Does $sqrt{-1cdot-1}=1$ or $-1$? [duplicate]












-2












$begingroup$



This question already has an answer here:




  • Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?

    9 answers




Let's define $x=sqrt{ab}$, where $a=-1$ and $b=-1$.



Does $x=1$, as $-1cdot-1=1impliessqrt{-1cdot-1}=sqrt{1}=1$?



Or maybe $x=-1$, as $sqrt{a^2}=aimpliessqrt{-1cdot-1}=sqrt{(-1)^2}=-1$?



I assume that at least one of the following is true:




  1. The former solution is true, and the fallacy is in assuming $sqrt{a^2}=a$ for a negative base $a$. This doesn't seem to hold, given that this is equivalent to saying that $a^{2^{frac12}}not=a^{2cdotfrac12}=a$ for a negative $a$, where the value of $a$ seemingly has no bearing on the arithmetic performed with the exponents.

  2. The former solution is true, and the fallacy is in assuming the positive root is meant in the latter solution; in fact, $sqrt{-1cdot-1}=-sqrt{(-1)^2}=--1=1$. But this implies that, depending on how the radicand is factored, one is forced to take the positive root for one factoring and the negative root for another factoring to get the same answer, which doesn't seem right.

  3. The latter solution is true, and the fallacy is in assuming the positive root is meant in the former solution; in fact, $sqrt{-1cdot-1}=-sqrt1=-1$. But, in addition to the problem with the previous solution, this additionally means that $sqrt{ab}=sqrt{a}sqrt{b}$ holds even for negative $a,b$.


Which of my reasonings is incorrect? Or are all of them correct, and there's an additional solution that I'm not seeing?










share|cite|improve this question









$endgroup$



marked as duplicate by KReiser, JMoravitz, Adrian Keister, metamorphy, Community Jan 4 at 1:47


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




    $begingroup$
    $sqrt{x^2} = |x|$
    $endgroup$
    – krirkrirk
    Jan 3 at 20:57










  • $begingroup$
    Every complex number, apart from zero, has two square roots.
    $endgroup$
    – Lord Shark the Unknown
    Jan 3 at 21:01










  • $begingroup$
    You can do the same thing with any square: $sqrt{25}$ = $sqrt{-5 cdot -5}$. for example. But we take the positive branch and define $sqrt{25}$ = 5.
    $endgroup$
    – Joel Pereira
    Jan 3 at 21:03












  • $begingroup$
    @JoelPereira I know, I thought I'd take the most basic case, though.
    $endgroup$
    – DonielF
    Jan 3 at 21:04






  • 1




    $begingroup$
    I don't understand the downvotes on this. It's perfectly reasonable to ask where the fallacious step is in a well-known fallacy, and it's clear enough what's being ssked. Also the questioner's own thoughts are included in quite a lot of detail
    $endgroup$
    – timtfj
    Jan 4 at 1:38


















-2












$begingroup$



This question already has an answer here:




  • Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?

    9 answers




Let's define $x=sqrt{ab}$, where $a=-1$ and $b=-1$.



Does $x=1$, as $-1cdot-1=1impliessqrt{-1cdot-1}=sqrt{1}=1$?



Or maybe $x=-1$, as $sqrt{a^2}=aimpliessqrt{-1cdot-1}=sqrt{(-1)^2}=-1$?



I assume that at least one of the following is true:




  1. The former solution is true, and the fallacy is in assuming $sqrt{a^2}=a$ for a negative base $a$. This doesn't seem to hold, given that this is equivalent to saying that $a^{2^{frac12}}not=a^{2cdotfrac12}=a$ for a negative $a$, where the value of $a$ seemingly has no bearing on the arithmetic performed with the exponents.

  2. The former solution is true, and the fallacy is in assuming the positive root is meant in the latter solution; in fact, $sqrt{-1cdot-1}=-sqrt{(-1)^2}=--1=1$. But this implies that, depending on how the radicand is factored, one is forced to take the positive root for one factoring and the negative root for another factoring to get the same answer, which doesn't seem right.

  3. The latter solution is true, and the fallacy is in assuming the positive root is meant in the former solution; in fact, $sqrt{-1cdot-1}=-sqrt1=-1$. But, in addition to the problem with the previous solution, this additionally means that $sqrt{ab}=sqrt{a}sqrt{b}$ holds even for negative $a,b$.


