When to apply negative sign when number is squared












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I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?






algebra-precalculus recreational-mathematics







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  • 3




    $begingroup$
    because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
    $endgroup$
    – Luke
    Apr 21 at 22:01










  • $begingroup$
    Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
    $endgroup$
    – Henry
    Apr 21 at 22:14












  • $begingroup$
    Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
    $endgroup$
    – MCMastery
    Apr 21 at 23:40


















2












$begingroup$


I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?






algebra-precalculus recreational-mathematics







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
    $endgroup$
    – Luke
    Apr 21 at 22:01










  • $begingroup$
    Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
    $endgroup$
    – Henry
    Apr 21 at 22:14












  • $begingroup$
    Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
    $endgroup$
    – MCMastery
    Apr 21 at 23:40
















2












2








2





$begingroup$


I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?






algebra-precalculus recreational-mathematics







share|cite|improve this question









$endgroup$




I always had this confusion of when I need to apply the negative sign in the calculation.
I understand that $(-1)^2 = 1$ however why isn't $-1^2 = 1$?






algebra-precalculus recreational-mathematics




algebra-precalculus recreational-mathematics






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Apr 21 at 21:57









JohnJohnyPapaJohnJohnJohnyPapaJohn

707




707








  • 3




    $begingroup$
    because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
    $endgroup$
    – Luke
    Apr 21 at 22:01










  • $begingroup$
    Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
    $endgroup$
    – Henry
    Apr 21 at 22:14












  • $begingroup$
    Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
    $endgroup$
    – MCMastery
    Apr 21 at 23:40
















  • 3




    $begingroup$
    because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
    $endgroup$
    – Luke
    Apr 21 at 22:01










  • $begingroup$
    Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
    $endgroup$
    – Henry
    Apr 21 at 22:14












  • $begingroup$
    Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
    $endgroup$
    – MCMastery
    Apr 21 at 23:40










3




3




$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
Apr 21 at 22:01




$begingroup$
because $(-1)^2=(-1)*(-1)=1$, but $-1^2 =-(1^2)=-(1*1)=-(1)=-1$
$endgroup$
– Luke
Apr 21 at 22:01












$begingroup$
Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
$endgroup$
– Henry
Apr 21 at 22:14






$begingroup$
Though beware Excel and some similar cases, where =-1^2 gives 1 but =0-1^2 gives -1, because if interprets the former as $(-1)^2$ and the latter as $0-(1^2)$, i.e. the first - as a unary operation taking precedence over exponentiation and the second - as a binary operation with exponentiation taking precedence over it
$endgroup$
– Henry
Apr 21 at 22:14














$begingroup$
Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
$endgroup$
– MCMastery
Apr 21 at 23:40






$begingroup$
Just for an example, that's the same as writing $-1 times 1^2 = 1$, which probably is pretty clear that it's not true
$endgroup$
– MCMastery
Apr 21 at 23:40












3 Answers
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$begingroup$

When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)






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    1












    $begingroup$

    Unary minus has lower precedence than elevation to a power.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      As it is already in the previous answers:
      $(-x)^2neq-x^2$
      To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$






      share|cite|improve this answer








      New contributor




      user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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        3 Answers
        3






        active

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        3 Answers
        3






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        6












        $begingroup$

        When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)






        share|cite|improve this answer









        $endgroup$


















          6












          $begingroup$

          When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)






          share|cite|improve this answer









          $endgroup$
















            6












            6








            6





            $begingroup$

            When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)






            share|cite|improve this answer









            $endgroup$



            When we write $-x^2$, it means we square $x$ first, then take the negative of this. That is, $$-x^2 = -left(x^2right).$$ So $$-1^2 = -left(1^2right)=-1.$$ (And thus $-x^2$ means something different to $(-x)^2$.)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Apr 21 at 22:00









            Minus One-TwelfthMinus One-Twelfth

            3,653513




            3,653513























                1












                $begingroup$

                Unary minus has lower precedence than elevation to a power.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Unary minus has lower precedence than elevation to a power.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Unary minus has lower precedence than elevation to a power.






                    share|cite|improve this answer









                    $endgroup$



                    Unary minus has lower precedence than elevation to a power.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Apr 22 at 18:32









                    maumau

                    7,11523264




                    7,11523264























                        0












                        $begingroup$

                        As it is already in the previous answers:
                        $(-x)^2neq-x^2$
                        To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$






                        share|cite|improve this answer








                        New contributor




                        user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                        Check out our Code of Conduct.






                        $endgroup$


















                          0












                          $begingroup$

                          As it is already in the previous answers:
                          $(-x)^2neq-x^2$
                          To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$






                          share|cite|improve this answer








                          New contributor




                          user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                          Check out our Code of Conduct.






                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            As it is already in the previous answers:
                            $(-x)^2neq-x^2$
                            To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$






                            share|cite|improve this answer








                            New contributor




                            user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






                            $endgroup$



                            As it is already in the previous answers:
                            $(-x)^2neq-x^2$
                            To avoid confusion, it is better to use parentheses. $-x^2=-(x^2)$







                            share|cite|improve this answer








                            New contributor




                            user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            share|cite|improve this answer



                            share|cite|improve this answer






                            New contributor




                            user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.









                            answered Apr 21 at 22:29









                            user665960user665960

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                            155




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                            user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                            user665960 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                            Check out our Code of Conduct.






























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