How to you prove $|a|=|-a|$ is true












1












$begingroup$


Let $a in R$ prove that:



$|a|=|-a|$



I am new to proofs so this is my attempt:



Case 1: $|a|=|-a|$



$$(a)=-(-a)$$



$$a=a$$



Case 2: $|-a|=|-a|$



$$-(-a)=-(-a)$$



$$a=a$$



Case 3: $|a|=|a|$



$$a=a$$



Is this the correct way to approach a proof like this?






proof-verification proof-writing proof-explanation absolute-value







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I don't understand the cases you are distinguishing.
    $endgroup$
    – lulu
    Jan 18 at 2:23






  • 3




    $begingroup$
    Make your cases $a>0, a<0, a=0$ An alternative would be to say $|a| = sqrt {a^2} = sqrt {(-a)^2} = |-a|$
    $endgroup$
    – Doug M
    Jan 18 at 2:23








  • 1




    $begingroup$
    Hint: $|a|$ is the greater of $a,-a$.
    $endgroup$
    – lulu
    Jan 18 at 2:24






  • 2




    $begingroup$
    It's going to depend on how you define $|a|$. There are several ways to do it; which one are you using?
    $endgroup$
    – dbx
    Jan 18 at 2:25
















1












$begingroup$


Let $a in R$ prove that:



$|a|=|-a|$



I am new to proofs so this is my attempt:



Case 1: $|a|=|-a|$



$$(a)=-(-a)$$



$$a=a$$



Case 2: $|-a|=|-a|$



$$-(-a)=-(-a)$$



$$a=a$$



Case 3: $|a|=|a|$



$$a=a$$



Is this the correct way to approach a proof like this?






proof-verification proof-writing proof-explanation absolute-value







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I don't understand the cases you are distinguishing.
    $endgroup$
    – lulu
    Jan 18 at 2:23






  • 3




    $begingroup$
    Make your cases $a>0, a<0, a=0$ An alternative would be to say $|a| = sqrt {a^2} = sqrt {(-a)^2} = |-a|$
    $endgroup$
    – Doug M
    Jan 18 at 2:23








  • 1




    $begingroup$
    Hint: $|a|$ is the greater of $a,-a$.
    $endgroup$
    – lulu
    Jan 18 at 2:24






  • 2




    $begingroup$
    It's going to depend on how you define $|a|$. There are several ways to do it; which one are you using?
    $endgroup$
    – dbx
    Jan 18 at 2:25














1












1








1





$begingroup$


Let $a in R$ prove that:



$|a|=|-a|$



I am new to proofs so this is my attempt:



Case 1: $|a|=|-a|$



$$(a)=-(-a)$$



$$a=a$$



Case 2: $|-a|=|-a|$



$$-(-a)=-(-a)$$



$$a=a$$



Case 3: $|a|=|a|$



$$a=a$$



Is this the correct way to approach a proof like this?






proof-verification proof-writing proof-explanation absolute-value







share|cite|improve this question











$endgroup$




Let $a in R$ prove that:



$|a|=|-a|$



I am new to proofs so this is my attempt:



Case 1: $|a|=|-a|$



$$(a)=-(-a)$$



$$a=a$$



Case 2: $|-a|=|-a|$



$$-(-a)=-(-a)$$



$$a=a$$



Case 3: $|a|=|a|$



$$a=a$$



Is this the correct way to approach a proof like this?






proof-verification proof-writing proof-explanation absolute-value




proof-verification proof-writing proof-explanation absolute-value






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 7:31









Michael Rozenberg

103k1891195




103k1891195










asked Jan 18 at 2:21









JohnJohn

325




325








  • 1




    $begingroup$
    I don't understand the cases you are distinguishing.
    $endgroup$
    – lulu
    Jan 18 at 2:23






  • 3




    $begingroup$
    Make your cases $a>0, a<0, a=0$ An alternative would be to say $|a| = sqrt {a^2} = sqrt {(-a)^2} = |-a|$
    $endgroup$
    – Doug M
    Jan 18 at 2:23








  • 1




    $begingroup$
    Hint: $|a|$ is the greater of $a,-a$.
    $endgroup$
    – lulu
    Jan 18 at 2:24






  • 2




    $begingroup$
    It's going to depend on how you define $|a|$. There are several ways to do it; which one are you using?
    $endgroup$
    – dbx
    Jan 18 at 2:25














  • 1




    $begingroup$
    I don't understand the cases you are distinguishing.
    $endgroup$
    – lulu
    Jan 18 at 2:23






  • 3




    $begingroup$
    Make your cases $a>0, a<0, a=0$ An alternative would be to say $|a| = sqrt {a^2} = sqrt {(-a)^2} = |-a|$
    $endgroup$
    – Doug M
    Jan 18 at 2:23








  • 1




    $begingroup$
    Hint: $|a|$ is the greater of $a,-a$.
    $endgroup$
    – lulu
    Jan 18 at 2:24






  • 2




    $begingroup$
    It's going to depend on how you define $|a|$. There are several ways to do it; which one are you using?
    $endgroup$
    – dbx
    Jan 18 at 2:25








