Question

Let V be the set of all positive real numbers; define + by u+v = uv (+ is ordinary multiplication) and define • by c•v = . Prove that V is a vector space.

Answer 1

$Solution: If~u,v \in V~and ~c \in R,we~have~u+v=uv \in V,c.v=v^c \in V, So~ V~is~ \\closed~under~addition~and~scalar~multiplication. \\Now ~we ~prove ~ remaining ~axioms: \\1) We~have \\~~~~~~~~u+v=uv(\because addition) \\~~~~~~~~~~~~~~~~~~~=vu(\because multiplication~cummutative~in~R) \\~~~~~~~~~~~~~~~~~~~=v+u (\because addition) \\2) Note ~that \\~~~~~~~~~~(u+v)+w=(uv)+w(\because addition) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(uv)w(\because multiplication~cummutative~in~R) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=u(vw)(\because multiplication~ associative ~in~R) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=u+(vw) (\because addition) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=u+(v+w) (\because addition) \\3)Observe~that~1\in V~and~1+u=1u=u \\4) For~any~u \in V, note~that~u^{-1} \in V~so~that~u+u^{-1}=uu^{-1}=1 \\5) Note~that \\~~~~~~~~~~~~~~~~~~~~~1.u=u^1=u \\6)Let~c,d\in R.We~have~ \\~~~~~~~~~~~~~~~~~~~~~(cd).u=u^{cd}~~~~~(\because scalar ~multiplication) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(u^d)^c ~~~(\because~exponents ~in~R) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=c.(u^d)~~~~~(\because scalar ~multiplication) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=c.(d.u)~~~~~(\because scalar ~multiplication) \\7)Note~that \\~~~~~~~~~~~~~~~~~~~~~~c.(u+v)=c.(uv)~~~(\because addition) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(uv)^c~~(\because scalar ~multiplication) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=u^cv^c ~~~(\because~exponents ~in~R) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(u^c)+(v^c)~~(\because scalar ~multiplication) \\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=(c.u)+(c.v)~~(\because addition) \\8)We~have \\~~~~~~~~(c+d).u=u^{c+d} (\because scalar~multiplication) \\~~~~~~~~~~~~~~~~~~~~~~~~=u^c.u^d~~~~(\because exponents ~in~R) \\~~~~~~~~~~~~~~~~~~~~~~~~=(u^c)+(u^d)~~~~(\because~addition) \\~~~~~~~~~~~~~~~~~~~~~~~~=(c.u)+(d.u)~~(\because~Scalar~multiplication) \\All~axioms ~are~satisfied.Hence~V~is ~ a~vector~space.$