Exercise: R01 : Number System|āϏāĻāĻā§āϝāĻž āĻĒāĻĻā§āϧāϤāĻŋ
Question / āĻĒā§ā§°āĻļā§āύ : Find the value of â correct to two decimal places using the square root method. (āĻŦā§°ā§āĻāĻŽā§āϞ āĻĒāĻĻā§āϧāϤāĻŋ (Square Root Method) āĻŦā§āĻ¯ā§ąāĻšāĻžā§° āĻā§°āĻŋ ā§° āĻŽāĻžāύ āĻĻāĻļāĻŽāĻŋāĻā§° āĻĒāĻŋāĻā§° āĻĻā§āĻāĻž āϏā§āĻĨāĻžāύāϞā§āĻā§ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻžāĨ¤)
Solution
We know: 12=1, 22=4
So,
1.42 = 1.96
1.412 = 1.9881
1.422 = 2.0164
Since 2 lies between 1.9881 and 2.0164, and is closer to 1.9881,(āϝāĻŋāĻšā§āϤ⧠2, 1.9881 āĻā§°ā§ 2.0164-ā§° āĻŽāĻžāĻāϤ āĻ ā§ąāϏā§āĻĨāĻŋāϤ āĻā§°ā§ 1.9881-ā§° āĻ āϧāĻŋāĻ āĻāĻā§°āϤ, āϏā§āϝāĻŧā§āĻšā§ ā§° āĻŽāĻžāύ āĻĒā§ā§°āĻžāϝāĻŧ 1.41āĨ¤)
â≈ 1.41
Ans / āĻāϤā§āϤ⧰ : â≈ 1.41
(Correct to two decimal places / āĻĻāĻļāĻŽāĻŋāĻā§° āĻĒāĻŋāĻā§° āĻĻā§āĻāĻž āϏā§āĻĨāĻžāύāϞā§āĻā§)
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Question / āĻĒā§ā§°āĻļā§āύ : Write two irrational numbers whose sum and product are rational. (āĻāύ⧠āĻĻā§āĻāĻž āĻ āĻŽā§āϞāĻĻ āϏāĻāĻā§āϝāĻž (Irrational Numbers) āϞāĻŋāĻāĻž, āϝāĻžā§° āϝā§āĻāĻĢāϞ āĻā§°ā§ āĻā§āĻŖāĻĢāϞ āĻĻā§āϝāĻŧā§āĻāĻžāĻ āĻŽā§āϞāĻĻ āϏāĻāĻā§āϝāĻž (Rational Numbers) āĻšāϝāĻŧāĨ¤
Solution / āϏāĻŽāĻžāϧāĻžāύ
Take the numbers and . ( āĻā§°ā§ āϞāĻāĻāĨ¤)
Sum / āϝā§āĻāĻĢāϞ: + (−â) = 0 (0 is a rational number / 0 āĻāĻāĻž āĻŽā§āϞāĻĻ āϏāĻāĻā§āϝāĻž)
Product / āĻā§āĻŖāĻĢāϞ: x (−â) = -2
Ans / āĻāϤā§āϤ⧰ : and −â
These are irrational numbers whose sum and product are both rational. (āĻāϝāĻŧāĻžāϤ āĻā§°ā§ āĻĻā§āϝāĻŧā§āĻāĻžāĻ āĻ āĻŽā§āϞāĻĻ āϏāĻāĻā§āϝāĻž āĻā§°ā§ āĻāϝāĻŧāĻžā§° āϝā§āĻāĻĢāϞ āĻā§°ā§ āĻā§āĻŖāĻĢāϞ āĻĻā§āϝāĻŧā§āĻāĻžāĻ āĻŽā§āϞāĻĻ āϏāĻāĻā§āϝāĻžāĨ¤)
Note: Rational Number: A number that can be expressed in the form , where . āĻŽā§āϞāĻĻ āϏāĻāĻā§āϝāĻž (āĻāĻā§āύāĻžāĻāĻļā§° āĻāĻāĻžā§°āϤ (āϝ'āϤ ) āĻĒā§ā§°āĻāĻžāĻļ āĻā§°āĻŋāĻŦ āĻĒā§°āĻž āϏāĻāĻā§āϝāĻž)
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Question / āĻĒā§ā§°āĻļā§āύ: Write all integers between -5 and 5 with the help of a number line. (āϏāĻāĻā§āϝāĻž ā§°ā§āĻāĻžā§° (Number Line) āϏāĻšāĻžāϝāĻŧāϤ -5 āĻā§°ā§ 5-ā§° āĻŽāĻžāĻā§° āϏāĻāϞ⧠āĻĒā§ā§°ā§āĻŖāϏāĻāĻā§āϝāĻž āϞāĻŋāĻāĻžāĨ¤)
Solution / āϏāĻŽāĻžāϧāĻžāύ
Number Line / āϏāĻāĻā§āϝāĻž ā§°ā§āĻāĻž
←ââââ|ââââ|ââââ|ââââ|ââââ|ââââ|ââââ|ââââ|ââââ|ââââ|ââââ→
-5 -4 -3 -2 -1 0 1 2 3 4 5
Ans / āĻāϤā§āϤ⧰ :
Question / āĻĒā§ā§°āĻļā§āύ : In the property of indices am ÷ an = am−n taking = n, show that a0 1.(āϏā§āĻāĻā§° āĻā§āĻŖ am ÷ an = am−n-āϤ = n āϧ⧰āĻŋāϞā§, a0 1āĻĒā§ā§°āĻŽāĻžāĻŖ āĻā§°āĻžāĨ¤)
Solution / āϏāĻŽāĻžāϧāĻžāύ
Using the law of indices,(āϏā§āĻāĻā§° āĻā§āĻŖ āĻ āύā§āϏ⧰āĻŋ,)
am ÷ an = am−n
Take (āϧ⧰ā§āĻ) .
am ÷ an = am−n
1 = a0
[because (āĻāĻžā§°āĻŖ) ÷ an a ≠ 0]
a0 = 1
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Question / āĻĒā§ā§°āĻļā§āύ : In the property of indices am ÷ an = am−n, taking = 0, show that a−n = 1/ an. (āϏā§āĻāĻā§° āĻā§āĻŖ am ÷ an = am−n-āϤ āϧ⧰āĻŋāϞā§, a−n = 1/ an āĻĒā§ā§°āĻŽāĻžāĻŖ āĻā§°āĻžāĨ¤)
Solution / āϏāĻŽāĻžāϧāĻžāύ
Using the law of indices, (āϏā§āĻāĻā§° āĻā§āĻŖ āĻ āύā§āϏ⧰āĻŋ,)
am ÷ an = am−n
Put (āϧ⧰ā§āĻ) .
a0 ÷ an = a0−n
Since (āϝāĻŋāĻšā§āϤā§) a0 ,
1/ an â= a−n
Proved: a−n = 1ââ/an
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