Real Numbers Formula: Some Basic Properties of Real Numbers

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The infinite list of mathematical objects assisting humans to count, or measure inculcates in itself, another fascinating set of numbers called real numbers. Real numbers are numbers that are both rational and irrational numbers. They are represented by the symbol ‘R’. 

These numbers can also be plotted on a number line similar to the other types of numbers that we have studied. Also, real numbers can be used to perform various mathematical operations.

Examples of real numbers include -12, 5.2, 1/34, 0, 1, 3, etc.

Read Also:- Real Numbers

Keyterms: Rational numbers, irrational numbers, mathematical operations, subsets, Natural number, Whole number, Integers


Sets of Real Numbers

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The real numbers contain five subsets, namely-

1. Natural numbers

They contain all the positive counting numbers starting from one and are donated by the symbol ‘N’.

N= {1,2,3, 4…}

Examples- 1,1000, 6745, 99, etc.

2. Whole numbers

They contain all the positive counting numbers starting from zero and are denoted by the symbol ‘W’.

W= {0,1,2,3,4…}

Examples-0, 1, 344, 500, etc.

3. Integers

They contain all the positive and negative numbers along with zero. They are denoted by the symbol ‘Z’.

Z= {-4, -3, -2, -1,0,1,2,3,4…}

Examples- 0, 1, 2, -100, 100, etc.

Integers are further classified as positive and negative integers. Positive integers are represented by Z+ while negative integers are denoted by the symbol Z-

.  

Examples for Z+ =11, 27.6, 99, etc.

Examples for Z-= -11, -27.6, -99, etc.

4. Rational numbers

They can be expressed in the form a/b, where a and b are integers and b is non-zero. They are denoted by the symbol ‘Q’.

Examples -2/5, 1/80, 1, 7/8, etc.

5. Irrational numbers

They cannot be expressed in the form a/b, where a and b are integers and b is non-zero. They are denoted by the symbol ‘P’.

Examples- 3, 5,11, etc.

Real numbers are a subset of another type of number called complex numbers which are a combination of real and imaginary numbers.
The inter-relationship of natural numbers, whole numbers, integers, rational numbers, irrational numbers, and real numbers can be depicted by a Venn diagram as shown below:

Sets of Real Numbers

Sets of Real Numbers


Basic properties of real numbers

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Some of the basic properties of real numbers include commutative, distributive, identity, and associative properties. These properties help to ease the calculations to a greater extent and avoid basic human error.

The list of properties is given below:

1. Commutative properties

Let x and y are two real numbers. The commutative properties in addition and multiplication for x and y are:

x+y=y+x=real number

x.y=y.x=real number

Examples are 7+3=3+7=10, which is a real number.

7.3=3.7=21, which is a real number.

2. Associative properties

Let x, y, and z are three real numbers. The associative properties in addition and multiplication for x, y, and z are:

(x+y) +z=x+(y+z) =real number

(xy)z=x(yz) =real number

Examples are (2+3) +4=2+(3+4) =9, which is a real number

(2.3)4=2(3.4) =24, which is a real number

 3. Distributive properties

Let x, y, and z be three real numbers. The distributive properties in addition and multiplication for x, y, and z are:

x(y+z) =xy+xz

An example is 2(3+4) =2.3+2.4=14, which is a real number.

4. Identity properties

Let x is any real number. Identity properties in addition and multiplication for x are:

0 is the additive identity so x+0=0+x=x

1 is the multiplicative identity so x.1=1.x=x

Also Read:


Real Numbers Formulas

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  • Theorem 1: Euclid’s Division Lemma

It states that for given positive integers p and r, there exist unique integers x and y such that p=ry+x and 0≤x

Euclid’s division algorithm is based on Euclid’s division lemma.

  • Theorem 2: Euclid’s Division Algorithm

It states that to obtain the highest common factor (HCF) of two positive integers, say a and b, such that a>b, the following method has to be adopted:

Step 1: Use Euclid’s division lemma on the two positive integers, a and b. Next, we find the whole numbers x and y such that a=by+x, 0≤x

Step 2: b is the HCF if x is zero. Apply Euclid’s division lemma to b and x if x is non-zero.

Step 3: The process continues till zero is obtained as remainder. The HCF will be the divisor once the process terminates.

Here, one can see that the HCF of a and b is equal to the HCF of b and x.

Example: Show that every positive even integer is of the form 2q and that every positive odd integer is of the form 2q+1, where q is some integer.

Solution: let, the positive integer be a and b=2.

According to Euclid’s division lemma, we can say; a=bq+r

So, a=2q+r such that 0≤r<2.

So, case 1

When r=0

a=2q+0

a=2q

since, 2x integer= even integer

therefore, a=2q represents an even positive integer.

Case 2

When r=1

a=2q+1

so, positive integers= even positive integers + odd positive integers

Therefore, 2q+1 represents an odd positive integer.

Example: Use Euclid’s division algorithm to find HCF of 441, 567, and 693.

Solution: HCF (441,567,693)

The HCF of 441 and 567 is 63 by Euclid’s division algorithm and by step division method.

Now HCF of 63 and 693 is 11.

Since, the remainder is 0, the combined HCF of 441, 567, and 693 is 63.

  • Theorem 3: Fundamental Theorem of Arithmetic

It states that every composite number can be given as a product of prime numbers and this factorization is unique other than the order of the prime factors occurrence.

The prime factorization of a natural number is unique, except for the order of the factors.

Theorem 4: for a prime number, say, a. If a divides square of p, where p is a positive integer, then a divides p.

