Find-b-in-terms-of-a-if-a-a-b-a-b-1-a-where-a-and-b-are-whole-numbers-

Question Number 5226 by sanusihammed last updated on 02/May/16

F i n d b i n t e r m s o f a i f \sqrt{a \frac{a}{b}} = {(\frac{a}{b})}^{\frac{1}{a}} . w h e r e a a n d b a r e

w h o l e n u m b e r s .

Commented by prakash jain last updated on 02/May/16

{(\frac{a b + a}{b})}^{1 / 2} = {(\frac{a}{b})}^{1 / a}

{(\frac{a b + a}{b})}^{a} = \frac{a^{2}}{b^{2}}

\frac{{(a b + a)}^{a}}{a^{2}} = \frac{b^{a}}{b^{2}}

a^{a - 2} {(b + 1)}^{a} = b^{a - 2}

Assume we are only dealing when both

LHS and RHS are real \Rightarrow \frac{a^{2}}{b} > 0 \Rightarrow b > 0

also a \neq 0 and b \neq 0

c a s e a > 0

{(b + 1)}^{a} = {(\frac{b}{a})}^{a - 2}

LHS is a + ve interger so RHS must be + ve integer

c o n t i n u e

Answered by Yozzii last updated on 02/May/16

\frac{a}{\sqrt{b}} = \frac{a^{1 / a}}{b^{1 / a}}

a^{1 - \frac{1}{a}} = b^{0.5 - \frac{1}{a}}

\Rightarrow b = {(a^{\frac{a - 1}{a}})}^{\frac{1}{\frac{1}{2} - \frac{1}{a}}} (a \neq 2)

b = a^{\frac{a - 1}{a} \times \frac{2 a}{a - 2}} = a^{\frac{2 (a - 1)}{a - 2}} a \in (W - {2})

Commented by FilupSmith last updated on 02/May/16

What is W ?

Commented by sanusihammed last updated on 02/May/16

S o r r y i t i s n o t a \times \frac{a}{b} i t i s a \frac{a}{b} j u s t 3 \frac{1}{3} a s i n m i x e d f r a c t i o n .

T h a n k s f o r y o u r h e l p

Commented by Yozzii last updated on 02/May/16

W = {w h o l e n u m b e r s}

Commented by FilupSmith last updated on 02/May/16

{Isn}^{'} t that the same as Z ?

Commented by Yozzii last updated on 02/May/16

Answered by Yozzii last updated on 02/May/16

\sqrt{a + \frac{a}{b}} = \frac{a^{1 / a}}{b^{1 / a}} a, b \in {w h o l e n u m b e r s}

a (1 + \frac{1}{b}) = \frac{a^{2 / a}}{b^{2 / a}}

a^{a} {(1 + \frac{1}{b})}^{a} = a^{2} / b^{2}

L e t u = 1 / b

a^{a} {(1 + u)}^{a} = a^{2} u^{2}

L e t l = 1 + u

\Rightarrow l^{a} = a^{2 - a} {(l - 1)}^{2}

l^{a} = a^{2 - a} (l^{2} - 2 l + 1)

l^{a} - a^{2 - a} l^{2} + 2 a^{2 - a} l - a^{2 - a} = 0 (*)

F o r a b e i n g a w h o l e n u m b e r, i f

a ⩾ 5 t h e n n o g e n e r a l a l g e b r a i c

s o l u t i o n e x i s t s f o r (*), a c c o r d i n g t o

t h e A b e l - R u f f i n i t h e o r e m .

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