NCERT Solutions for Class 9 Math Chapter 1.5 NUMBER SYSTEM ALL SOLUTION

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EXERCISE 1.1 EXERCISE 1.2 EXERCISE 1.3 EXERCISE 1.4 EXERCISE 1.6

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1. Classify the following numbers as rational or irrational:

(i)   `\2-\sqrt{5}\ `

(ii)   `\(3+\sqrt{23})-\sqrt{23}\ `

(iii)   `\frac{2\sqrt7}{7\sqrt7}\ `

(iv)   `\frac{1}{\sqrt{2}\ `

(v)   `\2pi\ `

solution :

   (i)    `\2-\sqrt{5}\ `

   `\2-\sqrt{5}=2-2.2360679774997..........\ `.

   So, `\sqrt{5}=2.236067977499789696409173..........\ `.

   = - 0.2360679774997.........

   Thus, we say that `\2-\sqrt{5}\ ` is a irrational number.

   (ii)    `\(3+\sqrt{23})-\sqrt{23}\ `

   `\(3+\sqrt{23})-\sqrt{23}\ `

   `\=3+\sqrt{23}-\sqrt{23}\ `

    = 3

    Thus, we say that `\(3+\sqrt{23})-\sqrt{23}\ ` is a rational number.

   (iii)    `\frac{2\sqrt{7}}{7\sqrt{7}}\ `

   `\frac{2\sqrt{7}}{7\sqrt{7}}\ `

   `\=\frac{2\sqrt{7}}{\sqrt{7}\sqrt{7}\sqrt{7}}\ `

   `\=\frac{2}{\sqrt{7}\sqrt{7}}\ `

   `\=\frac{2}{7}\ `

   Thus, we say that `\frac{2\sqrt{7}}{7\sqrt{7}}\ ` is a rational number.

   (iv)    `\frac{1}{\sqrt{2}}\ `

   `\= 0.\overline{230769}\ ` non-terminating repeating.

   (v)    `\2pi\ `

   `\= 0.\overline{18}\ ` non-terminating repeating.

2.   Simplify each of the following expressions:

(i)   `\(3+\sqrt{3}) (2+\sqrt{2})\ `

(ii)   `\(3+\sqrt{3}) (3-\sqrt{3})\ `

(iii)   `\(\sqrt{5}+\sqrt{2})^2\ `

(iv)   `\(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})\ `

solution :

(i)   `\(3+\sqrt{3}) (2+\sqrt{2})\ `

   `\(3+\sqrt{3}) (2+\sqrt{2})\ `

   `\=(3×2)+(3×\sqrt{2})\ `+`\(\sqrt{3}×2)+(\sqrt{3}×\sqrt{2})\ `

   `\=6+3\sqrt{2}+2\sqrt{3}+\sqrt{6}\ `

(ii)   `\(3+\sqrt{3}) (3-\sqrt{3})\ `

   `\(3+\sqrt{3}) (3-\sqrt{3})\ `

   `\=(3 × 3) - (3 × \sqrt{3})\ ` + `\(\sqrt{3} × 3) - (\sqrt{3} ×\sqrt{3})\ `)

   `\= 9 - 3\sqrt{3}+ 3\sqrt{3}- 3\ `

   `\= 9 - 3\ `

   `\= 6\ `

(iii)   `\(\sqrt{5}+\sqrt{2})^2\ `

   `\(\sqrt{5}+\sqrt{2})^2\ `

   `\= (\sqrt{5}+\sqrt{2})(\sqrt{5}+\sqrt{2})\ `

   `\= (\sqrt{5}×\sqrt{5})+(\sqrt{5}×\sqrt{2})\ `+`\(\sqrt{2}×\sqrt{5})+(\sqrt{2}×\sqrt{2})\ `

   `\= (5 +\sqrt{10}+\sqrt{10}+2)\ `

   `\= (7 + 2\sqrt{10})\ `

(iv)   `\(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})\ `

   `\(\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})\ `

   `\= (\sqrt{5}-\sqrt{2})(\sqrt{5}+\sqrt{2})\ `

   `\= (\sqrt{5}×\sqrt{5})+(\sqrt{5}×\sqrt{2})\ `−`\(\sqrt{2}×\sqrt{5})-(\sqrt{2}×\sqrt{2})\ `

   `\= (5 +\sqrt{10}-\sqrt{10}-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, `\pi = \frac{c}{d}\ `. This seems to contradict the fact that `\pi\ ` is irrational. How will you resolve this contradiction?

solution :

There is no contradiction. Remember that when you measure a length with a scale or any other device, you only get an approximate rational value So, you may not realise that either c or d is irrational.

4.    Represent `\(\sqrt{9.3})\ ` on the number line.

solution :

    Step 1: Draw a 9.3 units long line segment AB. Extend AB to C such that BC = 1 unit.

    Step 2: Now, AC = 10.3 units. Let the centre of AC be O.

    Step 3: Draw a semi-circle of radius OC with centre O.

