Is my proof of Commutativity of addition in a Vector space correct?












0












$begingroup$


Adding vectors is commutative because adding coordinates is commutative and vector addition is merely two applications of that same law vectors $A$, $B$



$$A + B = (overrightarrow{a_1,b_1}) + (overrightarrow{a_2, b_2}) = (a_1 + b_1, a_2 + b_2) + (c_1 + d_1, c_2 + d_2)$$



which implies Commutativity be default due to the fact that we are merely adding real numbers.



Edit: Mistook commutativity for associativity!






linear-algebra vectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    I am seriously confused. What are $A,B$? What exactly are you trying to show?
    $endgroup$
    – BigbearZzz
    Dec 23 '18 at 4:51






  • 2




    $begingroup$
    associativity of vector addition operation -- axiom of vector space
    $endgroup$
    – Just do it
    Dec 23 '18 at 4:52










  • $begingroup$
    Just for context this problem is from "A Survey of Modern Algebra" - Garrett Birkoff, and Saunders Mac Lane, p.169.
    $endgroup$
    – PossiblyDakota
    Dec 23 '18 at 5:10






  • 1




    $begingroup$
    Not sure if it's not showing everything on my phone, but it seems like you have some c's and d's coning out of nowhere; you have to show $(A+B)+C=A+(B+C)$, so you should pick a side, start from it, and arrive at the other.
    $endgroup$
    – Ovi
    Dec 23 '18 at 5:16








  • 2




    $begingroup$
    @PossiblyDakota Commutativity of addition is also taken as an axiom... Can you quote the problem from "A Survey of Modern Algebra"?
    $endgroup$
    – Dair
    Dec 23 '18 at 5:23


















0












$begingroup$


Adding vectors is commutative because adding coordinates is commutative and vector addition is merely two applications of that same law vectors $A$, $B$



$$A + B = (overrightarrow{a_1,b_1}) + (overrightarrow{a_2, b_2}) = (a_1 + b_1, a_2 + b_2) + (c_1 + d_1, c_2 + d_2)$$



which implies Commutativity be default due to the fact that we are merely adding real numbers.



Edit: Mistook commutativity for associativity!






linear-algebra vectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    I am seriously confused. What are $A,B$? What exactly are you trying to show?
    $endgroup$
    – BigbearZzz
    Dec 23 '18 at 4:51






  • 2




    $begingroup$
    associativity of vector addition operation -- axiom of vector space
    $endgroup$
    – Just do it
    Dec 23 '18 at 4:52










  • $begingroup$
    Just for context this problem is from "A Survey of Modern Algebra" - Garrett Birkoff, and Saunders Mac Lane, p.169.
    $endgroup$
    – PossiblyDakota
    Dec 23 '18 at 5:10






  • 1




    $begingroup$
    Not sure if it's not showing everything on my phone, but it seems like you have some c's and d's coning out of nowhere; you have to show $(A+B)+C=A+(B+C)$, so you should pick a side, start from it, and arrive at the other.
    $endgroup$
    – Ovi
    Dec 23 '18 at 5:16








  • 2




    $begingroup$
    @PossiblyDakota Commutativity of addition is also taken as an axiom... Can you quote the problem from "A Survey of Modern Algebra"?
    $endgroup$
    – Dair
    Dec 23 '18 at 5:23
















0












0








0





$begingroup$


Adding vectors is commutative because adding coordinates is commutative and vector addition is merely two applications of that same law vectors $A$, $B$



$$A + B = (overrightarrow{a_1,b_1}) + (overrightarrow{a_2, b_2}) = (a_1 + b_1, a_2 + b_2) + (c_1 + d_1, c_2 + d_2)$$



which implies Commutativity be default due to the fact that we are merely adding real numbers.



Edit: Mistook commutativity for associativity!






linear-algebra vectors







share|cite|improve this question











$endgroup$




Adding vectors is commutative because adding coordinates is commutative and vector addition is merely two applications of that same law vectors $A$, $B$



$$A + B = (overrightarrow{a_1,b_1}) + (overrightarrow{a_2, b_2}) = (a_1 + b_1, a_2 + b_2) + (c_1 + d_1, c_2 + d_2)$$



which implies Commutativity be default due to the fact that we are merely adding real numbers.



