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Question Number 14119 by RasheedSindhi last updated on 28/May/17

For what values of n∈N, ω^(1/n) can be expressed as ω^m where m∈Z?

$$\mathrm{For}\:\mathrm{what}\:\mathrm{values}\:\mathrm{of}\:\mathrm{n}\in\mathbb{N}, \\ $$$$\omega^{\mathrm{1}/\mathrm{n}} \:\mathrm{can}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{as}\:\omega^{\mathrm{m}} \\ $$$$\mathrm{where}\:\mathrm{m}\in\mathbb{Z}? \\ $$

Commented by prakash jain last updated on 28/May/17

ω^(1/n) =w^m w=w^(mn) w^3 =w^(mn+2) 1=w^(mn+2) mn+2 needs to be multiple of 3.

$$\omega^{\mathrm{1}/{n}} ={w}^{{m}} \\ $$$${w}={w}^{{mn}} \\ $$$${w}^{\mathrm{3}} ={w}^{{mn}+\mathrm{2}} \\ $$$$\mathrm{1}={w}^{{mn}+\mathrm{2}} \\ $$$${mn}+\mathrm{2}\:\mathrm{needs}\:\mathrm{to}\:\mathrm{be}\:\mathrm{multiple}\:\mathrm{of}\:\mathrm{3}. \\ $$$$ \\ $$

Commented by RasheedSindhi last updated on 28/May/17

mn+2=3k m=((3k−2)/n) m∈Z⇒n∣3k−2 For what values of n,there exists such k that n ∣ 3k−2 ? For example for n=3,there is no such k So ω^(1/3) can′t be expressed as ω^m .

$$\mathrm{mn}+\mathrm{2}=\mathrm{3k} \\ $$$$\mathrm{m}=\frac{\mathrm{3k}−\mathrm{2}}{\mathrm{n}} \\ $$$$\mathrm{m}\in\mathbb{Z}\Rightarrow\mathrm{n}\mid\mathrm{3k}−\mathrm{2} \\ $$$$\mathrm{For}\:\mathrm{what}\:\mathrm{values}\:\mathrm{of}\:\mathrm{n},\mathrm{there} \\ $$$$\mathrm{exists}\:\mathrm{such}\:\mathrm{k}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{n}\:\mid\:\mathrm{3k}−\mathrm{2}\:\:? \\ $$$$\mathrm{For}\:\mathrm{example}\:\mathrm{for}\:\mathrm{n}=\mathrm{3},\mathrm{there} \\ $$$$\mathrm{is}\:\mathrm{no}\:\mathrm{such}\:\mathrm{k} \\ $$$$\mathrm{So}\:\omega^{\mathrm{1}/\mathrm{3}} \:\mathrm{can}'\mathrm{t}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{as}\:\omega^{\mathrm{m}} . \\ $$

Commented by RasheedSindhi last updated on 28/May/17

According to your method ω^(1/4) =(ω^3 ω)^(1/4) =ω ω^(1/5) =((ω^3 )^3 ω)^(1/5) =ω^2

$$\mathrm{According}\:\mathrm{to}\:\mathrm{your}\:\mathrm{method} \\ $$$$\omega^{\mathrm{1}/\mathrm{4}} =\left(\omega^{\mathrm{3}} \omega\right)^{\mathrm{1}/\mathrm{4}} =\omega \\ $$$$\omega^{\mathrm{1}/\mathrm{5}} =\left(\left(\omega^{\mathrm{3}} \right)^{\mathrm{3}} \omega\right)^{\mathrm{1}/\mathrm{5}} =\omega^{\mathrm{2}} \\ $$$$ \\ $$

