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Saturday, 17 September 2011

cbse maths class 9th Problem related to Number System

Numbers occur everywhere in our day-to-day life. “Numbers are in every thing”, said Pythagoras, an ancient Greek mathematician. If we understand more about numbers, then we know more about mathematics. “Mathematics is the Queen of the Sciences, and the Theory of Numbers is the Queen of Mathematics”, said Carl Friedrich Gauss, a German mathematician. Numbers possess very nice properties and the properties will help us to solve problems of other Sciences. “Numbers are my Friends”, said Srinivasa Ramanujan, an Indian mathematician of our modern times. “God created the natural numbers and all the rest is the work of man”, exclaimed Kronecker, a German mathematician.

Problem1.: Find the sum of the irrational numbers : 0.101001000100001… and 0.010110111011110… . If it is a rational number, find it in the integer- by-integer form. 

Solution: Let x = 0.1010010001… and y = 0.0101101110… . Then

x + y = 0.1010010001… + 0.0101101110…= 0.111111… = 0.1.

Since x + y has a non-terminating but repeating decimal expansion, x + y is a rational number. Let the rational number be a. Then a = 0.11111…

Multiplying by 10, we get 10a = 1.1111…

∴ 10a − a = 1.1111… − 0.1111 = 1.0000…. or 9a = 1 or a = 1/9

∴ x + y = 1/9

Problem 2. : Put π and 22/7 in order relation.

Solution: π = 3.141528…, 22/7 = 3.142857…

∴ 22/7 − π = 3.142857… − 3.141528… = 0.0013…

∴ 22/7 − π is positive.

∴ π < 22/7

Problem 3. : Insert any four rational numbers in between the rational numbers 1.201 and 1.202.

Solution: Here 1.202 − 1.201 = 0.001 > 0 and so 1.201 < 1.202. Consider the numbers 1.2011, 1.2012, 1.2013, 1.2014. Since these have terminating decimal expansions, they are rational numbers. We find 1.2011 − 1.201 = 0.0001 > 0 and so 1.201 < 1.2011.

Since 1.2012 − 1.2011 = 0.0001 > 0, we get 1.2011 < 1.2012.

Since 1.2013 − 1.2012 = 0.0001 > 0, we get 1.2012 < 1.2013.

Since 1.2014 − 1.2013 = 0.0001 > 0, we get 1.2013 < 1.2014.

Since 1.202 − 1.2014 = 0.0006 > 0, we get 1.2014 < 1.202.

∴ 1.201 < 1.2011 < 1.2012 < 1.2013 < 1.2014 < 1.202.

Problem 4. Insert any four irrational numbers between 1.201 and 1.202.

Solution: Since 1.202 − 1.201 = 0.001 > 0, we get 1.201 < 1.202.

Consider the real numbers

a = 1.2011010010001…,

b = 1.2012020020002…,

c = 1.2013030030003…,

d = 1.2014040040004….

Since these four real numbers have distinct non-terminating and non repeating decimal representations, they are distinct irrational numbers. We find

a − 1.201 = 1.2011010010001… − 1.2010000000000… = 0.0001010010001… > 0,

b − a = 1.2012020020002… − 1.2011010010001… = 0.0001010010001… > 0,

c − b = 1.2013030030003… − 1.2012020020002… = 0.0001010010001… > 0,

d − c = 1.2014040040004… − 1.2013030030003….= 0.0001010010001… > 0,

1.202 − d = 1.2020000000000… − 1.2014040040004… = 0.0005959959995… > 0.

∴ 1.201 < a < b < c < d < 1.202.

Problem 5 . Method of locating points for rational numbers on the number line.

Solution: Consider the rational number .2/3

To locate the point on the number line corresponding to 2/3, we proceed as follows:

Locate the point P corresponding to the positive integer 3 (denominator of 2/3).

Draw the line segment PQ of length 2 (numerator of 2/3) perpendicular to OP. Join OQ.

Consider the line through A parallel to PQ. This line meets OQ at the point R. Then the length of AR is 2/3 times that of OA. This is so because ΔOAR and ΔOPQ are similar and

so PQ/AR = OP/OA or, 2/AR = 3/1 or, AR= 2/3

Now, draw a circle with centre at O and radius equal to the length of AR. This circle cuts the real number line at a point on the right side of O. This point corresponds to the rational number -2/3 (see Figure). The same circle cuts the real number line at a point on the left side of O and this point corresponds to the rational number 32−.In the same way, we can represent any rational number on the real number line. The real number line is a straight line. Given any two distinct points P and Q on the line, however close they may be, we can find a point between P and Q different from P and Q. That is, there is no gap between any two points on the real number line. We say that the real number line is a continuum of points. We have shown that every rational number corresponds to a unique point on the real number line.

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