Why and When do we use the absolute value?












3












$begingroup$


I looked at a video from Khan Academy on limits at infinity and in the video, the instructor said $sqrt {x^2}$ is 'essentially taking the absolute value of x; |x|'. I couldn't understand why is the same thing. Is it because when we take the principal root of $x^2$, we are always going to have a positive outcome and absolute value does that as well? So, Why and When do we use the absolute value?



Thank you. P.S I am just learning the basics of calculus.






radicals absolute-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    $|x|$ is nice to write than $sqrt{x^2}$. Also, it’s just less confusing, as square roots and powers of twos have other uses.
    $endgroup$
    – Zachary Hunter
    Jan 6 at 7:06










  • $begingroup$
    @ZacharyHunter they are the samething? So my assumption is correct?
    $endgroup$
    – Fred Weasley
    Jan 6 at 7:08










  • $begingroup$
    Yes they are the same, $|x|$ is just more concise and all.
    $endgroup$
    – Zachary Hunter
    Jan 6 at 7:09










  • $begingroup$
    Yes, your reason is correct. $sqrt{x^2}=|x|ne xforall x <0$; for example, $sqrt{(-5)^2}=5=|-5|$
    $endgroup$
    – Shubham Johri
    Jan 6 at 7:28












  • $begingroup$
    Just a small nitpick: $sqrt{x^2} = vert xvert$ means the outcome is always non-negative since $x$ can be $0$.
    $endgroup$
    – KM101
    Jan 6 at 7:34


















3












$begingroup$


I looked at a video from Khan Academy on limits at infinity and in the video, the instructor said $sqrt {x^2}$ is 'essentially taking the absolute value of x; |x|'. I couldn't understand why is the same thing. Is it because when we take the principal root of $x^2$, we are always going to have a positive outcome and absolute value does that as well? So, Why and When do we use the absolute value?



Thank you. P.S I am just learning the basics of calculus.






radicals absolute-value







share|cite|improve this question









$endgroup$












  • $begingroup$
    $|x|$ is nice to write than $sqrt{x^2}$. Also, it’s just less confusing, as square roots and powers of twos have other uses.
    $endgroup$
    – Zachary Hunter
    Jan 6 at 7:06










  • $begingroup$
    @ZacharyHunter they are the samething? So my assumption is correct?
    $endgroup$
    – Fred Weasley
    Jan 6 at 7:08










  • $begingroup$
    Yes they are the same, $|x|$ is just more concise and all.
    $endgroup$
    – Zachary Hunter
    Jan 6 at 7:09










  • $begingroup$
    Yes, your reason is correct. $sqrt{x^2}=|x|ne xforall x <0$; for example, $sqrt{(-5)^2}=5=|-5|$
    $endgroup$
    – Shubham Johri
    Jan 6 at 7:28












  • $begingroup$
    Just a small nitpick: $sqrt{x^2} = vert xvert$ means the outcome is always non-negative since $x$ can be $0$.
    $endgroup$
    – KM101
    Jan 6 at 7:34
















3












3








3





$begingroup$


I looked at a video from Khan Academy on limits at infinity and in the video, the instructor said $sqrt {x^2}$ is 'essentially taking the absolute value of x; |x|'. I couldn't understand why is the same thing. Is it because when we take the principal root of $x^2$, we are always going to have a positive outcome and absolute value does that as well? So, Why and When do we use the absolute value?



Thank you. P.S I am just learning the basics of calculus.






radicals absolute-value







share|cite|improve this question









$endgroup$




I looked at a video from Khan Academy on limits at infinity and in the video, the instructor said $sqrt {x^2}$ is 'essentially taking the absolute value of x; |x|'. I couldn't understand why is the same thing. Is it because when we take the principal root of $x^2$, we are always going to have a positive outcome and absolute value does that as well? So, Why and When do we use the absolute value?



Thank you. P.S I am just learning the basics of calculus.






radicals absolute-value




radicals absolute-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 6 at 7:04









Fred WeasleyFred Weasley

16810




16810












  • $begingroup$
    $|x|$ is nice to write than $sqrt{x^2}$. Also, it’s just less confusing, as square roots and powers of twos have other uses.
    $endgroup$
    – Zachary Hunter
    Jan 6 at 7:06










  • $begingroup$
    @ZacharyHunter they are the samething? So my assumption is correct?
    $endgroup$
    – Fred Weasley
    Jan 6 at 7:08










  • $begingroup$
    Yes they are the same, $|x|$ is just more concise and all.
    $endgroup$
    – Zachary Hunter
    Jan 6 at 7:09