Which of my reasonings is incorrect? Or are all of them correct, and there's an additional solution that I'm not seeing?










share|cite|improve this question









$endgroup$



marked as duplicate by KReiser, JMoravitz, Adrian Keister, metamorphy, Community Jan 4 at 1:47


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




    $begingroup$
    $sqrt{x^2} = |x|$
    $endgroup$
    – krirkrirk
    Jan 3 at 20:57










  • $begingroup$
    Every complex number, apart from zero, has two square roots.
    $endgroup$
    – Lord Shark the Unknown
    Jan 3 at 21:01










  • $begingroup$
    You can do the same thing with any square: $sqrt{25}$ = $sqrt{-5 cdot -5}$. for example. But we take the positive branch and define $sqrt{25}$ = 5.
    $endgroup$
    – Joel Pereira
    Jan 3 at 21:03












  • $begingroup$
    @JoelPereira I know, I thought I'd take the most basic case, though.
    $endgroup$
    – DonielF
    Jan 3 at 21:04






  • 1




    $begingroup$
    I don't understand the downvotes on this. It's perfectly reasonable to ask where the fallacious step is in a well-known fallacy, and it's clear enough what's being ssked. Also the questioner's own thoughts are included in quite a lot of detail
    $endgroup$
    – timtfj
    Jan 4 at 1:38
















-2












-2








-2





$begingroup$



This question already has an answer here:




  • Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?

    9 answers




Let's define $x=sqrt{ab}$, where $a=-1$ and $b=-1$.



Does $x=1$, as $-1cdot-1=1impliessqrt{-1cdot-1}=sqrt{1}=1$?



Or maybe $x=-1$, as $sqrt{a^2}=aimpliessqrt{-1cdot-1}=sqrt{(-1)^2}=-1$?



I assume that at least one of the following is true:




  1. The former solution is true, and the fallacy is in assuming $sqrt{a^2}=a$ for a negative base $a$. This doesn't seem to hold, given that this is equivalent to saying that $a^{2^{frac12}}not=a^{2cdotfrac12}=a$ for a negative $a$, where the value of $a$ seemingly has no bearing on the arithmetic performed with the exponents.

  2. The former solution is true, and the fallacy is in assuming the positive root is meant in the latter solution; in fact, $sqrt{-1cdot-1}=-sqrt{(-1)^2}=--1=1$. But this implies that, depending on how the radicand is factored, one is forced to take the positive root for one factoring and the negative root for another factoring to get the same answer, which doesn't seem right.

  3. The latter solution is true, and the fallacy is in assuming the positive root is meant in the former solution; in fact, $sqrt{-1cdot-1}=-sqrt1=-1$. But, in addition to the problem with the previous solution, this additionally means that $sqrt{ab}=sqrt{a}sqrt{b}$ holds even for negative $a,b$.


Which of my reasonings is incorrect? Or are all of them correct, and there's an additional solution that I'm not seeing?










share|cite|improve this question









$endgroup$





This question already has an answer here:




  • Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?

    9 answers




Let's define $x=sqrt{ab}$, where $a=-1$ and $b=-1$.



Does $x=1$, as $-1cdot-1=1impliessqrt{-1cdot-1}=sqrt{1}=1$?



Or maybe $x=-1$, as $sqrt{a^2}=aimpliessqrt{-1cdot-1}=sqrt{(-1)^2}=-1$?



I assume that at least one of the following is true:




  1. The former solution is true, and the fallacy is in assuming $sqrt{a^2}=a$ for a negative base $a$. This doesn't seem to hold, given that this is equivalent to saying that $a^{2^{frac12}}not=a^{2cdotfrac12}=a$ for a negative $a$, where the value of $a$ seemingly has no bearing on the arithmetic performed with the exponents.

  2. The former solution is true, and the fallacy is in assuming the positive root is meant in the latter solution; in fact, $sqrt{-1cdot-1}=-sqrt{(-1)^2}=--1=1$. But this implies that, depending on how the radicand is factored, one is forced to take the positive root for one factoring and the negative root for another factoring to get the same answer, which doesn't seem right.

  3. The latter solution is true, and the fallacy is in assuming the positive root is meant in the former solution; in fact, $sqrt{-1cdot-1}=-sqrt1=-1$. But, in addition to the problem with the previous solution, this additionally means that $sqrt{ab}=sqrt{a}sqrt{b}$ holds even for negative $a,b$.