Edit: Mistook commutativity for associativity!






linear-algebra vectors




linear-algebra vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 5:29









Saad

19.7k92352




19.7k92352










asked Dec 23 '18 at 4:47









PossiblyDakotaPossiblyDakota

696




696












  • $begingroup$
    I am seriously confused. What are $A,B$? What exactly are you trying to show?
    $endgroup$
    – BigbearZzz
    Dec 23 '18 at 4:51






  • 2




    $begingroup$
    associativity of vector addition operation -- axiom of vector space
    $endgroup$
    – Just do it
    Dec 23 '18 at 4:52










  • $begingroup$
    Just for context this problem is from "A Survey of Modern Algebra" - Garrett Birkoff, and Saunders Mac Lane, p.169.
    $endgroup$
    – PossiblyDakota
    Dec 23 '18 at 5:10






  • 1




    $begingroup$
    Not sure if it's not showing everything on my phone, but it seems like you have some c's and d's coning out of nowhere; you have to show $(A+B)+C=A+(B+C)$, so you should pick a side, start from it, and arrive at the other.
    $endgroup$
    – Ovi
    Dec 23 '18 at 5:16








  • 2




    $begingroup$
    @PossiblyDakota Commutativity of addition is also taken as an axiom... Can you quote the problem from "A Survey of Modern Algebra"?
    $endgroup$
    – Dair
    Dec 23 '18 at 5:23




















  • $begingroup$
    I am seriously confused. What are $A,B$? What exactly are you trying to show?
    $endgroup$
    – BigbearZzz
    Dec 23 '18 at 4:51






  • 2




    $begingroup$
    associativity of vector addition operation -- axiom of vector space
    $endgroup$
    – Just do it
    Dec 23 '18 at 4:52










  • $begingroup$
    Just for context this problem is from "A Survey of Modern Algebra" - Garrett Birkoff, and Saunders Mac Lane, p.169.
    $endgroup$
    – PossiblyDakota
    Dec 23 '18 at 5:10






  • 1




    $begingroup$
    Not sure if it's not showing everything on my phone, but it seems like you have some c's and d's coning out of nowhere; you have to show $(A+B)+C=A+(B+C)$, so you should pick a side, start from it, and arrive at the other.
    $endgroup$
    – Ovi
    Dec 23 '18 at 5:16








  • 2




    $begingroup$
    @PossiblyDakota Commutativity of addition is also taken as an axiom... Can you quote the problem from "A Survey of Modern Algebra"?
    $endgroup$
    – Dair
    Dec 23 '18 at 5:23


















$begingroup$
I am seriously confused. What are $A,B$? What exactly are you trying to show?
$endgroup$
– BigbearZzz
Dec 23 '18 at 4:51




$begingroup$
I am seriously confused. What are $A,B$? What exactly are you trying to show?
$endgroup$
– BigbearZzz
Dec 23 '18 at 4:51




2




2




$begingroup$
associativity of vector addition operation -- axiom of vector space
$endgroup$
– Just do it
Dec 23 '18 at 4:52




$begingroup$
associativity of vector addition operation -- axiom of vector space
$endgroup$
– Just do it
Dec 23 '18 at 4:52












$begingroup$
Just for context this problem is from "A Survey of Modern Algebra" - Garrett Birkoff, and Saunders Mac Lane, p.169.
$endgroup$
– PossiblyDakota
Dec 23 '18 at 5:10




$begingroup$
Just for context this problem is from "A Survey of Modern Algebra" - Garrett Birkoff, and Saunders Mac Lane, p.169.
$endgroup$
– PossiblyDakota
Dec 23 '18 at 5:10




1




1




$begingroup$
Not sure if it's not showing everything on my phone, but it seems like you have some c's and d's coning out of nowhere; you have to show $(A+B)+C=A+(B+C)$, so you should pick a side, start from it, and arrive at the other.
$endgroup$
– Ovi
Dec 23 '18 at 5:16