Theorem 5: Let a is a rational number with terminating decimal expansion. So, a is expressed in the fractional form x/y, where x and y are coprime numbers, and the prime factorization of y is in the form 2m5n such that m and n are positive integers.

Theorem 6: let a=x/y is a rational number, and the prime factorization of y is in the form 2m5n such that m and n are positive integers, then a has a terminating decimal expansion.

Theorem 7: Let a=x/y where x and y are co-prime numbers and a is a rational number, such that the prime factorization of y is not in the form 2m5n such that m and n are positive integers, then a has a non-terminating, repeating decimal expansion.

Example: Find the simple form of 368/496

Solution: we find the HCF of 368 and 496 by step division method.

HCF=16

Now, dividing numerator and denominator with the HCF, we get 23/31.

The simplified form is 23/31.

Thus, for every rational number, the decimal expansion can be terminating or non-terminating repeating.

HCF is the highest common factor. It is the greatest number that can divide any two numbers exactly.

LCM is the least common multiple. It is the smallest number that is divisible by any two numbers considered.

Also, for any two positive integers x and y, the product of the HCF of x and y and the LCM of x and y are equal to the product of the two positive integers x and y.

Also Read: Complex Numbers and Quadratic Equations


Sample Questions

Ques. What is HCF (a, b) if two positive integers a and b are written as a=x3 y2 and b=xy3; x,y are prime numbers? (1 mark)

Ans. HCF=xy2 as minimum power is considered.

Ques. The decimal expansion of the rational number 14587/1250 will terminate after how many decimal places? (1 mark)

Ans. The factorisation of 1250 is 2x5x5x5x5x1 or 2x54. Since the maximum power of 2x54 is 4 so decimal expansion of the rational number 14587/1250 will terminate after four decimal places.

Ques. If HCF (336, 54) = 6, find LCM (336,54). (1 mark)

Ans. We know that the product of two numbers=HCF x LCM; so, 336x54=6XLCM.

LCM=336x54/6

LCM=3024

Ques. What is the HCF of the smallest composite number and the smallest prime number? (1 mark)

Ans. The smallest prime number is 2 and the smallest composite number is 4. So, the required HCF is (22, 2) =2.

Ques. The length, breadth, and height of a room are 8m 50cm, 6m 25cm, and 4m 75cm, respectively. Find the length of the longest rod that can measure the dimensions of the room exactly. (2 marks)

Ans. We have to determine the HCF of all lengths which can measure the dimensions.

Length= 8m 50cm= 850cm

Prime factorisation of 850=2x52 x17

Breadth= 6m 25cm= 625cm

Factorisation of 625=52 x52

Height= 4m 75cm= 475cm

Factorisation of 475=52 x19

So, HCF(length, breadth, height)=HCF(850, 625, 475)=52 =25.

Ques. Three bells toll at intervals of 9,12,15 minutes, respectively. If they start tolling together, after what time will they toll together next? (2 marks)

Ans. The LCM of 9,12 and 15 is to be considered. So, determining the prime factors, we have

9=32

12=22 x 3

15=3x5

So, LCM(9,12,15)=22 x32 x5=150 minutes

Ques. Prove that 5 is an irrational number. (3 marks)

Ans. Let that 5 be a rational number.

So, we find two integers p and q where q is non-zero such that 5 =p/q

p=5 q

p2 =5q2

therefore, p2 is divisible by 5. So, p is divisible by 5.

Let p=5k where k is any integer

On squaring both sides,

p2 =25k2

and p2 =5q2

so, 25k2 =5q2

q2=5 k2

This means that q2 is divisible by 5.

And hence, q is divisible by 5.

This implies that p and q have a common factor of 5.

This is contrary to the fact that p and q are co-prime numbers.

Thus, 5 cannot be expressed in the form p/q and is irrational.

Ques. Express 429 as a product of its prime factors.(1 mark)

Ans. 429=3x11x13

Ques. There is a circular path around a sports field, ram takes 18 minutes to drive one round of the field while ria takes 12 minutes for the same. Suppose they are both starting at the same point and at the same time and go in the same direction. After how many minutes will they meet at the starting point? (2 marks)

Ans. We consider the LCM of 12 and 18. The prime factors of 12 and 18 are:

12=2x2x3

18=2x3x3

So, LCM=2x2x3x3=36

They meet at the starting point after 36 minutes.

Ques. Check whether 6n  can end with the digit 0 for any natural number n. (2 marks)

Ans. We begin with finding factors of 6.

6n =2n x3n

For any number, 5 must be a factor of it to end with 0.

Prime factorization is unique. So, according to the fundamental theorem of arithmetic, 6n  cannot end with the digit 0. 

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CBSE X Related Questions

1.
Find the sums given below :
  1. \(7 + 10\frac 12+ 14 + ....... + 84\)
  2. \(34 + 32 + 30 + ....... + 10\)
  3. \(–5 + (–8) + (–11) + ....... + (–230)\)

      2.
      Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

          3.
          Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) \(x + y = 5\),\( 2x + 2y = 10\) (ii)\( x – y = 8 , 3x – 3y = 16\) (iii) \(2x + y – 6 = 0\) , \(4x – 2y – 4 = 0\) (iv) \(2x – 2y – 2 = 0,\) \( 4x – 4y – 5 = 0\)

              4.

              The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them

              Monthly consumption 
              (in units)

               Number of consumers

              65 - 85 

              4

              85 - 105

              5

              105 - 125

              13

              125 - 145

              20

              145 - 165

              14

              165 - 185

              8

              185 - 205

              4

                  5.
                  The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

                      6.

                      A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

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