Which of my reasonings is incorrect? Or are all of them correct, and there's an additional solution that I'm not seeing?





This question already has an answer here:




  • Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?

    9 answers








radicals fake-proofs






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 20:55









DonielFDonielF

484515




484515




marked as duplicate by KReiser, JMoravitz, Adrian Keister, metamorphy, Community Jan 4 at 1:47


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.









marked as duplicate by KReiser, JMoravitz, Adrian Keister, metamorphy, Community Jan 4 at 1:47


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




    $begingroup$
    $sqrt{x^2} = |x|$
    $endgroup$
    – krirkrirk
    Jan 3 at 20:57










  • $begingroup$
    Every complex number, apart from zero, has two square roots.
    $endgroup$
    – Lord Shark the Unknown
    Jan 3 at 21:01










  • $begingroup$
    You can do the same thing with any square: $sqrt{25}$ = $sqrt{-5 cdot -5}$. for example. But we take the positive branch and define $sqrt{25}$ = 5.
    $endgroup$
    – Joel Pereira
    Jan 3 at 21:03












  • $begingroup$
    @JoelPereira I know, I thought I'd take the most basic case, though.
    $endgroup$
    – DonielF
    Jan 3 at 21:04






  • 1




    $begingroup$
    I don't understand the downvotes on this. It's perfectly reasonable to ask where the fallacious step is in a well-known fallacy, and it's clear enough what's being ssked. Also the questioner's own thoughts are included in quite a lot of detail
    $endgroup$
    – timtfj
    Jan 4 at 1:38
















  • 1




    $begingroup$
    $sqrt{x^2} = |x|$
    $endgroup$
    – krirkrirk
    Jan 3 at 20:57










  • $begingroup$
    Every complex number, apart from zero, has two square roots.
    $endgroup$
    – Lord Shark the Unknown
    Jan 3 at 21:01










  • $begingroup$
    You can do the same thing with any square: $sqrt{25}$ = $sqrt{-5 cdot -5}$. for example. But we take the positive branch and define $sqrt{25}$ = 5.
    $endgroup$
    – Joel Pereira
    Jan 3 at 21:03












  • $begingroup$
    @JoelPereira I know, I thought I'd take the most basic case, though.
    $endgroup$
    – DonielF
    Jan 3 at 21:04






  • 1




    $begingroup$
    I don't understand the downvotes on this. It's perfectly reasonable to ask where the fallacious step is in a well-known fallacy, and it's clear enough what's being ssked. Also the questioner's own thoughts are included in quite a lot of detail
    $endgroup$
    – timtfj
    Jan 4 at 1:38










1




1




$begingroup$
$sqrt{x^2} = |x|$
$endgroup$
– krirkrirk
Jan 3 at 20:57




$begingroup$
$sqrt{x^2} = |x|$
$endgroup$
– krirkrirk
Jan 3 at 20:57












$begingroup$
Every complex number, apart from zero, has two square roots.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 21:01




$begingroup$
Every complex number, apart from zero, has two square roots.
$endgroup$
– Lord Shark the Unknown
Jan 3 at 21:01












$begingroup$
You can do the same thing with any square: $sqrt{25}$ = $sqrt{-5 cdot -5}$. for example. But we take the positive branch and define $sqrt{25}$ = 5.
$endgroup$
– Joel Pereira
Jan 3 at 21:03






$begingroup$
You can do the same thing with any square: $sqrt{25}$ = $sqrt{-5 cdot -5}$. for example. But we take the positive branch and define $sqrt{25}$ = 5.
$endgroup$
– Joel Pereira
Jan 3 at 21:03














$begingroup$
@JoelPereira I know, I thought I'd take the most basic case, though.
$endgroup$
– DonielF
Jan 3 at 21:04




$begingroup$
@JoelPereira I know, I thought I'd take the most basic case, though.
$endgroup$
– DonielF
Jan 3 at 21:04




1




1




$begingroup$
I don't understand the downvotes on this. It's perfectly reasonable to ask where the fallacious step is in a well-known fallacy, and it's clear enough what's being ssked. Also the questioner's own thoughts are included in quite a lot of detail
$endgroup$
– timtfj
Jan 4 at 1:38






$begingroup$
I don't understand the downvotes on this. It's perfectly reasonable to ask where the fallacious step is in a well-known fallacy, and it's clear enough what's being ssked. Also the questioner's own thoughts are included in quite a lot of detail
$endgroup$
– timtfj
Jan 4 at 1:38












3 Answers
3






active

oldest

votes


















1












$begingroup$

For $a in mathbb{R}^+$, $sqrt a $ is defined as the positive real $b $ such that $b^2=a $. Thus $sqrt{x^2} = |x|$ because by definition a square root is positive. Hence in your case $sqrt {(-1)^2} =1$.



