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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