Commented by prakash jain last updated on 28/May/17

mn+2=3k m=((3k−2)/n) 3 ∤ (3k−2)⇒n≠3j So no m for n=3j n=3j+1 or n=3j+2 For n=3j+1 k=(j+1)⇒3k−2=3(j+1)−2=3j+1 For n=3j+2 k=(3j+2+2j+2) 3k−2=3(3j+2)+(6j+6)−2 =3(3j+2)+(6j+4) =5(3j+2) Summay n=3j⇒no solution n=3j+1⇒k=(j+1)⇒m=1 or (3l+1) n=3j+2⇒k=(5j+4)⇒m=5 or (3l+2)

$${mn}+\mathrm{2}=\mathrm{3}{k} \\ $$$${m}=\frac{\mathrm{3}{k}−\mathrm{2}}{{n}} \\ $$$$\mathrm{3}\:\nmid\:\left(\mathrm{3}{k}−\mathrm{2}\right)\Rightarrow{n}\neq\mathrm{3}{j} \\ $$$$\mathrm{So}\:\mathrm{no}\:{m}\:\mathrm{for}\:{n}=\mathrm{3}{j} \\ $$$${n}=\mathrm{3}{j}+\mathrm{1}\:{or}\:{n}=\mathrm{3}{j}+\mathrm{2} \\ $$$$\mathrm{For}\:{n}=\mathrm{3}{j}+\mathrm{1} \\ $$$${k}=\left({j}+\mathrm{1}\right)\Rightarrow\mathrm{3}{k}−\mathrm{2}=\mathrm{3}\left({j}+\mathrm{1}\right)−\mathrm{2}=\mathrm{3}{j}+\mathrm{1} \\ $$$$\mathrm{For}\:{n}=\mathrm{3}{j}+\mathrm{2} \\ $$$${k}=\left(\mathrm{3}{j}+\mathrm{2}+\mathrm{2}{j}+\mathrm{2}\right) \\ $$$$\mathrm{3}{k}−\mathrm{2}=\mathrm{3}\left(\mathrm{3}{j}+\mathrm{2}\right)+\left(\mathrm{6}{j}+\mathrm{6}\right)−\mathrm{2} \\ $$$$=\mathrm{3}\left(\mathrm{3}{j}+\mathrm{2}\right)+\left(\mathrm{6}{j}+\mathrm{4}\right) \\ $$$$=\mathrm{5}\left(\mathrm{3}{j}+\mathrm{2}\right) \\ $$$$\mathrm{Summay} \\ $$$${n}=\mathrm{3}{j}\Rightarrow{no}\:{solution} \\ $$$${n}=\mathrm{3}{j}+\mathrm{1}\Rightarrow{k}=\left({j}+\mathrm{1}\right)\Rightarrow{m}=\mathrm{1}\:\mathrm{or}\:\left(\mathrm{3}{l}+\mathrm{1}\right) \\ $$$${n}=\mathrm{3}{j}+\mathrm{2}\Rightarrow{k}=\left(\mathrm{5}{j}+\mathrm{4}\right)\Rightarrow{m}=\mathrm{5}\:\mathrm{or}\:\left(\mathrm{3}{l}+\mathrm{2}\right) \\ $$

Commented by RasheedSindhi last updated on 28/May/17

LOT_(of) THαnX Sir!

$$\mathcal{LOT}_{{of}} \mathcal{TH}\alpha{n}\mathcal{X}\:\:\mathcal{S}{ir}! \\ $$

Commented by RasheedSindhi last updated on 29/May/17

TH∀NKS Sir! Learnt one more thing from you. This is something like “Squaring both sides and extraneous roots”.

$$\mathrm{TH}\forall\mathrm{NKS}\:\:\mathrm{Sir}! \\ $$$$\:\mathrm{Learnt}\:\mathrm{one}\:\mathrm{more}\:\mathrm{thing}\:\mathrm{from} \\ $$$$\mathrm{you}. \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{something}\:\mathrm{like} \\ $$$$``\mathrm{Squaring}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{and} \\ $$$$\mathrm{extraneous}\:\mathrm{roots}''. \\ $$