Regarding your other points :




  • As you noticed $sqrt {ab} = sqrt a sqrt b $ only applies for $a,b $ positive so it cant be used here


  • $sqrt{a^2}=(a^2)^frac 12$ is true, but $(a^x)^y = a^{xy} $ for all $x,y in Bbb R $ is only true for $a>0$. This is because in general $a^b = e^{bln a} $ which only makes sense for $a>0$. With your example, you can see why this will not work : if $(a^x)^y = a^{xy} $ then $((-1)^2)^frac 12 = (-1)^1 iff 1=-1$







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you. Finally an answer here that actually explains the concepts, rather than just saying “it doesn’t work like that.”
    $endgroup$
    – DonielF
    Jan 4 at 1:44



















3












$begingroup$

The following statement is not true:



$$sqrt{x^2} = x$$



The square root function returns the non-negative square root, or the principal square root, so it’s actually



$$sqrt{x^2} = vert xvert = begin{cases} x; quad xgeq 0 \ -x; quad x < 0 end{cases}$$



So, in your case, $x = -1 < 0$, so $sqrt{(-1)^2} = vert -1vert = 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why should $sqrt{x}=|x|$ when $x^{2^{frac12}}=x^{2cdotfrac12}=xnot=|x|$?
    $endgroup$
    – DonielF
    Jan 3 at 21:02








  • 2




    $begingroup$
    $sqrt{x^2} = vert xvert$ because the square root function gives the non-negative value by convention. It’s just how it works.
    $endgroup$
    – KM101
    Jan 3 at 21:03










  • $begingroup$
    Then how can $sqrt{x}=x^frac12$?
    $endgroup$
    – DonielF
    Jan 3 at 21:05










  • $begingroup$
    Why shouldn’t that be the case?
    $endgroup$
    – KM101
    Jan 3 at 21:07






  • 1




    $begingroup$
    @DonielF $(a^n)^m = a^{nm} $ makes sense for $a>0$ because $ln (a) $ does.
    $endgroup$
    – krirkrirk
    Jan 3 at 21:13



















1












$begingroup$

Based on the definition of the square root which is positive, it should be 1. See here to know more about that.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what’s wrong with $sqrt{-1cdot-1}=sqrt{(-1)^2}=-1$? Either solution #1, with its problem, or solution #2, with its problem. How do you defend against those problems?
    $endgroup$
    – DonielF
    Jan 3 at 20:59










  • $begingroup$
    @DonielF it's not correct based on the definition of square root. just this.
    $endgroup$
    – OmG
    Jan 3 at 20:59










  • $begingroup$
    @DonielF $sqrt{x^2} = x$ is false.
    $endgroup$
    – KM101
    Jan 3 at 21:00










  • $begingroup$
    @DonielF: base on what do you think that $sqrt{(-1)^2}=-1?$
    $endgroup$
    – user587192
    Jan 3 at 22:54










  • $begingroup$
    @user587192 Read the immediately preceding line. Since $sqrt{a^2}=a$, shouldn’t $sqrt{(-1)^2}=-1$?
    $endgroup$
    – DonielF
    Jan 4 at 1:42


















3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

For $a in mathbb{R}^+$, $sqrt a $ is defined as the positive real $b $ such that $b^2=a $. Thus $sqrt{x^2} = |x|$ because by definition a square root is positive. Hence in your case $sqrt {(-1)^2} =1$.



Regarding your other points :




  • As you noticed $sqrt {ab} = sqrt a sqrt b $ only applies for $a,b $ positive so it cant be used here


  • $sqrt{a^2}=(a^2)^frac 12$ is true, but $(a^x)^y = a^{xy} $ for all $x,y in Bbb R $ is only true for $a>0$. This is because in general $a^b = e^{bln a} $ which only makes sense for $a>0$. With your example, you can see why this will not work : if $(a^x)^y = a^{xy} $ then $((-1)^2)^frac 12 = (-1)^1 iff 1=-1$







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you. Finally an answer here that actually explains the concepts, rather than just saying “it doesn’t work like that.”
    $endgroup$
    – DonielF
    Jan 4 at 1:44
















1












$begingroup$

For $a in mathbb{R}^+$, $sqrt a $ is defined as the positive real $b $ such that $b^2=a $. Thus $sqrt{x^2} = |x|$ because by definition a square root is positive. Hence in your case $sqrt {(-1)^2} =1$.