    Step 4: Draw a BD perpendicular to AC at point B intersecting the semicircle at D. Join OD.

    Step 5: OBD, obtained, is a right angled triangle.

    Here, `\OD\frac{10.3}{2}\ ` (radius of semi-circle), `\OC = \frac{10.3}{2}\ `, BC = 1

    OB = OC – BC

    ⇒ `\(\frac{10.3}{2}) - 1 = \frac{8.3}{2}\ `

    Using Pythagoras theorem,

    We get,

    `\OD^2 = BD^2+ OB^2\ `

    ⇒ `\(\frac{10.3}{2})^2= BD^2+ (\frac{8.3}{2})^2\ `

    ⇒ `\BD^2 = (\frac{10.3}{2})^2 - (\frac{8.3}{2})^2\ `

    ⇒ `\BD^2 = {(\frac{10.3}{2})-(\frac{8.3}{2})}\ `×`\{(\frac{10.3}{2})+(\frac{8.3}{2})}\ `

    ⇒ `\BD^2 = (\frac{10.3-8.3}{2})×(\frac{10.3+8.3}{2})\ `

    ⇒ `\BD^2 = (\frac{2}{2})×(\frac{18.6}{2})\ `

    ⇒ `\BD^2= 1 × 9.3\ `

    ⇒ `\BD^2 = 9.3\ `

    ⇒ `\BD = \sqrt{9.3}\ `

    Thus, the length of BD is `\sqrt{9.3}\ `.

    Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of `\sqrt{9.3}\ ` from O as shown in the figure.

5.    Rationalize the denominators of the following:

(i)    `\frac{1}{\sqrt7}\ `

(ii)    `\frac{1}{(\sqrt{7}-\sqrt{6})}\ `

(iii)    `\frac{1}{(\sqrt{5}+\sqrt{2})\ `

(iv)    `\frac{1}{(\sqrt{7}-2)}\ `

solution :

    (i)    `\frac{1}{\sqrt{7}}\ `

    Multiply and divide `\frac{1}{\sqrt{7}\ ` By `\sqrt{7}\ `.

    `\= (\frac{1}{\sqrt{7}} × (\frac{\sqrt{7}}{\sqrt{7}}))\ `.

    `\=\frac{\sqrt7}{7}\ `.

    (ii)    `\frac{1}{(\sqrt{7}-\sqrt{6})}\ `

    Multiply and divide `\frac{1}{(\sqrt{7}-\sqrt{6})}\ ` By `\(\sqrt{7}+\sqrt{6})\ `.

    `\=frac{1}{(\sqrt{7}-\sqrt{6})}×((\sqrt{7}+\sqrt{6}))/((\sqrt{7}+\sqrt{6}))\ `.

    `\=\frac{(\sqrt{7}+\sqrt{6})}{(\sqrt{7}-\sqrt{6})×(\sqrt{7}+\sqrt{6})}\ `

`\ [ (\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=(a-b) ]\ `.

    `\=\frac{(\sqrt{7}+\sqrt{6})}{(7-6)}\ `.

    `\=\frac{(\sqrt{7}+\sqrt{6})}{1}\ `.

    `\=(\sqrt{7}+\sqrt{6})\ ` .

    (iii)    `\frac{1}{(\sqrt{5}+\sqrt{2})\ `

    Multiply and divide `\frac{1}{(\sqrt{5}+\sqrt{2})\ ` By `\(\sqrt{5}-\sqrt{2})\ `.

    `\=\frac{1}{(\sqrt{5}+\sqrt{2})}×\frac{(\sqrt{5}-\sqrt{2})}{(\sqrt{5}-\sqrt{2})} \ `.

    `\=\frac{(\sqrt{5}-\sqrt{2})}{(\sqrt{5}+\sqrt{2})×(\sqrt{5}-\sqrt{2})}\ `

`\ [ (\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})=(a-b) ]\ `.

    `\=\frac{(\sqrt{5}-\sqrt{2})}{(5-2)}\ `.

    `\=\frac{(\sqrt{5}-\sqrt{2})}{1}\ `.

    `\=(\sqrt{5}-\sqrt{2})\ `.

    (iv)    `\frac{1}{(\sqrt{7}-2)}\ `

    Multiply and divide `\frac{1}{(\sqrt{7}-2)}\ ` By `\(\sqrt{7}+2)\ `.

    `\=\frac{1}{(\sqrt{7}-2)}\ ` ×`\frac{(\sqrt{7}+2)}{(\sqrt{7}+2)}\ ` .

    `\=\frac{(\sqrt{7}+2)}{(\sqrt{7}-2)(\sqrt{7}+2)}\ `

`\ [ (\sqrt{a}-b)(\sqrt{a}+b)=(a-b^2) ]\ `.

    `\=\frac{(\sqrt{7}+2)}{(7-4)}\ `.

    `\=\frac{(\sqrt{7}+2)}{3}\ ` .

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EXERCISE 1.1 EXERCISE 1.2 EXERCISE 1.3 EXERCISE 1.4 EXERCISE 1.6