$begingroup$
Not sure if it's not showing everything on my phone, but it seems like you have some c's and d's coning out of nowhere; you have to show $(A+B)+C=A+(B+C)$, so you should pick a side, start from it, and arrive at the other.
$endgroup$
– Ovi
Dec 23 '18 at 5:16






2




2




$begingroup$
@PossiblyDakota Commutativity of addition is also taken as an axiom... Can you quote the problem from "A Survey of Modern Algebra"?
$endgroup$
– Dair
Dec 23 '18 at 5:23






$begingroup$
@PossiblyDakota Commutativity of addition is also taken as an axiom... Can you quote the problem from "A Survey of Modern Algebra"?
$endgroup$
– Dair
Dec 23 '18 at 5:23












1 Answer
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1












$begingroup$

Your proof is confusing because $c_1, c_2, d_1$, and $d_2$ appear out of nowhere. But I think your general idea is correct. The proof is as follows.



The commutative property of two-dimensional real vectors is: For all two-dimensional real vectors $a = (a_1, a_2), b = (b_1, b_2)$, we must have $a + b = b + a$. This is true since



$$
a + b = (a_1 + b_1, a_2 + b_2) = (b_1 + a_1, b_2 + a_2) = b + a
$$



where the second equality follows from the commutative property of real numbers and the other equalities follow from the definition of vector addition.






share|cite|improve this answer











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    $begingroup$

    Your proof is confusing because $c_1, c_2, d_1$, and $d_2$ appear out of nowhere. But I think your general idea is correct. The proof is as follows.



    The commutative property of two-dimensional real vectors is: For all two-dimensional real vectors $a = (a_1, a_2), b = (b_1, b_2)$, we must have $a + b = b + a$. This is true since



    $$
    a + b = (a_1 + b_1, a_2 + b_2) = (b_1 + a_1, b_2 + a_2) = b + a
    $$



    where the second equality follows from the commutative property of real numbers and the other equalities follow from the definition of vector addition.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Your proof is confusing because $c_1, c_2, d_1$, and $d_2$ appear out of nowhere. But I think your general idea is correct. The proof is as follows.



      The commutative property of two-dimensional real vectors is: For all two-dimensional real vectors $a = (a_1, a_2), b = (b_1, b_2)$, we must have $a + b = b + a$. This is true since



      $$
      a + b = (a_1 + b_1, a_2 + b_2) = (b_1 + a_1, b_2 + a_2) = b + a
      $$



      where the second equality follows from the commutative property of real numbers and the other equalities follow from the definition of vector addition.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Your proof is confusing because $c_1, c_2, d_1$, and $d_2$ appear out of nowhere. But I think your general idea is correct. The proof is as follows.



        The commutative property of two-dimensional real vectors is: For all two-dimensional real vectors $a = (a_1, a_2), b = (b_1, b_2)$, we must have $a + b = b + a$. This is true since



        $$
        a + b = (a_1 + b_1, a_2 + b_2) = (b_1 + a_1, b_2 + a_2) = b + a
        $$



        where the second equality follows from the commutative property of real numbers and the other equalities follow from the definition of vector addition.






        share|cite|improve this answer











        $endgroup$



        Your proof is confusing because $c_1, c_2, d_1$, and $d_2$ appear out of nowhere. But I think your general idea is correct. The proof is as follows.



        The commutative property of two-dimensional real vectors is: For all two-dimensional real vectors $a = (a_1, a_2), b = (b_1, b_2)$, we must have $a + b = b + a$. This is true since



        $$
        a + b = (a_1 + b_1, a_2 + b_2) = (b_1 + a_1, b_2 + a_2) = b + a
        $$



        where the second equality follows from the commutative property of real numbers and the other equalities follow from the definition of vector addition.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 23 '18 at 5:29









        Aniruddh Agarwal

        1218




        1218










        answered Dec 23 '18 at 5:25









        ToriTori

        585




        585






























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