Regarding your other points :




  • As you noticed $sqrt {ab} = sqrt a sqrt b $ only applies for $a,b $ positive so it cant be used here


  • $sqrt{a^2}=(a^2)^frac 12$ is true, but $(a^x)^y = a^{xy} $ for all $x,y in Bbb R $ is only true for $a>0$. This is because in general $a^b = e^{bln a} $ which only makes sense for $a>0$. With your example, you can see why this will not work : if $(a^x)^y = a^{xy} $ then $((-1)^2)^frac 12 = (-1)^1 iff 1=-1$







share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    Thank you. Finally an answer here that actually explains the concepts, rather than just saying “it doesn’t work like that.”
    $endgroup$
    – DonielF
    Jan 4 at 1:44














1












1








1





$begingroup$

For $a in mathbb{R}^+$, $sqrt a $ is defined as the positive real $b $ such that $b^2=a $. Thus $sqrt{x^2} = |x|$ because by definition a square root is positive. Hence in your case $sqrt {(-1)^2} =1$.



Regarding your other points :




  • As you noticed $sqrt {ab} = sqrt a sqrt b $ only applies for $a,b $ positive so it cant be used here


  • $sqrt{a^2}=(a^2)^frac 12$ is true, but $(a^x)^y = a^{xy} $ for all $x,y in Bbb R $ is only true for $a>0$. This is because in general $a^b = e^{bln a} $ which only makes sense for $a>0$. With your example, you can see why this will not work : if $(a^x)^y = a^{xy} $ then $((-1)^2)^frac 12 = (-1)^1 iff 1=-1$







share|cite|improve this answer









$endgroup$



For $a in mathbb{R}^+$, $sqrt a $ is defined as the positive real $b $ such that $b^2=a $. Thus $sqrt{x^2} = |x|$ because by definition a square root is positive. Hence in your case $sqrt {(-1)^2} =1$.



Regarding your other points :




  • As you noticed $sqrt {ab} = sqrt a sqrt b $ only applies for $a,b $ positive so it cant be used here


  • $sqrt{a^2}=(a^2)^frac 12$ is true, but $(a^x)^y = a^{xy} $ for all $x,y in Bbb R $ is only true for $a>0$. This is because in general $a^b = e^{bln a} $ which only makes sense for $a>0$. With your example, you can see why this will not work : if $(a^x)^y = a^{xy} $ then $((-1)^2)^frac 12 = (-1)^1 iff 1=-1$








share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 1:01









krirkrirkkrirkrirk

1,483518




1,483518








  • 1




    $begingroup$
    Thank you. Finally an answer here that actually explains the concepts, rather than just saying “it doesn’t work like that.”
    $endgroup$
    – DonielF
    Jan 4 at 1:44














  • 1




    $begingroup$
    Thank you. Finally an answer here that actually explains the concepts, rather than just saying “it doesn’t work like that.”
    $endgroup$
    – DonielF
    Jan 4 at 1:44








1




1




$begingroup$
Thank you. Finally an answer here that actually explains the concepts, rather than just saying “it doesn’t work like that.”
$endgroup$
– DonielF
Jan 4 at 1:44




$begingroup$
Thank you. Finally an answer here that actually explains the concepts, rather than just saying “it doesn’t work like that.”
$endgroup$
– DonielF
Jan 4 at 1:44











3












$begingroup$

The following statement is not true:



$$sqrt{x^2} = x$$



The square root function returns the non-negative square root, or the principal square root, so it’s actually



$$sqrt{x^2} = vert xvert = begin{cases} x; quad xgeq 0 \ -x; quad x < 0 end{cases}$$



So, in your case, $x = -1 < 0$, so $sqrt{(-1)^2} = vert -1vert = 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why should $sqrt{x}=|x|$ when $x^{2^{frac12}}=x^{2cdotfrac12}=xnot=|x|$?
    $endgroup$
    – DonielF
    Jan 3 at 21:02








