Fdtxjr

What does Apple's new App Store requirement mean

The Digit Triangles

C++ check if statement can be evaluated constexpr

Permission on Database

How to draw a matrix with arrows in limited space

Multiplicative persistence

Non-trope happy ending?

How do I fix the group tension caused by my character stealing and possibly killing without provocation?

Quoting Keynes in a lecture

Is there a RAID 0 Equivalent for RAM?

Is there a nicer/politer/more positive alternative for "negates"?

Shouldn’t conservatives embrace universal basic income?

Can I turn my anal-retentiveness into a career?

Does an advisor owe his/her student anything? Will an advisor keep a PhD student only out of pity?

How to convince somebody that he is fit for something else, but not this job?

Why is the "ls" command showing permissions of files in a FAT32 partition?

Does the Linux kernel need a file system to run?

Is this part of the description of the Archfey warlock's Misty Escape feature redundant?

Has any country ever had 2 former presidents in jail simultaneously?

How do you make your own symbol when Detexify fails?

Why is it that I can sometimes guess the next note?

Which Article Helped Get Rid of Technobabble in RPGs?

A variation to the phrase "hanging over my shoulders"

Why Shazam when there is already Superman?

What went wrong in proving $i=1$ [duplicate]

Why $sqrt-1 times -1 neq sqrt-1^2$?Which step in this process allows me to erroneously conclude that $i = 1$$1/i=i$. I must be wrong but why?Simplifying $x^i$ to real numbersWhat is wrong with my proof: $-1 = 1$?What's wrong here?Proving a complex inequalityWhat's wrong about this limit?I am getting two answers for the fourth root of i. What am I doing wrong?What is wrong with this proof? -3 = 3Why the wrong result?What went wrong in this fake proof?

-3

$begingroup$

This question already has an answer here:

• 
  Why $sqrt-1 times -1 neq sqrt-1^2$?
  
  9 answers
  
  


• 
  Which step in this process allows me to erroneously conclude that $i = 1$
  
  4 answers
  
  

I started with

$$x=(-16)^frac12$$

$$x=(-16)^frac24$$

Since $$(a^m)^n=a^mn$$ we have:

$$x=((-16)^2)^frac14$$

$$x=((16^2)^frac14$$

$$x=sqrt16=4$$

Hence $$(-16)^frac12=4i=4$$

$$i=1$$

complex-numbers exponentiation radicals fake-proofs

share|cite|improve this question

edited Mar 14 at 15:23

Maria Mazur

47.9k1260120

asked Mar 14 at 15:21

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,61311429

$endgroup$

marked as duplicate by Morgan Rodgers, John Omielan, Dietrich Burde, Xander Henderson, Alex Provost Mar 14 at 19:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $m$ or $n$ are not integers and $a<0$ the formula $a^mn=(a^m)^n$ is false. In fact, for $a<0$ and $rin Bbb Q$ the meaning of $a^r$ is undefined.
  $endgroup$
  – ajotatxe
  Mar 14 at 15:24
  
  

  
  
  
  

add a comment | 

-3

$begingroup$

This question already has an answer here:

• 
  Why $sqrt-1 times -1 neq sqrt-1^2$?
  
  9 answers
  
  


• 
  Which step in this process allows me to erroneously conclude that $i = 1$
  
  4 answers
  
  

I started with

$$x=(-16)^frac12$$

$$x=(-16)^frac24$$

Since $$(a^m)^n=a^mn$$ we have:

$$x=((-16)^2)^frac14$$

$$x=((16^2)^frac14$$

$$x=sqrt16=4$$

Hence $$(-16)^frac12=4i=4$$

$$i=1$$

complex-numbers exponentiation radicals fake-proofs

share|cite|improve this question

edited Mar 14 at 15:23

Maria Mazur

47.9k1260120

asked Mar 14 at 15:21

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,61311429

$endgroup$

marked as duplicate by Morgan Rodgers, John Omielan, Dietrich Burde, Xander Henderson, Alex Provost Mar 14 at 19:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $m$ or $n$ are not integers and $a<0$ the formula $a^mn=(a^m)^n$ is false. In fact, for $a<0$ and $rin Bbb Q$ the meaning of $a^r$ is undefined.
  $endgroup$
  – ajotatxe
  Mar 14 at 15:24
  
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Why $sqrt-1 times -1 neq sqrt-1^2$?
  
  9 answers
  
  


• 
  Which step in this process allows me to erroneously conclude that $i = 1$
  
  4 answers
  
  

I started with

$$x=(-16)^frac12$$

$$x=(-16)^frac24$$

Since $$(a^m)^n=a^mn$$ we have:

$$x=((-16)^2)^frac14$$

$$x=((16^2)^frac14$$

$$x=sqrt16=4$$

Hence $$(-16)^frac12=4i=4$$

$$i=1$$

complex-numbers exponentiation radicals fake-proofs

share|cite|improve this question

edited Mar 14 at 15:23

Maria Mazur

47.9k1260120

asked Mar 14 at 15:21

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,61311429

$endgroup$

share|cite|improve this question

edited Mar 14 at 15:23

Maria Mazur

47.9k1260120

asked Mar 14 at 15:21

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,61311429

share|cite|improve this question

edited Mar 14 at 15:23

Maria Mazur

47.9k1260120

asked Mar 14 at 15:21

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,61311429

edited Mar 14 at 15:23

Maria Mazur

47.9k1260120

edited Mar 14 at 15:23

Maria Mazur

47.9k1260120

asked Mar 14 at 15:21

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,61311429

asked Mar 14 at 15:21

Ekaveera Kumar SharmaEkaveera Kumar Sharma

5,61311429

3 Answers
3

active

oldest

votes

3

$begingroup$

The rule $(a^b)^c = a^bc$ does not hold in general except in special situations such as:

• 
  $a$ is a positive real and $b,c$ are real, or


• 
  $a$ is arbitrary and $b,c$ are integers.

