For which $p$ is $(-1)^p$ real?

Clash Royale CLAN TAG#URR8PPP

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This should be a simple question but I am confused by it. For which real $pin mathbbR$ is $(-1)^p$ real?

Clearly it is real for any integer $p$ but is it true for any non-integer $p$?

algebra-precalculus

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asked Jul 22 at 10:55

Anush

190224

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  How do you define $(-1)^p$ for non-integer $p?$
  – gammatester
  Jul 22 at 11:05
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

This should be a simple question but I am confused by it. For which real $pin mathbbR$ is $(-1)^p$ real?

Clearly it is real for any integer $p$ but is it true for any non-integer $p$?

algebra-precalculus

share|cite|improve this question

asked Jul 22 at 10:55

Anush

190224

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  How do you define $(-1)^p$ for non-integer $p?$
  – gammatester
  Jul 22 at 11:05
  

  
  
  
  

add a comment | 

up vote
4
down vote

favorite

up vote
4
down vote

favorite

This should be a simple question but I am confused by it. For which real $pin mathbbR$ is $(-1)^p$ real?

Clearly it is real for any integer $p$ but is it true for any non-integer $p$?

algebra-precalculus

share|cite|improve this question

asked Jul 22 at 10:55

Anush

190224

This should be a simple question but I am confused by it. For which real $pin mathbbR$ is $(-1)^p$ real?

Clearly it is real for any integer $p$ but is it true for any non-integer $p$?

algebra-precalculus

share|cite|improve this question

asked Jul 22 at 10:55

Anush

190224

share|cite|improve this question

share|cite|improve this question

asked Jul 22 at 10:55

Anush

190224

asked Jul 22 at 10:55

Anush

190224

asked Jul 22 at 10:55

Anush

190224

190224

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  How do you define $(-1)^p$ for non-integer $p?$
  – gammatester
  Jul 22 at 11:05
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  How do you define $(-1)^p$ for non-integer $p?$
  – gammatester
  Jul 22 at 11:05
  

  
  
  
  

2

2

How do you define $(-1)^p$ for non-integer $p?$
– gammatester
Jul 22 at 11:05

How do you define $(-1)^p$ for non-integer $p?$
– gammatester
Jul 22 at 11:05

add a comment | 

2 Answers
2

active

oldest

votes

up vote
4
down vote



accepted

hint
$$left( -1 right)^p=left( e^ipi +2pi k right)^p=e^left( ipi +2pi k right)p=cos left( pleft( pi +2pi k right) right)+isin left( pleft( pi +2pi k right) right)$$

share|cite|improve this answer

edited Jul 22 at 11:26

answered Jul 22 at 11:09

user577488

1085

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You could also write $left( -1 right)^p=left( e^i(pi+2pi k) right)^p=e^i(pi+2pi k) p=cos( p(pi+2pi k))+isin( p(pi+2pi k))$ and that makes the difference.
  – Hugocito
  Jul 22 at 11:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks Hugocito i have edited
  – user577488
  Jul 22 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What about $k=1, p=1/3?$
  – gammatester
  Jul 22 at 11:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user577488 My question is a little ambiguous I realise. If $p$ is the reciprocal of an odd integer then $(-1)^p$ has a real root.
  – Anush
  Jul 22 at 11:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sin(frac13(pi+2times 1 times pi)) = 0$
  – gammatester
  Jul 22 at 11:33
  

  
  
  
  

 | 
show 6 more comments

up vote
3
down vote

By definition $(-1)^p = e^plog(-1)$ where $log(-1) = (pi+2pi k)i$ with $kin mathbb Z$. So, essentially, $(-1)^p$ is not a number, but a list of numbers, like for example $sqrt-1$, which is $i$ or $-i$.

So we have
$$(-1)^p = e^p(pi+2kpi)i quad forall kin mathbb Z.$$

Now, in order to all these numbers be real we need $p(pi+2pi k) in pimathbb Z$ for all $k$, which means $pin mathbb Z$.