Commented by RasheedSindhi last updated on 29/May/17

I now thought that although for n=3j, there is a solution: ω^(1/3j) =ω^(3/9j) =(ω^3 )^(1/9j) =1=ω^3 =ω^(3k) Is this correct?

$$\mathrm{I}\:\mathrm{now}\:\mathrm{thought}\:\mathrm{that}\:\mathrm{although} \\ $$$$\mathrm{for}\:\mathrm{n}=\mathrm{3j},\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}: \\ $$$$\:\omega^{\mathrm{1}/\mathrm{3j}} =\omega^{\mathrm{3}/\mathrm{9j}} =\left(\omega^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{9j}} =\mathrm{1}=\omega^{\mathrm{3}} =\omega^{\mathrm{3k}} \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{correct}? \\ $$

Commented by prakash jain last updated on 29/May/17

w^(1/2) =w^(3/6) =(w^3 )^(1/6) =1=w^3 ? If u notice w^(1/2) is one of 6^(th) root of unity. it is not correct to say w^(1/2) =1

$${w}^{\mathrm{1}/\mathrm{2}} ={w}^{\mathrm{3}/\mathrm{6}} =\left({w}^{\mathrm{3}} \right)^{\mathrm{1}/\mathrm{6}} =\mathrm{1}={w}^{\mathrm{3}} \:? \\ $$$$\mathrm{If}\:\mathrm{u}\:\mathrm{notice}\:{w}^{\mathrm{1}/\mathrm{2}} \:\mathrm{is}\:\mathrm{one}\:\mathrm{of}\:\mathrm{6}^{\mathrm{th}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{unity}. \\ $$$$\mathrm{it}\:\mathrm{is}\:\mathrm{not}\:\mathrm{correct}\:\mathrm{to}\:\mathrm{say}\:{w}^{\mathrm{1}/\mathrm{2}} =\mathrm{1} \\ $$

Commented by RasheedSindhi last updated on 29/May/17

Sir, no doubt we have reached at wrong result. The question is, where is the error in process? Do we break any rule or restriction of maths?What is that?

$$\mathrm{Sir},\:\mathrm{no}\:\mathrm{doubt}\:\mathrm{we}\:\mathrm{have}\:\mathrm{reached} \\ $$$$\mathrm{at}\:\mathrm{wrong}\:\mathrm{result}.\:\mathrm{The}\:\mathrm{question} \\ $$$$\mathrm{is},\:\mathrm{where}\:\mathrm{is}\:\mathrm{the}\:\mathrm{error}\:\mathrm{in}\:\mathrm{process}? \\ $$$$\mathrm{Do}\:\mathrm{we}\:\mathrm{break}\:\mathrm{any}\:\mathrm{rule}\:\mathrm{or} \\ $$$$\mathrm{restriction}\:\mathrm{of}\:\mathrm{maths}?\mathrm{What}\:\mathrm{is} \\ $$$$\mathrm{that}? \\ $$

Commented by prakash jain last updated on 29/May/17

a^(1/n) =b⇏(b^n )^(1/n) =a^(1/n) Since (b^n )^(1/n) will give n roots of b^n and one of them will be equal to a^(1/n) or b.

$${a}^{\mathrm{1}/{n}} ={b}\nRightarrow\left({b}^{{n}} \right)^{\mathrm{1}/{n}} ={a}^{\mathrm{1}/{n}} \\ $$$$\mathrm{Since}\:\left({b}^{{n}} \right)^{\mathrm{1}/{n}} \:\mathrm{will}\:\mathrm{give}\:{n}\:\mathrm{roots}\:\mathrm{of} \\ $$$${b}^{{n}} \:\mathrm{and}\:\mathrm{one}\:\mathrm{of}\:\mathrm{them}\:\mathrm{will}\:\:\mathrm{be}\:\mathrm{equal}\:\mathrm{to} \\ $$$${a}^{\mathrm{1}/{n}} \:\mathrm{or}\:{b}. \\ $$

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