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Question Number 5226 by sanusihammed last updated on 02/May/16
Find b in terms of a if (√(a(a/b))) = ((a/b))^(1/a) . where a and b are whole numbers.
$${Find}\:{b}\:{in}\:{terms}\:{of}\:{a}\:{if}\:\sqrt{{a}\frac{{a}}{{b}}}\:\:\:=\:\:\left(\frac{{a}}{{b}}\right)^{\frac{\mathrm{1}}{{a}}} \:\:\:\:.\:\:{where}\:{a}\:{and}\:{b}\:{are} \\ $$$${whole}\:{numbers}.\:\: \\ $$
Commented by prakash jain last updated on 02/May/16
(((ab+a)/b))^(1/2) =((a/b))^(1/a) (((ab+a)/b))^a =(a^2 /b^2 ) (((ab+a)^a )/a^2 )=(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⇒(a^2 /b)>0⇒b>0 also a≠0 and b≠0 case a>0 (b+1)^a =((b/a))^(a−2) LHS is a +ve interger so RHS must be +ve integer continue
$$\left(\frac{{ab}+{a}}{{b}}\right)^{\mathrm{1}/\mathrm{2}} =\left(\frac{{a}}{{b}}\right)^{\mathrm{1}/{a}} \\ $$$$\left(\frac{{ab}+{a}}{{b}}\right)^{{a}} =\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$$\frac{\left({ab}+{a}\right)^{{a}} }{{a}^{\mathrm{2}} }=\frac{{b}^{{a}} }{{b}^{\mathrm{2}} } \\ $$$${a}^{{a}−\mathrm{2}} \left({b}+\mathrm{1}\right)^{{a}} ={b}^{{a}−\mathrm{2}} \\ $$$$\mathrm{Assume}\:\mathrm{we}\:\mathrm{are}\:\mathrm{only}\:\mathrm{dealing}\:\mathrm{when}\:\mathrm{both} \\ $$$$\mathrm{LHS}\:\mathrm{and}\:\mathrm{RHS}\:\mathrm{are}\:\mathrm{real}\Rightarrow\frac{{a}^{\mathrm{2}} }{{b}}>\mathrm{0}\Rightarrow{b}>\mathrm{0} \\ $$$$\mathrm{also}\:{a}\neq\mathrm{0}\:\mathrm{and}\:{b}\neq\mathrm{0} \\ $$$${case}\:{a}>\mathrm{0} \\ $$$$\left({b}+\mathrm{1}\right)^{{a}} =\left(\frac{{b}}{{a}}\right)^{{a}−\mathrm{2}} \\ $$$$\mathrm{LHS}\:\mathrm{is}\:\mathrm{a}\:+\mathrm{ve}\:\mathrm{interger}\:\mathrm{so}\:\mathrm{RHS}\:\mathrm{must}\:\mathrm{be}\:+\mathrm{ve}\:\mathrm{integer} \\ $$$${continue} \\ $$
Answered by Yozzii last updated on 02/May/16
(a/( (√b)))=(a^(1/a) /b^(1/a) ) a^(1−(1/a)) =b^(0.5−(1/a)) ⇒b=(a^((a−1)/a) )^(1/((1/2)−(1/a))) (a≠2) b=a^(((a−1)/a)×((2a)/(a−2))) =a^((2(a−1))/(a−2)) a∈(W−{2})
$$\frac{{a}}{\:\sqrt{{b}}}=\frac{{a}^{\mathrm{1}/{a}} }{{b}^{\mathrm{1}/{a}} } \\ $$$${a}^{\mathrm{1}−\frac{\mathrm{1}}{{a}}} ={b}^{\mathrm{0}.\mathrm{5}−\frac{\mathrm{1}}{{a}}} \\ $$$$\Rightarrow{b}=\left({a}^{\frac{{a}−\mathrm{1}}{{a}}} \right)^{\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{{a}}}} \:\:\:\:\:\left({a}\neq\mathrm{2}\right) \\ $$$${b}={a}^{\frac{{a}−\mathrm{1}}{{a}}×\frac{\mathrm{2}{a}}{{a}−\mathrm{2}}} ={a}^{\frac{\mathrm{2}\left({a}−\mathrm{1}\right)}{{a}−\mathrm{2}}} \:\:\:{a}\in\left(\mathbb{W}−\left\{\mathrm{2}\right\}\right) \\ $$$$ \\ $$
Commented by FilupSmith last updated on 02/May/16
What is W?
