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Difference between $sqrt x $(square root)and $x^frac12$(half power)

1

$begingroup$

My teacher said that for $sqrtx$ X must belong to integer whereas in $x^frac12$ X belong to entire complex plane. Is there any source for that? How are $sqrtx$ and $x^frac12$ actually defined?

algebra-precalculus

share|cite|improve this question

edited 2 days ago

Fitz Watson

249112

asked 2 days ago

user654700user654700

584

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is ... wrong. $sqrt x equiv x^frac 12$ and both belong to the entire complex plane. Are you sure you haven't misunderstood your teacher?
  $endgroup$
  – Mohammad Zuhair Khan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Why is $x^frac12$ the same as $sqrt x $?
  $endgroup$
  – saket kumar
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now when I look back what he might mean is that $fracab sqrt X $ has no meaning if a/b don't simplify to natural number. Which again opens a new question. Sigh.
  $endgroup$
  – user654700
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saketkumar no they are completely different questions.
  $endgroup$
  – user654700
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure that you don’t mean real number instead of “integer” or “natural number?” The value of $sqrt2$ is neither, but it is a real number.
  $endgroup$
  – amd
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

1

$begingroup$

My teacher said that for $sqrtx$ X must belong to integer whereas in $x^frac12$ X belong to entire complex plane. Is there any source for that? How are $sqrtx$ and $x^frac12$ actually defined?

algebra-precalculus

share|cite|improve this question

edited 2 days ago

Fitz Watson

249112

asked 2 days ago

user654700user654700

584

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That is ... wrong. $sqrt x equiv x^frac 12$ and both belong to the entire complex plane. Are you sure you haven't misunderstood your teacher?
  $endgroup$
  – Mohammad Zuhair Khan
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  4
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Why is $x^frac12$ the same as $sqrt x $?
  $endgroup$
  – saket kumar
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Now when I look back what he might mean is that $fracab sqrt X $ has no meaning if a/b don't simplify to natural number. Which again opens a new question. Sigh.
  $endgroup$
  – user654700
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @saketkumar no they are completely different questions.
  $endgroup$
  – user654700
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you sure that you don’t mean real number instead of “integer” or “natural number?” The value of $sqrt2$ is neither, but it is a real number.
  $endgroup$
  – amd
  2 days ago
  

  
  
  
  

 | 
show 2 more comments

$begingroup$

My teacher said that for $sqrtx$ X must belong to integer whereas in $x^frac12$ X belong to entire complex plane. Is there any source for that? How are $sqrtx$ and $x^frac12$ actually defined?

algebra-precalculus

share|cite|improve this question

edited 2 days ago

Fitz Watson

249112

asked 2 days ago

user654700user654700

584

$endgroup$

share|cite|improve this question

edited 2 days ago

Fitz Watson

249112

asked 2 days ago

user654700user654700

584

share|cite|improve this question

edited 2 days ago

Fitz Watson

249112

asked 2 days ago

user654700user654700

584

edited 2 days ago

Fitz Watson

249112

edited 2 days ago

Fitz Watson

249112

asked 2 days ago

user654700user654700

584

asked 2 days ago

user654700user654700

584

2 Answers
2

active

oldest

votes

2

$begingroup$

There is no difference between $sqrt x$ and $x^1/2$.

However, for the second part of your question, sometimes $f=sqrtcdot$ is understood as a function, e.g. $f:mathbb R_0^+tomathbb R_0^+$ or $f:mathbb Ntomathbb R$. Sometimes (and more infrequently), however, it is understood as a relation. The latter is often needed in the complex plane, where there are multiple branches of the square root.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

st.mathst.math

1,05615

$endgroup$

add a comment | 

0

$begingroup$

I would say that without further specification, they both require that $x$ is a non-negative two number to be well-defined.

Ultimately, they are interchangeable. The difference is mainly about what you want to convey. For $sqrt x$, the $sqrtphantom x$ part is usually quite fixed, while for $x^1/2$, the $^1/2$ part is very much something that can partake in any arithmetic that might happen. At least that's how I feel about them.

share|cite|improve this answer

answered 2 days ago

ArthurArthur

120k7121206

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is $sqrt •$ fixed ? Can't we write $ fracab sqrt x $.
  $endgroup$
  – user654700
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user654700 What do you mean by $frac absqrt x$? Do you mean $sqrt[a/b]x$, the $frac ab$th root of $x$? Sure, we can write that. That's not what I mean. What I mean is that when I write $sqrt x$, I usually intend for it to stay as $sqrtphantom x$. I may change what's outside the root and what's inside the root, or the root might disappear, but I usually wouldn't change the root itself. If I intend for it to change, then I would use fractional exponents instead (for instance, I wouldn't write $sqrt xsqrt[3]x=sqrt[6/5]x$, but instead go for $x^1/2x^1/3=x^5/6$).
  $endgroup$
  – Arthur
  2 days ago
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

There is no difference between $sqrt x$ and $x^1/2$.