  • $begingroup$
    Yes, your reason is correct. $sqrt{x^2}=|x|ne xforall x <0$; for example, $sqrt{(-5)^2}=5=|-5|$
    $endgroup$
    – Shubham Johri
    Jan 6 at 7:28












  • $begingroup$
    Just a small nitpick: $sqrt{x^2} = vert xvert$ means the outcome is always non-negative since $x$ can be $0$.
    $endgroup$
    – KM101
    Jan 6 at 7:34




















  • $begingroup$
    $|x|$ is nice to write than $sqrt{x^2}$. Also, it’s just less confusing, as square roots and powers of twos have other uses.
    $endgroup$
    – Zachary Hunter
    Jan 6 at 7:06










  • $begingroup$
    @ZacharyHunter they are the samething? So my assumption is correct?
    $endgroup$
    – Fred Weasley
    Jan 6 at 7:08










  • $begingroup$
    Yes they are the same, $|x|$ is just more concise and all.
    $endgroup$
    – Zachary Hunter
    Jan 6 at 7:09










  • $begingroup$
    Yes, your reason is correct. $sqrt{x^2}=|x|ne xforall x <0$; for example, $sqrt{(-5)^2}=5=|-5|$
    $endgroup$
    – Shubham Johri
    Jan 6 at 7:28












  • $begingroup$
    Just a small nitpick: $sqrt{x^2} = vert xvert$ means the outcome is always non-negative since $x$ can be $0$.
    $endgroup$
    – KM101
    Jan 6 at 7:34


















$begingroup$
$|x|$ is nice to write than $sqrt{x^2}$. Also, it’s just less confusing, as square roots and powers of twos have other uses.
$endgroup$
– Zachary Hunter
Jan 6 at 7:06




$begingroup$
$|x|$ is nice to write than $sqrt{x^2}$. Also, it’s just less confusing, as square roots and powers of twos have other uses.
$endgroup$
– Zachary Hunter
Jan 6 at 7:06












$begingroup$
@ZacharyHunter they are the samething? So my assumption is correct?
$endgroup$
– Fred Weasley
Jan 6 at 7:08




$begingroup$
@ZacharyHunter they are the samething? So my assumption is correct?
$endgroup$
– Fred Weasley
Jan 6 at 7:08












$begingroup$
Yes they are the same, $|x|$ is just more concise and all.
$endgroup$
– Zachary Hunter
Jan 6 at 7:09




$begingroup$
Yes they are the same, $|x|$ is just more concise and all.
$endgroup$
– Zachary Hunter
Jan 6 at 7:09












$begingroup$
Yes, your reason is correct. $sqrt{x^2}=|x|ne xforall x <0$; for example, $sqrt{(-5)^2}=5=|-5|$
$endgroup$
– Shubham Johri
Jan 6 at 7:28






$begingroup$
Yes, your reason is correct. $sqrt{x^2}=|x|ne xforall x <0$; for example, $sqrt{(-5)^2}=5=|-5|$
$endgroup$
– Shubham Johri
Jan 6 at 7:28














$begingroup$
Just a small nitpick: $sqrt{x^2} = vert xvert$ means the outcome is always non-negative since $x$ can be $0$.
$endgroup$
– KM101
Jan 6 at 7:34






$begingroup$
Just a small nitpick: $sqrt{x^2} = vert xvert$ means the outcome is always non-negative since $x$ can be $0$.
$endgroup$
– KM101
Jan 6 at 7:34












2 Answers
2






active

oldest

votes


















4












$begingroup$

Yes, $vert xvert$ means the outcome is always non-negative. This requires clarification because $x$ itself may be negative. Here are a few examples:



$$sqrt{color{blue}{+4}^2} = sqrt{16} = color{blue}{+4}$$



$$sqrt{color{blue}{+9}^2} = sqrt{81} = color{blue}{+9}$$



$$sqrt{(color{blue}{-4}^2)} = sqrt{16} = color{blue}{+4} = -(color{blue}{-4})$$



$$sqrt{(color{blue}{-9}^2)} = sqrt{81} = color{blue}{+9} = -(color{blue}{-9})$$



So, when $x > 0$, then $sqrt{x^2}$ will return $x > 0$, so it equals $x$. However, when $x < 0$, then $sqrt{x^2}$ will return $x > 0$ as well, so it equals $-x$. Hence,



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



This is all just a way of saying $sqrt{x^2}$ will always return a non-negative value.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks! and so in concluding, the absolute value sign is used just to simplify stuff? Also, in addition to this are there specific rules that i have to obey when using this as well as does this comes up often in calculus or other topics?
    $endgroup$
    – Fred Weasley
    Jan 6 at 8:15










  • $begingroup$
    It’s really just a reiteration if the definition of the principal root, which is always non-negative. It can be used for simplification I suppose. As for your second question, I’m not sure, as I’m quite new to calculus myself.
    $endgroup$
    – KM101
    Jan 6 at 8:45





















0












$begingroup$

I agree with KM 101. I haven't seen the video, but what they try to say is that by taking the square root you get a distance on the number line (which is what the absolute value is), which is the same in both directions.