$$\mathrm{What}\:\mathrm{is}\:\mathbb{W}? \\ $$
Commented by sanusihammed last updated on 02/May/16
Sorry it is not a ×(a/b) it is a(a/b) just 3(1/3) as in mixed fraction. Thanks for your help
$${Sorry}\:{it}\:{is}\:{not}\:{a}\:×\frac{{a}}{{b}}\:{it}\:{is}\:{a}\frac{{a}}{{b}}\:\:{just}\:\:\mathrm{3}\frac{\mathrm{1}}{\mathrm{3}}\:{as}\:{in}\:{mixed}\:{fraction}. \\ $$$${Thanks}\:{for}\:{your}\:{help} \\ $$
Commented by Yozzii last updated on 02/May/16
W={whole numbers}
$$\mathbb{W}=\left\{{whole}\:{numbers}\right\} \\ $$
Commented by FilupSmith last updated on 02/May/16
Isn′t that the same as Z?
$$\mathrm{Isn}'\mathrm{t}\:\mathrm{that}\:\mathrm{the}\:\mathrm{same}\:\mathrm{as}\:\mathbb{Z}? \\ $$
Commented by Yozzii last updated on 02/May/16
Answered by Yozzii last updated on 02/May/16
(√(a+(a/b)))=(a^(1/a) /b^(1/a) ) a,b∈{whole numbers} a(1+(1/b))=(a^(2/a) /b^(2/a) ) a^a (1+(1/b))^a =a^2 /b^2 Let u=1/b a^a (1+u)^a =a^2 u^2 Let l=1+u ⇒l^a =a^(2−a) (l−1)^2 l^a =a^(2−a) (l^2 −2l+1) l^a −a^(2−a) l^2 +2a^(2−a) l−a^(2−a) =0 (∗) For a being a whole number, if a≥5 then no general algebraic solution exists for (∗), according to the Abel−Ruffini theorem.
$$\sqrt{{a}+\frac{{a}}{{b}}}=\frac{{a}^{\mathrm{1}/{a}} }{{b}^{\mathrm{1}/{a}} }\:\:\:\:\:\:\:{a},{b}\in\left\{{whole}\:{numbers}\right\} \\ $$$${a}\left(\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)=\frac{{a}^{\mathrm{2}/{a}} }{{b}^{\mathrm{2}/{a}} } \\ $$$${a}^{{a}} \left(\mathrm{1}+\frac{\mathrm{1}}{{b}}\right)^{{a}} ={a}^{\mathrm{2}} /{b}^{\mathrm{2}} \\ $$$${Let}\:{u}=\mathrm{1}/{b} \\ $$$${a}^{{a}} \left(\mathrm{1}+{u}\right)^{{a}} ={a}^{\mathrm{2}} {u}^{\mathrm{2}} \\ $$$${Let}\:{l}=\mathrm{1}+{u} \\ $$$$\Rightarrow{l}^{{a}} ={a}^{\mathrm{2}−{a}} \left({l}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$${l}^{{a}} ={a}^{\mathrm{2}−{a}} \left({l}^{\mathrm{2}} −\mathrm{2}{l}+\mathrm{1}\right) \\ $$$${l}^{{a}} −{a}^{\mathrm{2}−{a}} {l}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}−{a}} {l}−{a}^{\mathrm{2}−{a}} =\mathrm{0}\:\left(\ast\right) \\ $$$${For}\:{a}\:{being}\:{a}\:{whole}\:{number},\:{if}\: \\ $$$${a}\geqslant\mathrm{5}\:{then}\:{no}\:{general}\:{algebraic} \\ $$$${solution}\:{exists}\:{for}\:\left(\ast\right),\:{according}\:{to} \\ $$$${the}\:{Abel}−{Ruffini}\:{theorem}. \\ $$$$ \\ $$

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