  • 2




    $begingroup$
    $sqrt{x^2} = vert xvert$ because the square root function gives the non-negative value by convention. It’s just how it works.
    $endgroup$
    – KM101
    Jan 3 at 21:03










  • $begingroup$
    Then how can $sqrt{x}=x^frac12$?
    $endgroup$
    – DonielF
    Jan 3 at 21:05










  • $begingroup$
    Why shouldn’t that be the case?
    $endgroup$
    – KM101
    Jan 3 at 21:07






  • 1




    $begingroup$
    @DonielF $(a^n)^m = a^{nm} $ makes sense for $a>0$ because $ln (a) $ does.
    $endgroup$
    – krirkrirk
    Jan 3 at 21:13
















3












$begingroup$

The following statement is not true:



$$sqrt{x^2} = x$$



The square root function returns the non-negative square root, or the principal square root, so it’s actually



$$sqrt{x^2} = vert xvert = begin{cases} x; quad xgeq 0 \ -x; quad x < 0 end{cases}$$



So, in your case, $x = -1 < 0$, so $sqrt{(-1)^2} = vert -1vert = 1$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why should $sqrt{x}=|x|$ when $x^{2^{frac12}}=x^{2cdotfrac12}=xnot=|x|$?
    $endgroup$
    – DonielF
    Jan 3 at 21:02








  • 2




    $begingroup$
    $sqrt{x^2} = vert xvert$ because the square root function gives the non-negative value by convention. It’s just how it works.
    $endgroup$
    – KM101
    Jan 3 at 21:03










  • $begingroup$
    Then how can $sqrt{x}=x^frac12$?
    $endgroup$
    – DonielF
    Jan 3 at 21:05










  • $begingroup$
    Why shouldn’t that be the case?
    $endgroup$
    – KM101
    Jan 3 at 21:07






  • 1




    $begingroup$
    @DonielF $(a^n)^m = a^{nm} $ makes sense for $a>0$ because $ln (a) $ does.
    $endgroup$
    – krirkrirk
    Jan 3 at 21:13














3












3








3





$begingroup$

The following statement is not true:



$$sqrt{x^2} = x$$



The square root function returns the non-negative square root, or the principal square root, so it’s actually



$$sqrt{x^2} = vert xvert = begin{cases} x; quad xgeq 0 \ -x; quad x < 0 end{cases}$$



So, in your case, $x = -1 < 0$, so $sqrt{(-1)^2} = vert -1vert = 1$.






share|cite|improve this answer











$endgroup$



The following statement is not true:



$$sqrt{x^2} = x$$



The square root function returns the non-negative square root, or the principal square root, so it’s actually



$$sqrt{x^2} = vert xvert = begin{cases} x; quad xgeq 0 \ -x; quad x < 0 end{cases}$$



So, in your case, $x = -1 < 0$, so $sqrt{(-1)^2} = vert -1vert = 1$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 21:05

























answered Jan 3 at 20:59









KM101KM101

5,9481524




5,9481524












  • $begingroup$
    Why should $sqrt{x}=|x|$ when $x^{2^{frac12}}=x^{2cdotfrac12}=xnot=|x|$?
    $endgroup$
    – DonielF
    Jan 3 at 21:02








  • 2




    $begingroup$
    $sqrt{x^2} = vert xvert$ because the square root function gives the non-negative value by convention. It’s just how it works.
    $endgroup$
    – KM101
    Jan 3 at 21:03










  • $begingroup$
    Then how can $sqrt{x}=x^frac12$?
    $endgroup$
    – DonielF
    Jan 3 at 21:05










  • $begingroup$
    Why shouldn’t that be the case?
    $endgroup$
    – KM101
    Jan 3 at 21:07






  • 1




    $begingroup$
    @DonielF $(a^n)^m = a^{nm} $ makes sense for $a>0$ because $ln (a) $ does.
    $endgroup$
    – krirkrirk
    Jan 3 at 21:13


















  • $begingroup$
    Why should $sqrt{x}=|x|$ when $x^{2^{frac12}}=x^{2cdotfrac12}=xnot=|x|$?
    $endgroup$
    – DonielF
    Jan 3 at 21:02








  • 2




    $begingroup$
    $sqrt{x^2} = vert xvert$ because the square root function gives the non-negative value by convention. It’s just how it works.
    $endgroup$
    – KM101
    Jan 3 at 21:03