If $p$ is even then $(-1)^p = 1$ and if $p$ is odd $(-1)^p = -1$.

share|cite|improve this answer

answered Jul 22 at 11:10

Hugocito

1,6451019

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote



accepted

hint
$$left( -1 right)^p=left( e^ipi +2pi k right)^p=e^left( ipi +2pi k right)p=cos left( pleft( pi +2pi k right) right)+isin left( pleft( pi +2pi k right) right)$$

share|cite|improve this answer

edited Jul 22 at 11:26

answered Jul 22 at 11:09

user577488

1085

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You could also write $left( -1 right)^p=left( e^i(pi+2pi k) right)^p=e^i(pi+2pi k) p=cos( p(pi+2pi k))+isin( p(pi+2pi k))$ and that makes the difference.
  – Hugocito
  Jul 22 at 11:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks Hugocito i have edited
  – user577488
  Jul 22 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What about $k=1, p=1/3?$
  – gammatester
  Jul 22 at 11:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user577488 My question is a little ambiguous I realise. If $p$ is the reciprocal of an odd integer then $(-1)^p$ has a real root.
  – Anush
  Jul 22 at 11:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sin(frac13(pi+2times 1 times pi)) = 0$
  – gammatester
  Jul 22 at 11:33
  

  
  
  
  

 | 
show 6 more comments

up vote
4
down vote



accepted

hint
$$left( -1 right)^p=left( e^ipi +2pi k right)^p=e^left( ipi +2pi k right)p=cos left( pleft( pi +2pi k right) right)+isin left( pleft( pi +2pi k right) right)$$

share|cite|improve this answer

edited Jul 22 at 11:26

answered Jul 22 at 11:09

user577488

1085

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You could also write $left( -1 right)^p=left( e^i(pi+2pi k) right)^p=e^i(pi+2pi k) p=cos( p(pi+2pi k))+isin( p(pi+2pi k))$ and that makes the difference.
  – Hugocito
  Jul 22 at 11:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks Hugocito i have edited
  – user577488
  Jul 22 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What about $k=1, p=1/3?$
  – gammatester
  Jul 22 at 11:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user577488 My question is a little ambiguous I realise. If $p$ is the reciprocal of an odd integer then $(-1)^p$ has a real root.
  – Anush
  Jul 22 at 11:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sin(frac13(pi+2times 1 times pi)) = 0$
  – gammatester
  Jul 22 at 11:33
  

  
  
  
  

 | 
show 6 more comments

up vote
4
down vote



accepted

up vote
4
down vote



accepted

hint
$$left( -1 right)^p=left( e^ipi +2pi k right)^p=e^left( ipi +2pi k right)p=cos left( pleft( pi +2pi k right) right)+isin left( pleft( pi +2pi k right) right)$$

share|cite|improve this answer

edited Jul 22 at 11:26

answered Jul 22 at 11:09

user577488

1085

hint
$$left( -1 right)^p=left( e^ipi +2pi k right)^p=e^left( ipi +2pi k right)p=cos left( pleft( pi +2pi k right) right)+isin left( pleft( pi +2pi k right) right)$$

share|cite|improve this answer

edited Jul 22 at 11:26

answered Jul 22 at 11:09

user577488

1085

share|cite|improve this answer

share|cite|improve this answer

edited Jul 22 at 11:26

edited Jul 22 at 11:26

edited Jul 22 at 11:26

answered Jul 22 at 11:09

user577488

1085

answered Jul 22 at 11:09

user577488

1085

answered Jul 22 at 11:09

user577488

1085

1085

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You could also write $left( -1 right)^p=left( e^i(pi+2pi k) right)^p=e^i(pi+2pi k) p=cos( p(pi+2pi k))+isin( p(pi+2pi k))$ and that makes the difference.
  – Hugocito
  Jul 22 at 11:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks Hugocito i have edited
  – user577488
  Jul 22 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What about $k=1, p=1/3?$
  – gammatester
  Jul 22 at 11:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user577488 My question is a little ambiguous I realise. If $p$ is the reciprocal of an odd integer then $(-1)^p$ has a real root.
  – Anush
  Jul 22 at 11:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sin(frac13(pi+2times 1 times pi)) = 0$
  – gammatester
  Jul 22 at 11:33
  

  
  
  
  

 | 
show 6 more comments

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  You could also write $left( -1 right)^p=left( e^i(pi+2pi k) right)^p=e^i(pi+2pi k) p=cos( p(pi+2pi k))+isin( p(pi+2pi k))$ and that makes the difference.
  – Hugocito
  Jul 22 at 11:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  thanks Hugocito i have edited
  – user577488
  Jul 22 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What about $k=1, p=1/3?$
  – gammatester
  Jul 22 at 11:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @user577488 My question is a little ambiguous I realise. If $p$ is the reciprocal of an odd integer then $(-1)^p$ has a real root.
  – Anush
  Jul 22 at 11:32
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sin(frac13(pi+2times 1 times pi)) = 0$
  – gammatester
  Jul 22 at 11:33
  

  
  
  
  

1

1

You could also write $left( -1 right)^p=left( e^i(pi+2pi k) right)^p=e^i(pi+2pi k) p=cos( p(pi+2pi k))+isin( p(pi+2pi k))$ and that makes the difference.
– Hugocito
Jul 22 at 11:17