For example: $((-1)^2)^1/2 = 1$, but $(-1)^2cdot 1/2=-1$.

Or: $(e^2pi i)^i = 1^i = 1$, but $e^2pi icdot i = e^-2pi approx 0.002$.

share|cite|improve this answer

edited Mar 14 at 15:41

answered Mar 14 at 15:24

Henning MakholmHenning Makholm

242k17308550

$endgroup$

add a comment | 

1

$begingroup$

The imaginary numbers invalidate some of the usual rules on the exponents.

In the first place,

$$i^2<0.$$

share|cite|improve this answer

answered Mar 14 at 15:27

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

0

$begingroup$

For positives a and b,
$sqrt-a sqrt-b not = sqrtab $
But,
$sqrt-a sqrt-b = - sqrtab $

share|cite|improve this answer

answered Mar 14 at 15:53

AdityaAditya

496

$endgroup$

add a comment | 

3

$begingroup$

The rule $(a^b)^c = a^bc$ does not hold in general except in special situations such as:

• 
  $a$ is a positive real and $b,c$ are real, or


• 
  $a$ is arbitrary and $b,c$ are integers.

For example: $((-1)^2)^1/2 = 1$, but $(-1)^2cdot 1/2=-1$.

Or: $(e^2pi i)^i = 1^i = 1$, but $e^2pi icdot i = e^-2pi approx 0.002$.

share|cite|improve this answer

edited Mar 14 at 15:41

answered Mar 14 at 15:24

Henning MakholmHenning Makholm

242k17308550

$endgroup$

add a comment | 

3

$begingroup$

The rule $(a^b)^c = a^bc$ does not hold in general except in special situations such as:

• 
  $a$ is a positive real and $b,c$ are real, or


• 
  $a$ is arbitrary and $b,c$ are integers.

For example: $((-1)^2)^1/2 = 1$, but $(-1)^2cdot 1/2=-1$.

Or: $(e^2pi i)^i = 1^i = 1$, but $e^2pi icdot i = e^-2pi approx 0.002$.

share|cite|improve this answer

edited Mar 14 at 15:41

answered Mar 14 at 15:24

Henning MakholmHenning Makholm

242k17308550

$endgroup$

add a comment | 

$begingroup$

The rule $(a^b)^c = a^bc$ does not hold in general except in special situations such as:

• 
  $a$ is a positive real and $b,c$ are real, or


• 
  $a$ is arbitrary and $b,c$ are integers.

For example: $((-1)^2)^1/2 = 1$, but $(-1)^2cdot 1/2=-1$.

Or: $(e^2pi i)^i = 1^i = 1$, but $e^2pi icdot i = e^-2pi approx 0.002$.

share|cite|improve this answer

edited Mar 14 at 15:41

answered Mar 14 at 15:24

Henning MakholmHenning Makholm

242k17308550

$endgroup$

share|cite|improve this answer

edited Mar 14 at 15:41

answered Mar 14 at 15:24

Henning MakholmHenning Makholm

242k17308550

answered Mar 14 at 15:24

Henning MakholmHenning Makholm

242k17308550

answered Mar 14 at 15:24

Henning MakholmHenning Makholm

242k17308550

1

$begingroup$

The imaginary numbers invalidate some of the usual rules on the exponents.

In the first place,

$$i^2<0.$$

share|cite|improve this answer

answered Mar 14 at 15:27

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

1

$begingroup$

The imaginary numbers invalidate some of the usual rules on the exponents.

In the first place,

$$i^2<0.$$

share|cite|improve this answer

answered Mar 14 at 15:27

Yves DaoustYves Daoust

131k676229

$endgroup$

add a comment | 

$begingroup$

The imaginary numbers invalidate some of the usual rules on the exponents.

In the first place,

$$i^2<0.$$

share|cite|improve this answer

answered Mar 14 at 15:27

Yves DaoustYves Daoust

131k676229

$endgroup$

share|cite|improve this answer

answered Mar 14 at 15:27

Yves DaoustYves Daoust

131k676229

answered Mar 14 at 15:27

Yves DaoustYves Daoust

131k676229

answered Mar 14 at 15:27

Yves DaoustYves Daoust

131k676229

0

$begingroup$

For positives a and b,
$sqrt-a sqrt-b not = sqrtab $
But,
$sqrt-a sqrt-b = - sqrtab $

share|cite|improve this answer

answered Mar 14 at 15:53

AdityaAditya

496

$endgroup$

add a comment | 

0

$begingroup$

For positives a and b,
$sqrt-a sqrt-b not = sqrtab $
But,
$sqrt-a sqrt-b = - sqrtab $

share|cite|improve this answer

answered Mar 14 at 15:53

AdityaAditya

496

$endgroup$

add a comment | 

$begingroup$

For positives a and b,
$sqrt-a sqrt-b not = sqrtab $
But,
$sqrt-a sqrt-b = - sqrtab $

share|cite|improve this answer

answered Mar 14 at 15:53

AdityaAditya

496

$endgroup$

share|cite|improve this answer

answered Mar 14 at 15:53

AdityaAditya

496

answered Mar 14 at 15:53

AdityaAditya

496

answered Mar 14 at 15:53

AdityaAditya

496