Indices and their Laws | Advanced Mathematics | Class 8
Indices and their Laws | Advanced Mathematics | Class 8
Whenever we add a number to itself multiple times, we are essentially performing a multiplication, e.g., \((2+2+2)\) is the same as \((2\times 3)\) as both the cases result \(6\). Multiplication in a generalized form is therefore the product of the number and the number of times it is being added. On the otherhand, if we multiply a particular number repeatedly, it is called exponentiation and the product is called power of the number. So, exponentiation ivolves raising a number to a certain power by multiplying it by itself a specified number of times. It is a fundamental mathematical operation where two numbers are necessary, one is written in the superscript of the other and is called index or exponent and the other number is called base of the number to which the power is raised using the index.
An index, also known as an exponent, in an expression, indicates how many times a number is multiplied by itself. In that case, the number which is multiplied repeatedly is the base of the expression.
(i) when \(\ a \) is a positive integer:
\(\ a^n = a\times a\times a\times a\times a\times ....\) up to \(n\) factor
Here, \(\ a \) is called the base of the number and \(\ n \) is called the index or exponent and \(\ a^n \) is called the power of the number.
For example, \(\ 5^3 = 5\times 5\times 5 \)
(ii) when \(\ n \) is a negative integer:
\(\ a^{-n} = \frac{1}{a^n}; a ≠ 0 \)
Here, \(\ a^{-n} \) is called the inverse of \(\ a^n \)
Laws of Indices
1. If \(\ a \) is some real number and \(\ m \) and \(\ n \) are some natural numbers,then \(\ {a^m}\times {a^n} = a^{m+n} \)
Proof:
LHS = \(\ {a^m}\times {a^n} \)
\(\ = (a\times a\times a\times a\times ....\) up to \(m\) factor)
\(\ \times (a\times a\times a\times ....\) up to \(n\) factor)
\(\ = a\times a\times a\times a....\) up to \((m+n)\) factor
\(\ = a^{m+n} \) = RHS
2. If \(\ a \) is real number and \(\ m \) and \(\ n \) are some natural numbers,
then \(\ {a^m}÷{a^n} = a^{m-n} \)
Proof:
Case I: when \(\ m > n \) and \(a ≠ 0\)
LHS = \(\ {a^m}÷{a^n} \)
\(\ = \frac{a^m}{a^n} \)
\(\ = a^m\times a^{-n} \)
\(\ = [a\times a\times a\times ....\) up to \(m\) factor]
\(\ \times [a\times a\times a\times ....\) up to \(-n\) factor]
\(\ = a\times a\times a\times a\times .....\) up to \([m+(-n)]\) factor
\(\ = a\times a\times a\times a\times a\times .....\) up to \((m-n)\) factor
\(\ = a^{m-n} \) = RHS
Case II: when \(\ m = n \)
LHS = \(\ \frac{a^m}{a^n} \)
\(\ = a^m\times a^{-n} \)
\(\ = [a\times a\times a\times ....\) up to \(m\) factor]
\(\ \times [a\times a\times a\times ....\) up to \(-n\) factor]
\(\ = a\times a\times a\times a\times .....\) up to \([m+(-n)]\) factor
\(\ = a\times a\times a\times a\times a\times .....\) up to \((m-n)\) factor
\(\ = a^{m-n} \) = RHS
\(\ ⇨ {a^m}÷{a^n} = a^{m-n}\) ...(1)
Since, \(\ m = n \)
Eqn (1) ⇨ \(\ \frac{a^m}{a^m} = a^{m-m} \) ⇨ \(\ 1 = a^{0} \)
And the result \(\ a^{0} = 1 \) is another law of indices, where \(\ a ≠ 0 \)
Case III: when \(\ m < n \)
LHS = \(\ {a^m}/{a^n} \)
\(\ = {1\over{a^na^{-m}}} \)
\(={1\over{[a\times a\times ..~up~to~n~factor]\times [a\times a\times .... up~to~-m~ factor]}}\)
\(\ = {1\over{a\times a\times a\times a\times ...~up~to~{n+(-m)}~factor}} \)
\(\ = {1\over{a^{n-m}}} \)
\(\ = a^{-(n-m)} \)
\(\ = a^{m-n} \) = RHS
3. If \(\ a \) is a natural number and \(\ m \) and \(\ n \) are some natural numbers, then \(\ {(a^m)}^n = a^{mn} \)
Proof:
LHS = \(\ {(a^m)}^n \)
\(\ = [{a\times a\times a\times a....~ up~ to~ m~ factor}]^n \)
\(\ = [a\times a\times a\times a....~ up~ to~ m~ factor] \)
\(\ \times [a\times a\times a\times a....~ up~ to~ m~ factor] \)
\(\ \times [a\times a\times a\times a....~ up~ to~ m~ factor] \)
\(\ .....~ up~ to~ n~ factor \)
\(\ = a\times a\times a\times a....~ up~ to~ m\times n~ factor \)
\(\ = a^{mn} \) = RHS
4. If \(\ a \) and \(\ b \) are two real numbers and \(\ n \) is some natural number, then \(\ \left(ab\right)^n = a^nb^n \)
Proof:
LHS = \(\ \left(ab\right)^n \)
\(\ = (ab)^n \)
\(\ = (ab)(ab)(ab)(ab).....~ n~ factor \)
\(\ = (a\times a\times a\times ....~ n~ factor) \)
\(\ (b\times b\times b\times ....~ n~ factor) \)
\(\ = a^nb^n \) = RHS
5. If \(\ a \) and \(\ b \) are two real numbers and \(\ n \) is some natural numbers, then \(\ \left(\frac{a}{b}\right)^n=\frac{a^n}{b^n} \)
Proof:
LHS = \(\ \left(\frac{a}{b}\right)^n \)
\(\ = \left(\frac{a}{b}\right)\left(\frac{a}{b}\right)\left(\frac{a}{b}\right)\left(\frac{a}{b}\right)......up~ to~ n~ factor \)
\(\ = \frac{a\times a\times a\times .... n~ factor}{b\times b\times b\times .... n~ factor} \)
\(\ = \frac{a^n}{b^n} \) = RHS
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