dividing 3/13, we get



While dividing 3 by 13 the remainder is 3,which will continue to be 3 after carrying out 6 continuous divisions.
Therefore, 3/13=0.230769….. or 3/13=0.?2?3?0?7?6?9 which is non-terminating and recurring decimal.
 

v) 2/11

On dividing 2/11, we get



We can observe that while dividing 2 by 11, first the remainder is 2 then 9, which will continue to be 2 and 9 alternately.

Therefore , 2/11 = 0.1818… or 2/11 = 0.?1?8 which is non-terminating decimal.

 v) 329/400

On dividing 329 by 400 , we get



While dividing 329 by 400 , the remainder is “0”.

Therefore , 329/400 = 0.8225 which is a terminating decimal.

 Question: 2. You know that 1/7 = 0.142857…. can you predict what the decimal expansions of 2/7 , 3/7 , 4/7 , 5/7 , 6/7 , are , without actually doing the long division ?If so , how ?

[Hint: Study the remainders while finding the value of 1/7 carefully.]

 Solution: 1/7 = 0.?1?4?2?8?5?7 or 1/7 = 0.142857………

find the value of 2/7 , 3/7 , 4/7 , 5/7 , and 6/7 without performing long division.
Here , we can write 2/7 , 3/7 , 4/7 , 5/7 and 6/7 in the form of 2×1/7 , 3×1/7 , 4×1/7 , 5×1/7 and 6×1/7.

On substituting value of 1/7 = 0.142857….. , we get

* 2×1/7 = 2 × 0.142857 = 0.285714.
* 3×1/7 = 3 × 0.142857 = 0.428571
* 4×1/7 = 4 × 0.142857 = 0.571428
* 5×1/7 = 5 × 0.142857 = 0.714285
* 6×1/7 = 6 × 0.142857 = 0.857142
Therefore , we conclude that the values of 2/7 , 3/7 , 4/7 , 5/7 and 6/7 without performing long division , we get

* 2/7 = 0.?2?8?5?7?1?4
* 3/7 = 0.?4?2?8?5?7?1
* 4/7 = 0.?5?7?1?4?2?8
* 5/7 = 0.?7?1?4?2?8?5
* 6/7 = 0.?8?5?7?1?4?2

Question: 3. Express the following in the form of p/q where p and q are integers and q ? 0.

 i) 0.?6

 ii) 0.?4?7

 iii) 0.?0?0?1

Solution: i) Let X = 0.?6 = X = 0.6666… (1)

  Multiply both sides by 10.
10X = 10 × 0.6666
10x = 6.6666….. (2)
Substracting (1) from (2) , we get
10x = 6.6666…..
-x = 0.6666……
9x = 6
x = 6/9 = 2/3.
Therefore on converting 0.?6 = 2/3 which is in the p/q form ,

 ii) Let x = 0.?4?7 = x = 0.47777….(a)
Multiply both sides by 10 , we get
10x = 4.7777….. (b)
Substract the equation (a) from (b) , we get
10x = 4.7777….
 –   x = 0.47777……
9x = 4.3
x = 4.3/9 ×10/10 = 43/90
x = 43/90
Therefore , on converting o.?4?7 = 43/90 in the p/q form.

 iii) Let X = 0.?0?0?1….(a)
Multiply oth sides by 1000 (because the number of recurrinng decimal number is 3)
1000 × = 1000 × 0.001001…
So , 1000x = 1.001001….(b)
Substract the equation (a) from (b)
1000x = 1.001001…..
   –     x = 0.001001……..
999x = 1
= x = 1/999
Therefore , on converting 0.?0?0?1 = 1/999 which is in the p/q form.

 Question: 4. Express 0.9999….. in the form p/q . Are you surprised by your answer ? Discuss why the answer makes sense with your teacher and classmates .

 Solution: Let x = 0.9999…….(a)
we need to multiply by 10 on both sides , we get
10x = 9.9999…….(b)
Substract the equation (a) from (b) , to get
10x = 9.99999…..
  –  x = 0.9999….
   9x = 9
   9x = 9 as x = 9/9 or x = 1
Therefore , on converting 0.9999….. = 1/1 which is in the p/q form .

yes , we are surprised with our answer
But , the answer makes sense we observe that 0.9999…. goes on forever . So , there is no gap between 1 and 0.9999….. and hence they are equal .

 Question: 5. What can the maximum number of digits be in the recurring block of digits in the decimal expansion of 1/77 ? Perform the division to check your answer.

 Solution: We need to find the number of digits in the recurring block of 1/17 .
Let us perform the long division to get the recurring block of 1/17 , to get



We can observe that while dividing 1/17 we get 16 number of digits in the repeating block of decimal expansion which will continue to be 1 after carrying out 16 continuous divisions .
Therefore , we conclude that 1/17 = 0.0588235294117647….. or 1/17 = 0.?5?8?8?2?3?5?2?9?4?1?1?7?6?4?7 , which is a non-terminating and recurring decimal .

 Question: 6. Look at everal examples of rational numbers in the form p/q (q ? 0) , where p and q are integers with no common factors other than 1 and having terminating decimal representations (expansions).Can you guess what property q must satisfy ?

 Solution: Let us take the examples 7/2 , 5/4 , 15/6 , 9/4 , 2/10 , of the form p/q are the terminating decimals .
 * 7/2 = 3.5
 * 5/4 = 1.25
 * 15/6 = 2.5
 * 9/4 = 2.25
 * 2/10 = 0

 We can observe that the denominators of the above rational numbers have powers of 2 , 5 or both.Therefore , q must satisfy in the form either 2m or 5n or both 2m × 5n(where m = 0 , 1 , 2 , 3 …. and n = 0 , 1 , 2 , 3…)in p/q form .

 Quetion: 7. Write three numbers whose decimal expansions are non-terminating non-recurring .
E.g.v2 = 1.41421….
       v3 = 1.73205….
       v7 = 2.645751….


 Question: 8. Find three different irrational numbers between the rational numbers 5/7 and 9/11.

 Solution: Let us convert 5/7 and 9/11 into decimal form , we get 5/7 = 0.714285….. and 9/11 = 0.818181……
Three irrational numbers that lie between 0.714285…. and 0.818181…. are :
0.73073007300073……
0.74074007400074……
0.76076007600076……
Irational numbers cannot be written in the form of p/q .


 Question: 9. Classify the following numbers as rational or irrational :
 i) v23
 ii)v225
 iii) 0.3796
 iv) 7.478478
 v) 1.101001000100001

 Solution: We know that

 i) v23

   v23 = 4.795831
    It is an irrational number .


 ii) v225 = 15
      Therefore ,v225 is a rational number .


 iii) 0.3796

It is terminating decimal . Therefore , It is a rationl number . 

  iv) 7.478478…..
The given numbers 7.478478….. is a non-terminating recurring decimal . which can be converted into p/q form .
While , converting 7.478478……. into p/q form , we get
 X = 7.478478…….(a)
 1000X = 7478.478478…….(b)
 While , substracting (a) from (b) , we get
 1000x = 7478.478478……
      –   x = 7.478478
    999x = 7471
           x = 7471/999
 Therefore , 7.478478….. is a rational number .
 v) 1.101001000100001……..
 We can observe that the number 1.101001000100001 is a non- terminating and non- recurring decimal . thus , non- terminating and non- recurring decimals cannot be converted into p/q form .
 Therefore , we conclude that 1.101001000100001….. is an irrational number .


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