  • $begingroup$
    Then how can $sqrt{x}=x^frac12$?
    $endgroup$
    – DonielF
    Jan 3 at 21:05










  • $begingroup$
    Why shouldn’t that be the case?
    $endgroup$
    – KM101
    Jan 3 at 21:07






  • 1




    $begingroup$
    @DonielF $(a^n)^m = a^{nm} $ makes sense for $a>0$ because $ln (a) $ does.
    $endgroup$
    – krirkrirk
    Jan 3 at 21:13
















$begingroup$
Why should $sqrt{x}=|x|$ when $x^{2^{frac12}}=x^{2cdotfrac12}=xnot=|x|$?
$endgroup$
– DonielF
Jan 3 at 21:02






$begingroup$
Why should $sqrt{x}=|x|$ when $x^{2^{frac12}}=x^{2cdotfrac12}=xnot=|x|$?
$endgroup$
– DonielF
Jan 3 at 21:02






2




2




$begingroup$
$sqrt{x^2} = vert xvert$ because the square root function gives the non-negative value by convention. It’s just how it works.
$endgroup$
– KM101
Jan 3 at 21:03




$begingroup$
$sqrt{x^2} = vert xvert$ because the square root function gives the non-negative value by convention. It’s just how it works.
$endgroup$
– KM101
Jan 3 at 21:03












$begingroup$
Then how can $sqrt{x}=x^frac12$?
$endgroup$
– DonielF
Jan 3 at 21:05




$begingroup$
Then how can $sqrt{x}=x^frac12$?
$endgroup$
– DonielF
Jan 3 at 21:05












$begingroup$
Why shouldn’t that be the case?
$endgroup$
– KM101
Jan 3 at 21:07




$begingroup$
Why shouldn’t that be the case?
$endgroup$
– KM101
Jan 3 at 21:07




1




1




$begingroup$
@DonielF $(a^n)^m = a^{nm} $ makes sense for $a>0$ because $ln (a) $ does.
$endgroup$
– krirkrirk
Jan 3 at 21:13




$begingroup$
@DonielF $(a^n)^m = a^{nm} $ makes sense for $a>0$ because $ln (a) $ does.
$endgroup$
– krirkrirk
Jan 3 at 21:13











1












$begingroup$

Based on the definition of the square root which is positive, it should be 1. See here to know more about that.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what’s wrong with $sqrt{-1cdot-1}=sqrt{(-1)^2}=-1$? Either solution #1, with its problem, or solution #2, with its problem. How do you defend against those problems?
    $endgroup$
    – DonielF
    Jan 3 at 20:59










  • $begingroup$
    @DonielF it's not correct based on the definition of square root. just this.
    $endgroup$
    – OmG
    Jan 3 at 20:59










  • $begingroup$
    @DonielF $sqrt{x^2} = x$ is false.
    $endgroup$
    – KM101
    Jan 3 at 21:00










  • $begingroup$
    @DonielF: base on what do you think that $sqrt{(-1)^2}=-1?$
    $endgroup$
    – user587192
    Jan 3 at 22:54










  • $begingroup$
    @user587192 Read the immediately preceding line. Since $sqrt{a^2}=a$, shouldn’t $sqrt{(-1)^2}=-1$?
    $endgroup$
    – DonielF
    Jan 4 at 1:42
















1












$begingroup$

Based on the definition of the square root which is positive, it should be 1. See here to know more about that.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So what’s wrong with $sqrt{-1cdot-1}=sqrt{(-1)^2}=-1$? Either solution #1, with its problem, or solution #2, with its problem. How do you defend against those problems?
    $endgroup$
    – DonielF
    Jan 3 at 20:59










  • $begingroup$
    @DonielF it's not correct based on the definition of square root. just this.
    $endgroup$
    – OmG
    Jan 3 at 20:59










  • $begingroup$
    @DonielF $sqrt{x^2} = x$ is false.
    $endgroup$
    – KM101
    Jan 3 at 21:00










  • $begingroup$
    @DonielF: base on what do you think that $sqrt{(-1)^2}=-1?$
    $endgroup$
    – user587192
    Jan 3 at 22:54










  • $begingroup$
    @user587192 Read the immediately preceding line. Since $sqrt{a^2}=a$, shouldn’t $sqrt{(-1)^2}=-1$?
    $endgroup$
    – DonielF
    Jan 4 at 1:42