But once the number line enters the picture to explain the absolute values, the type of numbers for which the line exists also enters the picture, and that's the real numbers, which is a separate topic.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Yes, $vert xvert$ means the outcome is always non-negative. This requires clarification because $x$ itself may be negative. Here are a few examples:



    $$sqrt{color{blue}{+4}^2} = sqrt{16} = color{blue}{+4}$$



    $$sqrt{color{blue}{+9}^2} = sqrt{81} = color{blue}{+9}$$



    $$sqrt{(color{blue}{-4}^2)} = sqrt{16} = color{blue}{+4} = -(color{blue}{-4})$$



    $$sqrt{(color{blue}{-9}^2)} = sqrt{81} = color{blue}{+9} = -(color{blue}{-9})$$



    So, when $x > 0$, then $sqrt{x^2}$ will return $x > 0$, so it equals $x$. However, when $x < 0$, then $sqrt{x^2}$ will return $x > 0$ as well, so it equals $-x$. Hence,



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



    This is all just a way of saying $sqrt{x^2}$ will always return a non-negative value.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks! and so in concluding, the absolute value sign is used just to simplify stuff? Also, in addition to this are there specific rules that i have to obey when using this as well as does this comes up often in calculus or other topics?
      $endgroup$
      – Fred Weasley
      Jan 6 at 8:15










    • $begingroup$
      It’s really just a reiteration if the definition of the principal root, which is always non-negative. It can be used for simplification I suppose. As for your second question, I’m not sure, as I’m quite new to calculus myself.
      $endgroup$
      – KM101
      Jan 6 at 8:45


















    4












    $begingroup$

    Yes, $vert xvert$ means the outcome is always non-negative. This requires clarification because $x$ itself may be negative. Here are a few examples:



    $$sqrt{color{blue}{+4}^2} = sqrt{16} = color{blue}{+4}$$



    $$sqrt{color{blue}{+9}^2} = sqrt{81} = color{blue}{+9}$$



    $$sqrt{(color{blue}{-4}^2)} = sqrt{16} = color{blue}{+4} = -(color{blue}{-4})$$



    $$sqrt{(color{blue}{-9}^2)} = sqrt{81} = color{blue}{+9} = -(color{blue}{-9})$$



    So, when $x > 0$, then $sqrt{x^2}$ will return $x > 0$, so it equals $x$. However, when $x < 0$, then $sqrt{x^2}$ will return $x > 0$ as well, so it equals $-x$. Hence,



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



    This is all just a way of saying $sqrt{x^2}$ will always return a non-negative value.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thanks! and so in concluding, the absolute value sign is used just to simplify stuff? Also, in addition to this are there specific rules that i have to obey when using this as well as does this comes up often in calculus or other topics?
      $endgroup$
      – Fred Weasley
      Jan 6 at 8:15










    • $begingroup$
      It’s really just a reiteration if the definition of the principal root, which is always non-negative. It can be used for simplification I suppose. As for your second question, I’m not sure, as I’m quite new to calculus myself.
      $endgroup$
      – KM101
      Jan 6 at 8:45
















    4












    4








    4





    $begingroup$

    Yes, $vert xvert$ means the outcome is always non-negative. This requires clarification because $x$ itself may be negative. Here are a few examples:



    $$sqrt{color{blue}{+4}^2} = sqrt{16} = color{blue}{+4}$$



    $$sqrt{color{blue}{+9}^2} = sqrt{81} = color{blue}{+9}$$



    $$sqrt{(color{blue}{-4}^2)} = sqrt{16} = color{blue}{+4} = -(color{blue}{-4})$$



    $$sqrt{(color{blue}{-9}^2)} = sqrt{81} = color{blue}{+9} = -(color{blue}{-9})$$



    So, when $x > 0$, then $sqrt{x^2}$ will return $x > 0$, so it equals $x$. However, when $x < 0$, then $sqrt{x^2}$ will return $x > 0$ as well, so it equals $-x$. Hence,



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



    This is all just a way of saying $sqrt{x^2}$ will always return a non-negative value.






