Real Numbers

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The collection of numbers including all rational numbers and irrational numbers is called Real number.

➧ It is denoted by R.

➧ Every Real number is either a Rational number or an Irrational number.

➧ Every real number is represented by a unique point on the number line. Also, every point on the number line represents a unique real number. This is why we call the number line, the real number line.

➧ In 1870s two German mathematicians, Canton and Dedekind, showed that: Corresponding to every real number, there is a point on the real number line, and corresponding to every point on the number line, there exists a unique real number.

Representing Real numbers on Number line

➧ It is easy to represent integers on number line and we know how to do it including fractions and decimals up to few decimal places.

How can we represent numbers with more than two digits after decimal point on the number line?

How can we represent non-terminating decimal numbers on the number line?

Successive magnification 🔍

The process of visualisation or representation of numbers on the number line, through a magnifying glass, is known as the process of successive magnification.

It is possible by sufficient successive magnifications to visualise the position (or representation) of a real number with a terminating and non-terminating decimal expansion on the number line.

➢ Let us locate 3.765 on the number line.

↬ Integral part of the number is positive 3, therefor the number lie between 3 and 4. Locate 3 and 4 on the number line.

↬ Suppose we divide the portion between 3 & 4 into ten equal divisions & mark each point of division as in figure, then first mark is 3.1, the second mark is 3.2 & so on.

↬ Suppose we are using magnifying glass which will give us the clear view of subdivisions as in figure.

↬ The number at first decimal place is 7, therefore we look at the portion between 3.7 & 3.8. Again suppose this portion is divided into ten equal parts then the first mark is 3.71, the second mark is 3.72 & so on.

real number line

↬ Now, the number at second decimal place is 6, so the number will lie between 3.76 & 3.77, therefore we look at the portion between 3.76 & 3.77. Again suppose this portion is divided into ten equal parts then the first mark is 3.761, the second is 3.762 & so on.

↬ Therefore, 3.765 is the 5th mark in these subdivisions.

Properties of Real numbers

🔵 Rational numbers satisfy the commutative, associative and distributive laws for addition and multiplication.

a + b = b + a
a \times b = b\times a

(a+b)+c = a+(b+c)
(a\times b)\times c = a\times (b\times c)

a(b\pm c) = ab\pm ac

🔵 Rational numbers are ‘closed’ with respect to addition, subtraction, multiplication and division (except by zero).

🔴 Irrational numbers also satisfy the commutative, associative and distributive laws for addition and multiplication.

🔴 The sum, difference, product and division of irrational numbers are not always irrational i.e., not closed.

If we add, subtract, multiply or divide two irrationals, the result may be rational or irrational.

2√3 + √3 = (2+1)√3 = 3√3, (= irrational )

√3 – √3 = 0, (= rational)

√3 × √3 = 3, (= rational)

⚫ (i) The sum and difference of a rational number and an irrational number is irrational.

2 + \sqrt{3} = 2 + 1.71... = 3.71...\\  \ \sqrt{2} - 1 = 1.41... - 1 = 0.41...

(ii) The product and quotient of a non-zero rational number with an irrational number is irrational.

2\times \sqrt{3} = 2\sqrt{3}

\frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3}

Finding Roots of a Positive Real number x geometrically.

(Square root of a Negative real number does not exist as square of a real number is always positive).

➢ Let’s find \sqrt{3.5} geometrically,

→ Mark the distance 3.5 units from a fixed point A on a given line to obtain a point B such that AB = 3.5 units

→ From B, mark a distance of 1 unit and mark the new point as C. BC = 1 unit

→ Find the mid-point of AC and mark that point as O. Draw a semicircle with centre O and radius OC.

→ Draw a line BD perpendicular to AC passing through B and intersecting the semicircle at D.
Then, BD = \sqrt{3.5}.

➢ More generally, to find \sqrt{x}, for any positive real number x, we mark B so that AB = x units, and, as above, mark C so that BC = 1 unit. Then, as we have done for the case x = 3.5, we find BD= \sqrt{x} .

Proof – We can prove this result using the Pythagoras Theorem.

→ The radius of the circle is \frac{x+1}{2} units.

OC = OD = OA = \frac{x+1}{2} units.

OB = x-(\frac{x+1}{2}) = \frac{x-1}{2}

→ In right-angled triangle ∆OBD,

BD2 = OD2 – OB2 (by the Pythagoras Theorem)

\implies \,BD^2 \,= (\frac{x+1}{2})^2 - (\frac{x-1}{2})^2
= \frac{(x^2+1^2+2x)-(x^2+1^2-2x)}{4} = \frac{4x}{4} = x

\implies \,BD \,= \sqrt{x}

This construction gives us a visual, and geometric way of showing that √x exists for all real numbers \sqrt{x} > 0.

