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**REAL NUMBERS**

A number whose square is non- negative, is called a real number.

In fact, all rational and all irrational numbers form the collection of all real numbers.

Every real number is either rational or irrational.

Consider a real number.

(i) If it is an integer or it has a terminating or repeating decimal representation then it is rational.

(ii) If it has a nonterminating and nonrepeating decimal representation then it is irrational.

The totality of rational and irrationals form the collection of all real number.

**COMPLETENESS PROPERTY-: **On the number line, each point correspond to a unique real number. And, every real number can be represented by a unique point on the real line.

**DENSITY PROPERTY-: **Between any two real number, there exist infinitely many real number.

**ADDITION PROPERTIES OF REAL NUMBERS**

(i) Closure Property-: The sum of two real numbers is always a real number.

(ii) Associative Law-: (ab)c = a(bc) for all real numbers a,b,c.

(iii) Commutative Law-: ab =bs for all real numbers a and b.

(iv) Existence Of Multiplicative Identity-: Clearly, is a number such that 1 \times a = a \times 1 = a for every real number a.

1 is called the multiplicative identity for real numbers.

(v) Existence Of Multiplicative Inverse-: For each nonzero real number a, there exists a real number \left( \frac { 1 }{ a } \right) such that a\times\frac { 1 }{ a } = \frac { 1 }{ a } \times a =1.

A and \frac { 1 }{ a } are called multiplicative inverse (or reciprocal) of each other.

(vi) Distributive Law Of Multiplication Over Addition-: We have a(b + c) =ab + ac and (a + b)c= ac + bc for all real numbers a,b,c.

**SOME MORE RESULTS ON REAL NUMBERS**

For all positive real numbers a and b, we have:

(i) \sqrt { ab } =\sqrt { a } \times \sqrt { b }

(ii) \sqrt { \frac { a }{ b } } =\frac { \sqrt { a } }{ \sqrt { b } }

**Absolute Value of a Real Number**

The absolute value of a given number x denoted by mod of (\left| x \right|) is defined as

\left| x \right| = \left[ x\quad if\quad x\ge 0\\ x\quad if\quad x<0 \right]For example, \left| 7 \right| = +7 , \left| -5 \right| = -(-5) =5, \left| 0 \right| =0.

**Distance Between Two Real Numbers p and q on the Number Line**

\left| P-Q \right| means the distance between two points represented by P and Q on the number line.

So, the distance between \left| x-3 \right| means the distance between the points represented by x and 3 on the number line.

Suppose we have to find the distance between (i) 7 and -3 (ii) -4 and 6 on the number line.

It is given by:

(i) \left| 7-(-3) \right| =10 units.

(ii) \left| -4-(6) \right| =10 units

**Laws of Exponent for Real Number**

We know that when two or more numbers are multiplied together the result is called the continued product. Each number separately is called a **factor **of the product.

But when a product consists of the same factor repeated several times, it is called **a power of the factor. **For example:

(i) In the product a\times b\times c ; a, b, c are **called the factor of the product.**

** **(ii) The product a \times a \times a \times a \times a ={ a }^{ 5 }, is called the fifth power of a.

**Definition of **{ a }^{ m }** : **When m is a positive integer, **{ a }^{ m }** is defined to the product of m factors each equal to a. Thus, { a }^{ m }= a \times a \times a \times ...\times a.

## Existence of \sqrt { x } for a given positive real number x

Let x be a given positive real number.

We shall prove the existence of \sqrt { x } in the following three steps.

Step1-: Finding \sqrt { x } geometrically.

Draw a line segment AB= x units and extend it to c such that

BC= 1 unit.

Find the midpointO of AC.

With O as centre and OA ad radius, draw a semicircle.

Now, draw BD \bot AC, intersecting the semicircle at D.

We shall show that BD = \sqrt { x }.

Step 2-: Proving that BD= \sqrt { x }.

\Rightarrow We have, AB= x units and BC= 1 unit. So,AC= (x+1) units.

\Rightarrow Now,OD= OC (radii of same semicircle)

\Rightarrow \frac { 1 }{ 2 } AC=\frac { 1 }{ 2 } \left( x+1 \right) units.

And, OB=(OC – BC) =\left( \frac { x+1 }{ 2 } -1 \right) units=\frac { 1 }{ 2 } \left( x-1 \right)units.

\therefore { BD }^{ 2 }= \left( { OD }^{ 2 }-{ OB }^{ 2 } \right) = \frac { 1 }{ 4 } { \left( x+a \right) }^{ 2 }-\frac { 1 }{ 4 } { \left( x-1 \right) }^{ 2 }.

\Rightarrow \frac { 1 }{ 4 } \times { \left( x+1 \right) }^{ 2 }-{ \left( x-1 \right) }^{ 2 }=\left( \frac { 1 }{ 4 } \times 4x \right) =x.

\Rightarrow BD= \sqrt { X }.

Thus, we have proved that BD= \sqrt { X }.

Step 3-: Representing \sqrt { X } on the real line.

Take BC produced as the number line with the origin at b.

Now, BC= 1 unit.

So, B and C denote 0 and 1 respectively.

With B as centre and BD a radius, draw an arc, meeting BC produced at E. Then, BE= BD= \sqrt { X }.

Hence, the point E represents \sqrt { X }.

Thus, for every positive real number x, \sqrt { X } is also a positive real number and hence \sqrt { X } exists.

## Final Wordings:

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Source-: Ts aggarwal & R.L Arora References

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