You could also write $left( -1 right)^p=left( e^i(pi+2pi k) right)^p=e^i(pi+2pi k) p=cos( p(pi+2pi k))+isin( p(pi+2pi k))$ and that makes the difference.
– Hugocito
Jul 22 at 11:17

thanks Hugocito i have edited
– user577488
Jul 22 at 11:26

thanks Hugocito i have edited
– user577488
Jul 22 at 11:26

What about $k=1, p=1/3?$
– gammatester
Jul 22 at 11:28

What about $k=1, p=1/3?$
– gammatester
Jul 22 at 11:28

1

1

@user577488 My question is a little ambiguous I realise. If $p$ is the reciprocal of an odd integer then $(-1)^p$ has a real root.
– Anush
Jul 22 at 11:32

@user577488 My question is a little ambiguous I realise. If $p$ is the reciprocal of an odd integer then $(-1)^p$ has a real root.
– Anush
Jul 22 at 11:32

1

1

$sin(frac13(pi+2times 1 times pi)) = 0$
– gammatester
Jul 22 at 11:33

$sin(frac13(pi+2times 1 times pi)) = 0$
– gammatester
Jul 22 at 11:33

 | 
show 6 more comments

up vote
3
down vote

By definition $(-1)^p = e^plog(-1)$ where $log(-1) = (pi+2pi k)i$ with $kin mathbb Z$. So, essentially, $(-1)^p$ is not a number, but a list of numbers, like for example $sqrt-1$, which is $i$ or $-i$.

So we have
$$(-1)^p = e^p(pi+2kpi)i quad forall kin mathbb Z.$$

Now, in order to all these numbers be real we need $p(pi+2pi k) in pimathbb Z$ for all $k$, which means $pin mathbb Z$.

If $p$ is even then $(-1)^p = 1$ and if $p$ is odd $(-1)^p = -1$.

share|cite|improve this answer

answered Jul 22 at 11:10

Hugocito

1,6451019

add a comment | 

up vote
3
down vote

By definition $(-1)^p = e^plog(-1)$ where $log(-1) = (pi+2pi k)i$ with $kin mathbb Z$. So, essentially, $(-1)^p$ is not a number, but a list of numbers, like for example $sqrt-1$, which is $i$ or $-i$.

So we have
$$(-1)^p = e^p(pi+2kpi)i quad forall kin mathbb Z.$$

Now, in order to all these numbers be real we need $p(pi+2pi k) in pimathbb Z$ for all $k$, which means $pin mathbb Z$.

If $p$ is even then $(-1)^p = 1$ and if $p$ is odd $(-1)^p = -1$.

share|cite|improve this answer

answered Jul 22 at 11:10

Hugocito

1,6451019

add a comment | 

up vote
3
down vote

up vote
3
down vote

By definition $(-1)^p = e^plog(-1)$ where $log(-1) = (pi+2pi k)i$ with $kin mathbb Z$. So, essentially, $(-1)^p$ is not a number, but a list of numbers, like for example $sqrt-1$, which is $i$ or $-i$.

So we have
$$(-1)^p = e^p(pi+2kpi)i quad forall kin mathbb Z.$$

Now, in order to all these numbers be real we need $p(pi+2pi k) in pimathbb Z$ for all $k$, which means $pin mathbb Z$.

If $p$ is even then $(-1)^p = 1$ and if $p$ is odd $(-1)^p = -1$.

share|cite|improve this answer

answered Jul 22 at 11:10

Hugocito

1,6451019

By definition $(-1)^p = e^plog(-1)$ where $log(-1) = (pi+2pi k)i$ with $kin mathbb Z$. So, essentially, $(-1)^p$ is not a number, but a list of numbers, like for example $sqrt-1$, which is $i$ or $-i$.

So we have
$$(-1)^p = e^p(pi+2kpi)i quad forall kin mathbb Z.$$

Now, in order to all these numbers be real we need $p(pi+2pi k) in pimathbb Z$ for all $k$, which means $pin mathbb Z$.

If $p$ is even then $(-1)^p = 1$ and if $p$ is odd $(-1)^p = -1$.

share|cite|improve this answer

answered Jul 22 at 11:10

Hugocito

1,6451019

share|cite|improve this answer

share|cite|improve this answer

answered Jul 22 at 11:10

Hugocito

1,6451019

answered Jul 22 at 11:10

Hugocito

1,6451019

answered Jul 22 at 11:10

Hugocito

1,6451019

1,6451019

add a comment | 

add a comment | 

 

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