Students can go through AP Board 9th Class Maths Notes Chapter 1 Real Numbers to understand and remember the concepts easily.

## AP State Board Syllabus 9th Class Maths Notes Chapter 1 Real Numbers

→ Numbers of the form \(\frac{p}{q}\) where p and q are integers and q ≠ 0 are called rational numbers, represented by ‘Q’.

→ There are infinitely many rational numbers between any two integers.

E.g.: 3 < \(\frac{19}{6}\), \(\frac{20}{6}\), \(\frac{21}{6}\), \(\frac{22}{6}\), \(\frac{23}{6}\), ……. < 4

→ There are infinitely many rational numbers between any two rational numbers.

E.g.: \(\frac{3}{4}\) < \(\frac{29}{8}\) < \(\frac{71}{16}\) < \(\frac{81}{14}\) ……. < \(\frac{13}{2}\)

→ To find the decimal representation of a rational number we divide the numerator of a rational number by its denominator.

E.g.: The decimal representation of \(\frac{5}{6}\) is

∴ \(\frac{5}{6}\) = 0.833 …. = 0.8 \(\overline{3}\)

→ Every rational number can be expressed as a terminating decimal or as a non-terminating repeating decimal. Conversely every terminating decimal or non¬terminating recurring decimal can be expressed as a rational number.

E.g.: 1.6 \(\overline{2}\) = \(\frac{161}{99}\)

→ A rational number whose denominator consists of only 2’s or 5’s or a combination of 2’s and 5’s can be expressed as a terminating decimal.

E.g. : \(\frac{13}{32}\) can be expressed as a terminating decimal (∵ 32 = 2 × 2 × 2 × 2 × 2)

\(\frac{7}{125}\) can be expressed as a terminating decimal (∵ 125 = 5 × 5 × 5)

\(\frac{24}{40}\) can be expressed as a terminating decimal (∵ 40 = 2 × 2 × 2 × 5)

→ Numbers which can’t written in the form \(\frac{p}{q}\) where p and q are integers and q ≠ 0, are called irrational numbers.

E.g.: √2, √3, √5,….. etc.

The decimal form of an irrational number is neither terminating nor recurring decimal.

→ Irrational numbers can be represented on a number line using Pythagoras theorem.

E.g.: Represent √2 on a number line.

→ If ‘n’ is a natural number which is not a perfect square, then √n is always an irrational number.

E.g.: 2, 3, 5, 7, 8, …… etc., are not perfect squares.

∴ √2, √3, √5, √7 and √8 are irrational numbers.

→ We often write π as \(\frac{22}{7}\) there by π seems to be a rational number; but π is not a rational number.

→ The collection of rational numbers together with irrational numbers is called set of Real numbers.

R = Q ∪ S

→ If a and b are two positive rational numbers such that ab is not a perfect square, then , √ab is an irrational number between ‘a’ and b’.

E.g.: Consider any two rational numbers 7 and 4.

7 × 4 = 28 is not a perfect square; then √28 lies between 4 and 7.

i.e., 4 < √28 < 1

→ If ‘a’ is a rational number and ‘b’ is arty irrational number then a + b, a – b, a.b or \(\frac{a}{b}\) is an irrational number.

E.g.: Consider 7 and √5 then 7 + √5, 7 – √5, 7√5 and \(\frac{7}{\sqrt{5}}\)= are all irrational numbers.

→ If the product of any two irrational numbers is a rational number, then they are said to be the rationalising factor of each other.

E.g.: Consider any two irrational number 7√3 and 5√3.

7√3 × 5√3 = 7 × 5 × 3 = 105 a rational number.

Also 7√3 × √3 = 21 – a rational number.

5√3 × √3 = 15 – a rational number.

So the rationalising factor of an irrational number is not unique.

→ The general form of rationalising factor (R.F.) of (a ± √b} is (a ∓ √b). They are called conjugates to each other.

→ Laws of exponents:

i) a^{m} × a^{n} = a^{m+n}

e.g.: 5^{4} . 5^{-3} = 5^{4+(-3)} = 5^{1} = 5

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

e.g.: (4^{3})^{2} = 4^{3×2} = 4^{6}

iii)

= a^{m-n} if (m > n)

= 1 if m = n

= \(\frac{1}{a^{n-m}}\) if (m < n)

iv) a^{m} . b^{m} = (ab)^{m}

e.g.:(-5)^{3} . (2)^{3} = (-5 × 2)^{3} =(-10)^{3}

v) \(\frac{1}{a^{n}}\) = a^{-n}

e.g.: (6)^{-3} = \(\frac{1}{6^{3}}\) = \(\frac{1}{216}\)

vi) a^{0} = 1

e.g.: \(\left(\frac{-3}{4}\right)^{0}\) = 1

Where a, b are rationals and m, n are integers.

→ Let a, b be any two rational numbers such that a = b^{n} then b = \(\sqrt[n]{a}\) = \((\mathrm{a})^{1 / \mathrm{n}}\)

Here ‘b’ is called n^{th} root of a.

e.g.: 4^{2} = 16 then \(16^{1 / 2}\) or \(\sqrt[2]{16}\)

3^{4} = 81 then 3 = \(\sqrt[4]{81}\) or \((81)^{1 / 4}\)

→ Let ‘a’ be a positive number and n > 1 then \(\sqrt[n]{a}\) i.e., n^{th} root of a is called a surd.