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If $V$ is a vector space over the field $F$ then verify that

$$(\alpha_1+ \alpha_2)+(\alpha_3+\alpha_4)=[\alpha_2+(\alpha_3+\alpha_1)]+\alpha_4$$

for all the vectors $\alpha_1,\alpha_2,\alpha_3,\alpha_4$ in $V$?

i think we will use properties of vector spaces like associative,cummutative,etc,,,, but i dont how to get it

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Welcome to SE David. You might want to better format your question, make it more readable. Use the tex code \alpha_1 between dollar sings to produce $\alpha _1$. Also, you might want to give some indication of what you had tried so far. – Ittay Weiss Mar 31 '13 at 12:11

This is just the associativity and commutativity of the addition: In your example you would use the associativity in the following way: $(\alpha_1+\alpha_2)+(\alpha_3+\alpha_4)=((\alpha_1+\alpha_2)+\alpha_3)+\alpha_4=[\alpha_1+(\alpha_2+\alpha_3)]+\alpha_4=(\ast)$. Now we use commutativity and get $(\ast)=[\alpha_1+(\alpha_3+\alpha_2)]+\alpha_4=(\ast\ast)$ and associativity to get $(\ast\ast)=[(\alpha_1+\alpha_3)+\alpha_2]+\alpha_4=(\ast\ast\ast)$. Now we just have to use commutativity twice to get the final result $(\ast\ast\ast)=[(\alpha_3+\alpha_1)+\alpha_2]+\alpha_4=[\alpha_2+(\alpha_3+\alpha_1)]+\alpha_4$. I hope that helps a bit.

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$(\alpha_1+\alpha_2)+(\alpha_3+\alpha_4)$ = $[(\alpha_1+\alpha_2)+\alpha_3]+\alpha_4$ = $[(\alpha_2+\alpha_1)+\alpha_3]+\alpha_4$ = $[\alpha_2+(\alpha_1+\alpha_3)]+\alpha_4$ = $[\alpha_2+(\alpha_3+\alpha_1)]+\alpha_4$

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