1












1








1





$begingroup$

Based on the definition of the square root which is positive, it should be 1. See here to know more about that.






share|cite|improve this answer











$endgroup$



Based on the definition of the square root which is positive, it should be 1. See here to know more about that.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 3 at 20:59

























answered Jan 3 at 20:56









OmGOmG

2,492722




2,492722












  • $begingroup$
    So what’s wrong with $sqrt{-1cdot-1}=sqrt{(-1)^2}=-1$? Either solution #1, with its problem, or solution #2, with its problem. How do you defend against those problems?
    $endgroup$
    – DonielF
    Jan 3 at 20:59










  • $begingroup$
    @DonielF it's not correct based on the definition of square root. just this.
    $endgroup$
    – OmG
    Jan 3 at 20:59










  • $begingroup$
    @DonielF $sqrt{x^2} = x$ is false.
    $endgroup$
    – KM101
    Jan 3 at 21:00










  • $begingroup$
    @DonielF: base on what do you think that $sqrt{(-1)^2}=-1?$
    $endgroup$
    – user587192
    Jan 3 at 22:54










  • $begingroup$
    @user587192 Read the immediately preceding line. Since $sqrt{a^2}=a$, shouldn’t $sqrt{(-1)^2}=-1$?
    $endgroup$
    – DonielF
    Jan 4 at 1:42


















  • $begingroup$
    So what’s wrong with $sqrt{-1cdot-1}=sqrt{(-1)^2}=-1$? Either solution #1, with its problem, or solution #2, with its problem. How do you defend against those problems?
    $endgroup$
    – DonielF
    Jan 3 at 20:59










  • $begingroup$
    @DonielF it's not correct based on the definition of square root. just this.
    $endgroup$
    – OmG
    Jan 3 at 20:59










  • $begingroup$
    @DonielF $sqrt{x^2} = x$ is false.
    $endgroup$
    – KM101
    Jan 3 at 21:00










  • $begingroup$
    @DonielF: base on what do you think that $sqrt{(-1)^2}=-1?$
    $endgroup$
    – user587192
    Jan 3 at 22:54










  • $begingroup$
    @user587192 Read the immediately preceding line. Since $sqrt{a^2}=a$, shouldn’t $sqrt{(-1)^2}=-1$?
    $endgroup$
    – DonielF
    Jan 4 at 1:42
















$begingroup$
So what’s wrong with $sqrt{-1cdot-1}=sqrt{(-1)^2}=-1$? Either solution #1, with its problem, or solution #2, with its problem. How do you defend against those problems?
$endgroup$
– DonielF
Jan 3 at 20:59




$begingroup$
So what’s wrong with $sqrt{-1cdot-1}=sqrt{(-1)^2}=-1$? Either solution #1, with its problem, or solution #2, with its problem. How do you defend against those problems?
$endgroup$
– DonielF
Jan 3 at 20:59












$begingroup$
@DonielF it's not correct based on the definition of square root. just this.
$endgroup$
– OmG
Jan 3 at 20:59




$begingroup$
@DonielF it's not correct based on the definition of square root. just this.
$endgroup$
– OmG
Jan 3 at 20:59












$begingroup$
@DonielF $sqrt{x^2} = x$ is false.
$endgroup$
– KM101
Jan 3 at 21:00




$begingroup$
@DonielF $sqrt{x^2} = x$ is false.
$endgroup$
– KM101
Jan 3 at 21:00












$begingroup$
@DonielF: base on what do you think that $sqrt{(-1)^2}=-1?$
$endgroup$
– user587192
Jan 3 at 22:54




$begingroup$
@DonielF: base on what do you think that $sqrt{(-1)^2}=-1?$
$endgroup$
– user587192
Jan 3 at 22:54












$begingroup$
@user587192 Read the immediately preceding line. Since $sqrt{a^2}=a$, shouldn’t $sqrt{(-1)^2}=-1$?
$endgroup$
– DonielF
Jan 4 at 1:42




$begingroup$
@user587192 Read the immediately preceding line. Since $sqrt{a^2}=a$, shouldn’t $sqrt{(-1)^2}=-1$?
$endgroup$
– DonielF
Jan 4 at 1:42



Popular posts from this blog

Human spaceflight

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

File:DeusFollowingSea.jpg