For-what-values-of-n-N-1-n-can-be-expressed-as-m-where-m-Z-

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

For what values of n \in N,

ω^{1 / n} can be expressed as ω^{m}

where m \in Z ?

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

ω^{1 / n} = w^{m}

w = w^{m n}

w^{3} = w^{m n + 2}

1 = w^{m n + 2}

m n + 2 needs to be multiple of 3 .

Commented by RasheedSindhi last updated on 28/May/17

mn + 2 = 3 k

m = \frac{3 k - 2}{n}

m \in Z \Rightarrow n ∣ 3 k - 2

For what values of n, there

exists such k that

n ∣ 3 k - 2 ?

For example for n = 3, there

is no such k

So ω^{1 / 3} {can}^{'} t be expressed as ω^{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}

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

m n + 2 = 3 k

m = \frac{3 k - 2}{n}

3 ∤ (3 k - 2) \Rightarrow n \neq 3 j

So no m for n = 3 j

n = 3 j + 1 o r n = 3 j + 2

For n = 3 j + 1

k = (j + 1) \Rightarrow 3 k - 2 = 3 (j + 1) - 2 = 3 j + 1

For n = 3 j + 2

k = (3 j + 2 + 2 j + 2)

3 k - 2 = 3 (3 j + 2) + (6 j + 6) - 2

= 3 (3 j + 2) + (6 j + 4)

= 5 (3 j + 2)

Summay

n = 3 j \Rightarrow n o s o l u t i o n

n = 3 j + 1 \Rightarrow k = (j + 1) \Rightarrow m = 1 or (3 l + 1)

n = 3 j + 2 \Rightarrow k = (5 j + 4) \Rightarrow m = 5 or (3 l + 2)

Commented by RasheedSindhi last updated on 28/May/17

{LOT}_{o f} TH α n X S i r!

Commented by RasheedSindhi last updated on 29/May/17

TH \forall NKS Sir!

Learnt one more thing from

you .

This is something like

“ Squaring both sides and

extraneous {roots}^{″} .

Commented by RasheedSindhi last updated on 29/May/17

I now thought that although

for n = 3 j, there is a solution :

ω^{1 / 3 j} = ω^{3 / 9 j} = {(ω^{3})}^{1 / 9 j} = 1 = ω^{3} = ω^{3 k}

Is this 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

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 ?

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 .