A for each u in V there is an object u in V such that uuuu0

A.) for each u in V, there is an object -u in V, such that u+(-u)=(-u)+u=0.

B.) if k is any scalr and u is any object in V, then ku is in V.

Solution

If V=(x₁ , y₁ ) and U=(x₂ ,y₂) in R2.

Then the vector addition    V+U = ( x₁+x₂, y₁+y₂ )

and the scalar multiplication ,

    for a scalar c and the vector V then   cV = ( cx₁, cy₁ )

A) \"For each u in v, there is an object -u in V , such that u=(-u)=(-u)+u=0\" which axiom do not hold , because given that \"V be the set of pairs of real numbers (x,y) such that x>=0 and y>=0\". Here (-u)<0 so (-u) do not belongs toV.

B) \"If k is any scalar and u is any object in V , then ku is in V\" which axiom also do not hold, because given that k is any scalar it may be negative or posative. if it is negative, then ku is a negative number which contradict the given condition \"V be the set of pairs of real numbers (x,y) such that x>=0 and y>=0\".