To determine whether the given set, which is the set of all pairs of real numbers of the form $(1,x)$, equipped with the given operations, forms a vector space, we need to check if it satisfies the vector space axioms.
Let's denote an arbitrary element in the set as $(1,x)$ and $(1,y)$, and a scalar as $k$.
1. Closure under addition:
For any $(1,x)$ and $(1,y)$ in the set, the sum $(1,x)+(1,y)=(1,x+y)$ is also in the set. This satisfies closure under addition.
2. Closure under scalar multiplication:
For any $(1,x)$ in the set and scalar $k$, the scalar multiple $k(1,x)=(1,kx)$ is also in the set. This satisfies closure under scalar multiplication.
3. Commutativity of addition:
$(1,x)+(1,y)=(1,x+y)$ and $(1,y)+(1,x)=(1,y+x)$. Since addition of real numbers is commutative, $(1,x+y)=(1,y+x)$. This satisfies commutativity of addition.
4. Associativity of addition:
$((1,x)+(1,y))+(1,z)=(1,(x+y))+(1,z)=(1,(x+y)+z)=(1,x+(y+z))$. Since addition of real numbers is associative, $(1,(x+y)+z)=(1,x+(y+z))$. This satisfies associativity of addition.
5. Identity element of addition:
There should exist an identity element, denoted as $\mathbf{0}$, such that for any $(1,x)$ in the set, $(1,x)+\mathbf{0}=(1,x)$. However, in the given set, there is no pair $(1,x)$ that satisfies this condition. Therefore, the set does not have an identity element for addition, and it fails the identity element axiom.
6. Existence of additive inverse:
For any $(1,x)$ in the set, there should exist an element, denoted as $-(1,x)$, such that $(1,x)+(-(1,x))=\mathbf{0}$. However, since the set does not have an identity element for addition, it also does not have an additive inverse, and it fails the existence of additive inverse axiom.
7. Distributivity of scalar multiplication over vector addition:
For any scalar $k$ and $(1,x)$ and $(1,y)$ in the set, we have $k·((1,x)+(1,y))=k·(1,x+y)=(1,k(x+y))$. Similarly, $k·(1,x)+k·(1,y)=(1,kx)+(1,ky)=(1,kx+ky)$. Since scalar multiplication and addition of real numbers distribute, $(1,k(x+y))=(1,kx+ky)$. This satisfies distributivity of scalar multiplication over vector addition.
8. Distributivity of scalar multiplication over scalar addition:
For any scalars $k$ and $k^{\prime }$ and $(1,x)$ in the set, we have $(k+k^{\prime })·(1,x)=(1,(k+k^{\prime })x)$. Similarly, $k·(1,x)+k^{\prime }·(1,x)=(1,kx)+(1,k^{\prime }x)=(1,(k+k^{\prime })x)$. Since scalar multiplication and addition of real numbers distribute, $(1,(k+k^{\prime })x)=(1,kx+k^{\prime }x)$. This satisfies distributivity of scalar multiplication over scalar addition.
The given set does not form a vector space because it fails the identity element axiom and the existence of additive inverse axiom.