However, for the second part of your question, sometimes $f=sqrtcdot$ is understood as a function, e.g. $f:mathbb R_0^+tomathbb R_0^+$ or $f:mathbb Ntomathbb R$. Sometimes (and more infrequently), however, it is understood as a relation. The latter is often needed in the complex plane, where there are multiple branches of the square root.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

st.mathst.math

1,05615

$endgroup$

add a comment | 

2

$begingroup$

There is no difference between $sqrt x$ and $x^1/2$.

However, for the second part of your question, sometimes $f=sqrtcdot$ is understood as a function, e.g. $f:mathbb R_0^+tomathbb R_0^+$ or $f:mathbb Ntomathbb R$. Sometimes (and more infrequently), however, it is understood as a relation. The latter is often needed in the complex plane, where there are multiple branches of the square root.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

st.mathst.math

1,05615

$endgroup$

add a comment | 

$begingroup$

There is no difference between $sqrt x$ and $x^1/2$.

However, for the second part of your question, sometimes $f=sqrtcdot$ is understood as a function, e.g. $f:mathbb R_0^+tomathbb R_0^+$ or $f:mathbb Ntomathbb R$. Sometimes (and more infrequently), however, it is understood as a relation. The latter is often needed in the complex plane, where there are multiple branches of the square root.

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

st.mathst.math

1,05615

$endgroup$

share|cite|improve this answer

edited 2 days ago

answered 2 days ago

st.mathst.math

1,05615

answered 2 days ago

st.mathst.math

1,05615

answered 2 days ago

st.mathst.math

1,05615

0

$begingroup$

I would say that without further specification, they both require that $x$ is a non-negative two number to be well-defined.

Ultimately, they are interchangeable. The difference is mainly about what you want to convey. For $sqrt x$, the $sqrtphantom x$ part is usually quite fixed, while for $x^1/2$, the $^1/2$ part is very much something that can partake in any arithmetic that might happen. At least that's how I feel about them.

share|cite|improve this answer

answered 2 days ago

ArthurArthur

120k7121206

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is $sqrt •$ fixed ? Can't we write $ fracab sqrt x $.
  $endgroup$
  – user654700
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user654700 What do you mean by $frac absqrt x$? Do you mean $sqrt[a/b]x$, the $frac ab$th root of $x$? Sure, we can write that. That's not what I mean. What I mean is that when I write $sqrt x$, I usually intend for it to stay as $sqrtphantom x$. I may change what's outside the root and what's inside the root, or the root might disappear, but I usually wouldn't change the root itself. If I intend for it to change, then I would use fractional exponents instead (for instance, I wouldn't write $sqrt xsqrt[3]x=sqrt[6/5]x$, but instead go for $x^1/2x^1/3=x^5/6$).
  $endgroup$
  – Arthur
  2 days ago
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I would say that without further specification, they both require that $x$ is a non-negative two number to be well-defined.

Ultimately, they are interchangeable. The difference is mainly about what you want to convey. For $sqrt x$, the $sqrtphantom x$ part is usually quite fixed, while for $x^1/2$, the $^1/2$ part is very much something that can partake in any arithmetic that might happen. At least that's how I feel about them.

share|cite|improve this answer

answered 2 days ago

ArthurArthur

120k7121206

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How is $sqrt •$ fixed ? Can't we write $ fracab sqrt x $.
  $endgroup$
  – user654700
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user654700 What do you mean by $frac absqrt x$? Do you mean $sqrt[a/b]x$, the $frac ab$th root of $x$? Sure, we can write that. That's not what I mean. What I mean is that when I write $sqrt x$, I usually intend for it to stay as $sqrtphantom x$. I may change what's outside the root and what's inside the root, or the root might disappear, but I usually wouldn't change the root itself. If I intend for it to change, then I would use fractional exponents instead (for instance, I wouldn't write $sqrt xsqrt[3]x=sqrt[6/5]x$, but instead go for $x^1/2x^1/3=x^5/6$).
  $endgroup$
  – Arthur
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

I would say that without further specification, they both require that $x$ is a non-negative two number to be well-defined.

Ultimately, they are interchangeable. The difference is mainly about what you want to convey. For $sqrt x$, the $sqrtphantom x$ part is usually quite fixed, while for $x^1/2$, the $^1/2$ part is very much something that can partake in any arithmetic that might happen. At least that's how I feel about them.

share|cite|improve this answer

answered 2 days ago

ArthurArthur

120k7121206

$endgroup$

share|cite|improve this answer

answered 2 days ago

ArthurArthur

120k7121206

answered 2 days ago

ArthurArthur

120k7121206

answered 2 days ago

ArthurArthur

120k7121206