1




1




$begingroup$
I don't understand the cases you are distinguishing.
$endgroup$
– lulu
Jan 18 at 2:23




$begingroup$
I don't understand the cases you are distinguishing.
$endgroup$
– lulu
Jan 18 at 2:23




3




3




$begingroup$
Make your cases $a>0, a<0, a=0$ An alternative would be to say $|a| = sqrt {a^2} = sqrt {(-a)^2} = |-a|$
$endgroup$
– Doug M
Jan 18 at 2:23






$begingroup$
Make your cases $a>0, a<0, a=0$ An alternative would be to say $|a| = sqrt {a^2} = sqrt {(-a)^2} = |-a|$
$endgroup$
– Doug M
Jan 18 at 2:23






1




1




$begingroup$
Hint: $|a|$ is the greater of $a,-a$.
$endgroup$
– lulu
Jan 18 at 2:24




$begingroup$
Hint: $|a|$ is the greater of $a,-a$.
$endgroup$
– lulu
Jan 18 at 2:24




2




2




$begingroup$
It's going to depend on how you define $|a|$. There are several ways to do it; which one are you using?
$endgroup$
– dbx
Jan 18 at 2:25




$begingroup$
It's going to depend on how you define $|a|$. There are several ways to do it; which one are you using?
$endgroup$
– dbx
Jan 18 at 2:25










3 Answers
3






active

oldest

votes


















1












$begingroup$

Because the distances from $a$ and from $-a$ to the zero are equal.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The definition of absolute value of a real number $a$ is $|a|=sqrt{x^2}$. Since $a^2=(-a)^2$ and square roots only take positive values $|a|=|-a|$






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      If you are going to do a case by case proof, you only need to consider two cases:



      Case 1:
      $a ge 0 \ Rightarrow |a|=a, |-a|=a \ Rightarrow |a|=|-a|$



      Case 2:
      $a < 0 \ Rightarrow |a|=-a, |-a|=-a \ Rightarrow |a|=|-a|$






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        Because the distances from $a$ and from $-a$ to the zero are equal.






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Because the distances from $a$ and from $-a$ to the zero are equal.






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Because the distances from $a$ and from $-a$ to the zero are equal.






            share|cite|improve this answer









            $endgroup$



            Because the distances from $a$ and from $-a$ to the zero are equal.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 18 at 7:21









            Michael RozenbergMichael Rozenberg

            103k1891195




            103k1891195























                0












                $begingroup$

                The definition of absolute value of a real number $a$ is $|a|=sqrt{x^2}$. Since $a^2=(-a)^2$ and square roots only take positive values $|a|=|-a|$






                share|cite|improve this answer









                $endgroup$


















                  0












                  $begingroup$

                  The definition of absolute value of a real number $a$ is $|a|=sqrt{x^2}$. Since $a^2=(-a)^2$ and square roots only take positive values $|a|=|-a|$






                  share|cite|improve this answer









                  $endgroup$
















                    0












                    0








                    0





                    $begingroup$

                    The definition of absolute value of a real number $a$ is $|a|=sqrt{x^2}$. Since $a^2=(-a)^2$ and square roots only take positive values $|a|=|-a|$






                    share|cite|improve this answer









                    $endgroup$



                    The definition of absolute value of a real number $a$ is $|a|=sqrt{x^2}$. Since $a^2=(-a)^2$ and square roots only take positive values $|a|=|-a|$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Jan 18 at 4:41









                    Juan123Juan123

                    948




                    948























                        0












                        $begingroup$

                        If you are going to do a case by case proof, you only need to consider two cases:



                        Case 1:
                        $a ge 0 \ Rightarrow |a|=a, |-a|=a \ Rightarrow |a|=|-a|$



                        Case 2:
                        $a < 0 \ Rightarrow |a|=-a, |-a|=-a \ Rightarrow |a|=|-a|$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          If you are going to do a case by case proof, you only need to consider two cases:



                          Case 1:
                          $a ge 0 \ Rightarrow |a|=a, |-a|=a \ Rightarrow |a|=|-a|$



                          Case 2:
                          $a < 0 \ Rightarrow |a|=-a, |-a|=-a \ Rightarrow |a|=|-a|$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            If you are going to do a case by case proof, you only need to consider two cases:



                            Case 1:
                            $a ge 0 \ Rightarrow |a|=a, |-a|=a \ Rightarrow |a|=|-a|$



                            Case 2:
                            $a < 0 \ Rightarrow |a|=-a, |-a|=-a \ Rightarrow |a|=|-a|$






                            share|cite|improve this answer









                            $endgroup$



                            If you are going to do a case by case proof, you only need to consider two cases:



                            Case 1:
                            $a ge 0 \ Rightarrow |a|=a, |-a|=a \ Rightarrow |a|=|-a|$



                            Case 2:
                            $a < 0 \ Rightarrow |a|=-a, |-a|=-a \ Rightarrow |a|=|-a|$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 18 at 10:03









                            gandalf61gandalf61

                            8,719725




                            8,719725






























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