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The Defining Moment



Why isn't the definition of absolute value applied when squaring a radical containing a variable?


Different ways of defining Absolute ValueAdding $2$ absolute values together: $|x+2| + |x-3| =5.$Why does $|x_1| = |x_2| implies x_1 = pm x_2$Why does the integral $intfrac{1}{x+i}dx$ not require the absolute value in the logarithm?The Definition of the Absolute ValueSpivak's Calculus 4th ed: Absolute value definition p. 11Solving Equations with Roots and Absolute ValuesAbsolute Value. Real AnalysisSolving Radical InequalitiesHow to get rid of the absolute value when solving a system of complex equations?













1












$begingroup$


I recently learned about the following definition of absolute value:




$|a| = sqrt{a^2}$




Then I came across a solution to a problem that had the following step:




$5 geq sqrt{5 - x}$




In order to proceed, we had to square both sides:




$5^2 geq (sqrt{5 - x})^2$




With the aforementioned definition of absolute value in mind, I wrote:




$25 geq |5 - x|$




But the actual solution turned out to be:




$25 geq 5 - x$




I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?






algebra-precalculus radicals absolute-value







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
    $endgroup$
    – user647486
    1 hour ago


















1












$begingroup$


I recently learned about the following definition of absolute value:




$|a| = sqrt{a^2}$




Then I came across a solution to a problem that had the following step:




$5 geq sqrt{5 - x}$




In order to proceed, we had to square both sides:




$5^2 geq (sqrt{5 - x})^2$




With the aforementioned definition of absolute value in mind, I wrote:




$25 geq |5 - x|$




But the actual solution turned out to be:




$25 geq 5 - x$




I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?






algebra-precalculus radicals absolute-value







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
    $endgroup$
    – user647486
    1 hour ago
















1












1








1





$begingroup$


I recently learned about the following definition of absolute value:




$|a| = sqrt{a^2}$




Then I came across a solution to a problem that had the following step:




$5 geq sqrt{5 - x}$




In order to proceed, we had to square both sides:




$5^2 geq (sqrt{5 - x})^2$




With the aforementioned definition of absolute value in mind, I wrote:




$25 geq |5 - x|$




But the actual solution turned out to be:




$25 geq 5 - x$




I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?






algebra-precalculus radicals absolute-value







share|cite|improve this question









$endgroup$




I recently learned about the following definition of absolute value:




$|a| = sqrt{a^2}$




Then I came across a solution to a problem that had the following step:




$5 geq sqrt{5 - x}$




In order to proceed, we had to square both sides:




$5^2 geq (sqrt{5 - x})^2$




With the aforementioned definition of absolute value in mind, I wrote:




$25 geq |5 - x|$




But the actual solution turned out to be:




$25 geq 5 - x$




I don't understand why the absolute value definition wasn't applied here. Can anyone tell me why?






algebra-precalculus radicals absolute-value




algebra-precalculus radicals absolute-value






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 1 hour ago









CalculemusCalculemus

392217




392217








  • 1




    $begingroup$
    It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
    $endgroup$
    – user647486
    1 hour ago
















  • 1




    $begingroup$
    It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
    $endgroup$
    – user647486
    1 hour ago










1




1




$begingroup$
It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
$endgroup$
– user647486
1 hour ago






$begingroup$
It is applied. What you need to note is that the operation of squaring both sides of an inequality is not an equivalent transformation, like adding the same amount on both sides is. The inequality before squaring implies the inequality after squaring, but the reverse implication is not satisfied. So, when you square the new inequality has all solutions of the original, but it might have additional solutions. So, when you square you do apply the absolute value, since this is in the definition of $sqrt{ }$, but after solving the resulting inequality, you need to check for false solutions.
$endgroup$
– user647486
1 hour ago












4 Answers
4






active

oldest

votes


















3












$begingroup$

$$left(sqrt aright)^2nesqrt{a^2}.$$



Try with $a=-1$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Indeed, $sqrt{a^2}=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      The inequality has the domain $$5geq x$$, so we can square it and we get $$25geq 5-x$$






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        From the fact that you can take $sqrt {5-x}$ you know that $5-x ge 0$ so you don't need the absolute value signs.






        share|cite|improve this answer









        $endgroup$









        • 1




          $begingroup$
          You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
          $endgroup$
          – steven gregory
          1 hour ago












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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        $$left(sqrt aright)^2nesqrt{a^2}.$$



        Try with $a=-1$.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          $$left(sqrt aright)^2nesqrt{a^2}.$$



          Try with $a=-1$.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            $$left(sqrt aright)^2nesqrt{a^2}.$$



            Try with $a=-1$.






            share|cite|improve this answer









            $endgroup$



            $$left(sqrt aright)^2nesqrt{a^2}.$$



            Try with $a=-1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Yves DaoustYves Daoust

            134k676232




            134k676232























                2












                $begingroup$

                Indeed, $sqrt{a^2}=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Indeed, $sqrt{a^2}=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Indeed, $sqrt{a^2}=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.






                    share|cite|improve this answer









                    $endgroup$



                    Indeed, $sqrt{a^2}=lvert arvert$. But $sqrt a^2=a$ (assuming that $ageqslant0$), not $lvert arvert$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    José Carlos SantosJosé Carlos Santos

                    178k24139251




                    178k24139251























                        1












                        $begingroup$

                        The inequality has the domain $$5geq x$$, so we can square it and we get $$25geq 5-x$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          The inequality has the domain $$5geq x$$, so we can square it and we get $$25geq 5-x$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            The inequality has the domain $$5geq x$$, so we can square it and we get $$25geq 5-x$$






                            share|cite|improve this answer









                            $endgroup$



                            The inequality has the domain $$5geq x$$, so we can square it and we get $$25geq 5-x$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            79.9k42867




                            79.9k42867























                                1












                                $begingroup$

                                From the fact that you can take $sqrt {5-x}$ you know that $5-x ge 0$ so you don't need the absolute value signs.






                                share|cite|improve this answer









                                $endgroup$









                                • 1




                                  $begingroup$
                                  You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                  $endgroup$
                                  – steven gregory
                                  1 hour ago
















                                1












                                $begingroup$

                                From the fact that you can take $sqrt {5-x}$ you know that $5-x ge 0$ so you don't need the absolute value signs.






                                share|cite|improve this answer









                                $endgroup$









                                • 1




                                  $begingroup$
                                  You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                  $endgroup$
                                  – steven gregory
                                  1 hour ago














                                1












                                1








                                1





                                $begingroup$

                                From the fact that you can take $sqrt {5-x}$ you know that $5-x ge 0$ so you don't need the absolute value signs.






                                share|cite|improve this answer









                                $endgroup$



                                From the fact that you can take $sqrt {5-x}$ you know that $5-x ge 0$ so you don't need the absolute value signs.







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered 1 hour ago









                                Ross MillikanRoss Millikan

                                302k24201375




                                302k24201375








                                • 1




                                  $begingroup$
                                  You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                  $endgroup$
                                  – steven gregory
                                  1 hour ago














                                • 1




                                  $begingroup$
                                  You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                  $endgroup$
                                  – steven gregory
                                  1 hour ago








                                1




                                1




                                $begingroup$
                                You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                $endgroup$
                                – steven gregory
                                1 hour ago




                                $begingroup$
                                You don't really "know that" $5-x ge 0$, rather you "must have" $5-x ge 0$; in either case, it follows that $|5-x| = 5-x$. So $25 ge |5-x|$ becomes $25 ge 5-x$.
                                $endgroup$
                                – steven gregory
                                1 hour ago


















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