Position of √x on the real number line.

→ To know the position of √x on the number line, let us treat the line AC (in previous example) as the number line, with B as zero, then C = 1, and so on.

→ Draw an arc with centre B and radius BD, which intersects the number line at E.
Then, E represents \sqrt{x} on the number line.

➧ We can extend the idea of square roots to cube roots, fourth roots, and in general nth roots, where n is a positive integer.

→ We can define \sqrt[n]{a} for a real number a > 0 and a positive integer n,
let a > 0 be a real number and n be a positive integer, then
\sqrt[n]{a} =b, if\, b^n = a and b>0.

→ The symbol ‘\sqrt’ used in \sqrt{2} , \sqrt[3]{8}, \sqrt[n]{a}, etc. is called the radical sign.

Identities relating to square roots of Real numbers.

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

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

(iii)\, (\sqrt{a} + \sqrt{b}) (\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2

= a - b

(iv)\,(a + \sqrt{b})( a - \sqrt{b}) = (a)^2 - (\sqrt{b})^2

= a^2 - b

(v)\, \left(\sqrt{a} + \sqrt{b}\right)\left(\sqrt{c} + \sqrt{d}\right)
= \sqrt{a}\left(\sqrt{c} + \sqrt{d}\right) + \sqrt{b}\leff(\sqrt{c} + \sqrt{d}\right)
= \sqrt{a}\sqrt{c} + \sqrt{a}\sqrt{d} + \sqrt{b}\sqrt{c} + \sqrt{b}\sqrt{d}

(vi)\, (\sqrt{a} + \sqrt{b})^2 = (\sqrt{a})^2 + 2\sqrt{a}\sqrt{b} +(\sqrt{b})^2

= a + 2\sqrt{ab} + b

Rationalising the denominator

The method of converting a fraction with irrational (roots) denominator to an equivalent fraction with rational denominator is called rationalisation.

→ Thus, rationalization is a process of eliminating radicals from the denominator of a real fraction.

→ When denominator has single term – Multiply the numerator and the denominator with the irrational factor of the denominator.
e.g., \frac{1}{\sqrt{2}} = \frac{1\times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} = \frac{\sqrt{2}}{2}

→ When denominator has two terms (binomial expression) – Multiply the numerator and the denominator with the conjugate of the expression in the denominator.
Conjugate of an expression is the expression with opposite sign in between the terms. e.g., a+b and a-b are conjugate of each other.

After multiplying with conjugate we use the following idenditity
\, (\sqrt{a} + \sqrt{b}) (\sqrt{a} - \sqrt{b}) = (\sqrt{a})^2 - (\sqrt{b})^2 = a-b

\frac{2\sqrt{3}}{4+\sqrt{3}}=\frac{2\sqrt{3} \times (4-\sqrt{3})}{(4+ \sqrt{3}) \times (4-\sqrt{3})} (Rationalising)

= \frac{8\sqrt{3}-6}{13}

Laws of Exponents for Real numbers

➧ We have learned about the following laws of exponents from previous topics, (Here a, n and m are integers. a is the base and m and n are the exponents.)

(i)\, a^{m} \times a^{n} = a^{m+n}

(ii)\, (a^m)^n = a^{mn}

(iii)\, \frac{a^m}{a^n} = a^{m-n}, m > n

(iv)\, a^mb^m = (ab)^m

(v)\, a^0= 1

(vi)\, \frac{1}{a^n} = a^{−n} (putting m= 0 in (iii))

➧ Let, a > 0 be a real number and n be a positive integer. Then \sqrt [n]{a}=b,\, if\, b^n = a and b > 0.

In the language of exponents, we define

\sqrt[n]{a} = a^{\frac{1}{n}}

➧ Let, a > 0 be a real number. Let, m and n be integers such that m and n have no common factors other than 1, and n > 0. Then,

a^{\frac{m}{n}} = (\sqrt[n]{a})^m = \sqrt[n]{a^m}

➧ We now have the following extended laws of exponents for real numbers:

Let a > 0 be a real number and p and q be rational numbers. Then, we have

(i)\, a^p . a^q = a^{p+q}

(ii)\,(a^p)^q = a^{pq}

(iii)\, \frac{a^p}{a^q} = a^{p-q}

(iv)\, a^p . b^p = (ab)^p

(v)\,a^0= 1

(vi)\,a^{-p} = \frac{1}{a^p}

(vii)\,a^{\frac{1}{q}} = \sqrt[q]{a}

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