please prove (1/x) = x^(−1)


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Question Number 213725 by zakirullah last updated on 14/Nov/24

$p l e a s e p r o v e \frac{1}{x} = x^{- 1}$
	Answered by Rasheed.Sindhi last updated on 14/Nov/24

	$\frac{1}{x} = \frac{x^{0}}{x^{1}} = x^{0 - 1} = x^{- 1}$
	Answered by Rasheed.Sindhi last updated on 14/Nov/24

	$x^{- 1} = x^{0 - 1} = \frac{x^{0}}{x^{1}} = \frac{1}{x}$
	Commented by zakirullah last updated on 15/Nov/24

	$v e r y t r u e b u t s o m e s t u d e n t s a s k e d a b o u t$ $t h e p o w e r o r e x p o n e n t t h a t w h y t h e y c h a n g e i t s$ $s i g n w h e n i t g o f r o m n o m e r a t o r t o d e n o m i n a t o r a n d$ $f r o m d e n o m i n a t o r t o n o m e r a t o r ?$ $p l e a s e g i v e m e a n y s u g g i s t i o n .$
	Commented by Rasheed.Sindhi last updated on 15/Nov/24
	$(1) a^0 =1 (definition) (1/a)=((1.a^(−1) )/(a.a^(−1) )) (numerator and denominator may be multiplied by same number) =(a^(−1) /a^(1+(−1)) )=(a^(−1) /a^0 )=(a^(−1) /1)=a^(−1) (2) { ((a^x ×a^y =a^(x+y) )),(((a^x /a^y )=a^(x−y) (∗))) :}(As Rules) 1=(a/a)=a^(1−1) =a^0 ⇒a^0 =1 Now, (1/a^n )=(a^0 /a^n )=a^(0−n) =a^(−n) (∗) a^x ×a^y =a^(x+y) ⇒a^x =(a^(x+y) /a^y )=a^((x+y)−y) ⇒(a^m /a^n )=a^(m−n)$
	$(1)$ $a^{0} = 1 (d e f i n i t i o n)$ $\frac{1}{a} = \frac{1 . a^{- 1}}{a . a^{- 1}}$ $(n u m e r a t o r a n d d e n o m i n a t o r m a y$ $b e m u l t i p l i e d b y s a m e n u m b e r)$ $= \frac{a^{- 1}}{a^{1 + (- 1)}} = \frac{a^{- 1}}{a^{0}} = \frac{a^{- 1}}{1} = a^{- 1}$ $(2)$ ${\begin{cases} a^{x} \times a^{y} = a^{x + y} \\ \frac{a^{x}}{a^{y}} = a^{x - y} () \end{cases} (A s R u l e s)$ $1 = \frac{a}{a} = a^{1 - 1} = a^{0} \Rightarrow a^{0} = 1$ $N o w,$ $\frac{1}{a^{n}} = \frac{a^{0}}{a^{n}} = a^{0 - n} = a^{- n}$ $() a^{x} \times a^{y} = a^{x + y} \Rightarrow a^{x} = \frac{a^{x + y}}{a^{y}} = a^{(x + y) - y}$ $\Rightarrow \frac{a^{m}}{a^{n}} = a^{m - n}$
	Answered by Ghisom last updated on 14/Nov/24

	$we define the symbol a^{n}$ $a^{n} := \underset{n times}{\underset{⏟}{a \times a \times a \times . . .}}$ $a \times b = c \Leftrightarrow a = b \div c$ $\Rightarrow a^{m} \times a^{n} = a^{m + n} \Leftrightarrow a^{m + n} \div a^{n} = a^{m}$ $this can be written as$ $a^{m + n} \div a^{n} = a^{(m + n) - n}$ $generally$ $a^{m} \div a^{n} = a^{m - n}$ $what happens when m = n ?$ $a^{m} \div a^{m} = a^{m - m} = a^{0}$ $but a^{m} \div a^{m} = 1 \Rightarrow a^{0} = 1$ $what happens when m < n ?$ $let n = m + k$ $a^{m} \div a^{m + k} = a^{m - (m + k)} = a^{- k}$ $an example :$ $2^{3} \div 2^{5} = 2^{- 2} (?)$ $2^{3} \div 2^{5} = 8 \div 32 = \frac{1}{4} = \frac{1}{2^{2}}$ $\Rightarrow 2^{- 2} = \frac{1}{2^{2}}$ $\Rightarrow$ $a^{- m} = \frac{1}{a^{m}}$
	Commented by zakirullah last updated on 15/Nov/24

	$w e l l - d o n e s i r G$
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