NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4.
- Number Systems Class 9 Ex 1.1
- Number Systems Class 9 Ex 1.2
- Number Systems Class 9 Ex 1.3
- Number Systems Class 9 Ex 1.4
- Number Systems Class 9 Ex 1.5
- Number Systems Class 9 Ex 1.6
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 9 |
| Subject | Maths |
| Chapter | Chapter 1 |
| Chapter Name | Number Systems |
| Exercise | Ex 1.4 |
| Number of Questions Solved | 2 |
| Category | NCERT Solutions |
NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.4
Ex 1.4 Class 9 Maths Question 1.
Visualize 3.765 on the number line, using successive magnification.
Solution:
We know that 3.765 lies between 3 and 4. So, let us divide the part of the number line between 3 and 4 into 10 equal parts and look at the portion between 3.7 and 3.8 through a magnifying glass. Now 3.765 lies between 3.7 and 3.8 [Fig. (i)]. Now, we imagine dividing this again into ten equal parts. The first mark will represent 3.71, the next 3.72 and so on. To see this clearly, we magnify this as shown in
[Fig. (ii)].
Again 3.765 lies between 3.76 and 3.77 [Fig. (ii)]. So, let us focus on this portion of the number line [Fig. (iii)] and imagine to divide it again into ten equal parts [Fig. (iii)]. Here, we can visualize that 3.761 is the first mark and 3.765 is the 5th mark in these subdivisions. We call this process of visualization of representation of numbers on the number line through a magnifying glass as the process of successive magnification. So, we get seen that it is possible by sufficient successive magnifications of visualizing the position (or representation) of a real number with a terminating decimal expansion on the number line.
Ex 1.4 Class 9 Maths Question 2.
Visualise \( 4.\overline { 26 }\) on the number line, upto 4 decimal places.
Solution:
We adopt a process by successive magnification and successively decrease the lengths of the portion of the number line in which \( 4.\overline { 26 }\) is located. Since \( 4.\overline { 26 }\) is located between 4 and 5 and is divided into 10 equal parts [Fig. (i)]. In further, we locate \( 4.\overline { 26 }\) between 4.2 and 4.3 [Fig. (ii)].
To get more accurate visualization of the representation, we divide this portion into 10 equal parts and use a magnifying glass to visualize that \( 4.\overline { 26 }\) lies between 4.26 and 4.27. To visualise \( 4.\overline { 26 }\) more clearly we divide again between 4.26_and 4.27 into 10 equal parts and visualise the representation of \( 4.\overline { 26 }\) between 4.262 and 4.263 [Fig. (iii)].
Now, for a much better visualization between 4.262 and 4.263 is again divided into 10 equal parts
[Fig. (iv)]. Notice that \( 4.\overline { 26 }\) is located closer to 4.263 then to 4.262 at 4.2627.
Remark: We can adopt the process endlessly in this manner and simultaneously imagining the decrease in the length of the number line in which 4.26 is located.
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