# Number Systems : Exercise 1.4 (Mathematics NCERT Class 9th)

Q.1Â Â Â Â Â Â  Visualise 3.765 on the number line, using successive magnification.
Sol.

We know that 3.765 lies between 3 and 4, that is, in the interval [3, 4] to have a rough idea where it is. Now, we divide the interval [3,4] into 10 equal partsÂ and look at [3.7, 3.8] through a magnifying glass and realize that 3.765 lies between 3.7 and 3.8 (see figure (i)). Now, we imagine that each of the new Â intervals [3.1, 3.2], [3.2, 3.3], ... [3.9, 4] has been sub divided into 10 equal parts. As before , we can now visualize through the magnify glass that 3.765 Â lies in the interval [3.76, 3.77] (see figure (ii)).

So, we have seen that it is possible by sufficient successive magnifications to visualize the position (or representation of a real number with a terminatingÂ decimal expansion on the number line. Let us now try and visualize the position (or representation) of a real number with a non- terminating recurring Â decimal expansion on the number line. We can look at appropriate intervals through a magnifying glass and by successive magnifications visualize the Â position of the number on the number line.

Q.2Â Â Â Â Â Â  Visualise $4.\overline {26}$ on the number line, upto 4 decimal places.
Sol.

We proceed by successive magnifications, and successively decrease the lengths of the intervals in which $4.\overline {26}$ is located. $4.\overline {26}$Â  is located in the interval [4, 5] of length 1. We further locate $4.\overline {26}$ in the interval [4.2, 4.3] of length 0.1. To get more accurate visualization of the representation, we divide even this interval into 10 equal parts and use a magnifying glass to visualize that $4.\overline {26}$ lies in the interval [4.26, 4.27] of length 0.01. To visualize $4.\overline {26}$ in an interval of length 0.001, we again divide each of the new intervals Â into 10 equal parts and visualize the representation of $4.\overline {26}$ Â in the interval [4.262, 4.263] of length 0.001. Notice that $4.\overline {26}$Â is located closer to 4.263 than to 4.262.
Note : - We can proceed endlessly in this manner and simultaneously imagining the decrease in the length of the interval in which Â $4.\overline{26}$Â is located.