    share|cite|improve this answer









    $endgroup$



    Yes, $vert xvert$ means the outcome is always non-negative. This requires clarification because $x$ itself may be negative. Here are a few examples:



    $$sqrt{color{blue}{+4}^2} = sqrt{16} = color{blue}{+4}$$



    $$sqrt{color{blue}{+9}^2} = sqrt{81} = color{blue}{+9}$$



    $$sqrt{(color{blue}{-4}^2)} = sqrt{16} = color{blue}{+4} = -(color{blue}{-4})$$



    $$sqrt{(color{blue}{-9}^2)} = sqrt{81} = color{blue}{+9} = -(color{blue}{-9})$$



    So, when $x > 0$, then $sqrt{x^2}$ will return $x > 0$, so it equals $x$. However, when $x < 0$, then $sqrt{x^2}$ will return $x > 0$ as well, so it equals $-x$. Hence,



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



    This is all just a way of saying $sqrt{x^2}$ will always return a non-negative value.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 6 at 7:32









    KM101KM101

    6,0701525




    6,0701525












    • $begingroup$
      Thanks! and so in concluding, the absolute value sign is used just to simplify stuff? Also, in addition to this are there specific rules that i have to obey when using this as well as does this comes up often in calculus or other topics?
      $endgroup$
      – Fred Weasley
      Jan 6 at 8:15










    • $begingroup$
      It’s really just a reiteration if the definition of the principal root, which is always non-negative. It can be used for simplification I suppose. As for your second question, I’m not sure, as I’m quite new to calculus myself.
      $endgroup$
      – KM101
      Jan 6 at 8:45




















    • $begingroup$
      Thanks! and so in concluding, the absolute value sign is used just to simplify stuff? Also, in addition to this are there specific rules that i have to obey when using this as well as does this comes up often in calculus or other topics?
      $endgroup$
      – Fred Weasley
      Jan 6 at 8:15










    • $begingroup$
      It’s really just a reiteration if the definition of the principal root, which is always non-negative. It can be used for simplification I suppose. As for your second question, I’m not sure, as I’m quite new to calculus myself.
      $endgroup$
      – KM101
      Jan 6 at 8:45


















    $begingroup$
    Thanks! and so in concluding, the absolute value sign is used just to simplify stuff? Also, in addition to this are there specific rules that i have to obey when using this as well as does this comes up often in calculus or other topics?
    $endgroup$
    – Fred Weasley
    Jan 6 at 8:15




    $begingroup$
    Thanks! and so in concluding, the absolute value sign is used just to simplify stuff? Also, in addition to this are there specific rules that i have to obey when using this as well as does this comes up often in calculus or other topics?
    $endgroup$
    – Fred Weasley
    Jan 6 at 8:15












    $begingroup$
    It’s really just a reiteration if the definition of the principal root, which is always non-negative. It can be used for simplification I suppose. As for your second question, I’m not sure, as I’m quite new to calculus myself.
    $endgroup$
    – KM101
    Jan 6 at 8:45






    $begingroup$
    It’s really just a reiteration if the definition of the principal root, which is always non-negative. It can be used for simplification I suppose. As for your second question, I’m not sure, as I’m quite new to calculus myself.
    $endgroup$
    – KM101
    Jan 6 at 8:45













    0












    $begingroup$

    I agree with KM 101. I haven't seen the video, but what they try to say is that by taking the square root you get a distance on the number line (which is what the absolute value is), which is the same in both directions.



    But once the number line enters the picture to explain the absolute values, the type of numbers for which the line exists also enters the picture, and that's the real numbers, which is a separate topic.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      I agree with KM 101. I haven't seen the video, but what they try to say is that by taking the square root you get a distance on the number line (which is what the absolute value is), which is the same in both directions.



      But once the number line enters the picture to explain the absolute values, the type of numbers for which the line exists also enters the picture, and that's the real numbers, which is a separate topic.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        I agree with KM 101. I haven't seen the video, but what they try to say is that by taking the square root you get a distance on the number line (which is what the absolute value is), which is the same in both directions.



        But once the number line enters the picture to explain the absolute values, the type of numbers for which the line exists also enters the picture, and that's the real numbers, which is a separate topic.






        share|cite|improve this answer









        $endgroup$



        I agree with KM 101. I haven't seen the video, but what they try to say is that by taking the square root you get a distance on the number line (which is what the absolute value is), which is the same in both directions.



        But once the number line enters the picture to explain the absolute values, the type of numbers for which the line exists also enters the picture, and that's the real numbers, which is a separate topic.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 7:49









        Uncle VUncle V

        11




        11






























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