If $V$ is a vector space over the field $\mathbf F$, then it must satisfy two properties, namely closure under addition and closure under multiplication.

For closure under multiplication, we demand that if $u \in V$, $a \in \mathbf F$, then $a \mathbf F \in V$. Note that the 'multiplication' needs to be defined beforehand.

Specifically to your example, perhaps you are having trouble with what's meant by closure. According to the definition, you want to prove that $S:\{(a,b)|a,b \in \mathscr R\}$ is a vector space.

You have actually done it. You have shown it is closed under addition as well as multiplication.

Let's take your example,$–10 \times (1, –7) = (–10 \times 1, –10 \times –7) = (–10, 70)$, the original vector$(1,-7) \in S$,right? Also$(-10,70) \in S$ because let $a=-10,b=70$, then you see $(-10,70)$ satisfy the requirement $a,b \in R$. Thus you have proved closure under multiplication.

If $V$ is a vector space over the field $\mathbf F$, then it must satisfy two properties, namely closure under addition and closure under multiplication.

For closure under multiplication, we demand that if $u \in V$, $a \in \mathbf F$, then $a \mathbf F \in V$. Note that the 'multiplication' needs to be defined beforehand.

Specifically to your example, perhaps you are having trouble with what's meant by closure. According to the definition, you want to prove that $S:\{(a,b)|a,b \in \mathscr R\}$ is a vector space.

You have actually done it. You have shown it is closed under addition as well as multiplication.

Let's take your example,$–10 \times (1, –7) = (–10 \times 1, –10 \times –7) = (–10, 70)$, the original vector$(1,-7) \in S$,right? Also$(-10,70) \in S$ because let $a=-10,b=70$, then you see $(-10,70)$ satisfy the requirement $a,b \in R$. Thus you have proved closure under multiplication.

Edit: I will try to prove the statement

$\mathbb R^2$ is a vector space over the field $\mathbb R$

Proof: if $ a,b \in \mathbb R$, then $(a,b) \in \mathbb R^2$. Let $k \in \mathbb R$ be a scalar, then $k(a,b)=(ka,kb)$. As $k,a,b \in \mathbb R$, $ka,kb \in \mathbb R$, thus $(ka,kb) \in \mathbb R^2$. Therefore scalar multiplication on $\mathbb R^2$ is closed.

If $V$ is a vector space over the field $\mathbf F$, then it must satisfy two properties, namely closure under addition and closure under multiplication.

For closure under multiplication, we demand that if $u \in V$, $a \in \mathbf F$, then $a \mathbf F \in V$. Note that the 'multiplication' needs to be defined beforehand.

For a reference, see exercise 5 for chapter 2.1 in Hoffman's Linear Algebra, in case you have it.

If $V$ is a vector space over the field $\mathbf F$, then it must satisfy two properties, namely closure under addition and closure under multiplication.

For closure under multiplication, we demand that if $u \in V$, $a \in \mathbf F$, then $a \mathbf F \in V$. Note that the 'multiplication' needs to be defined beforehand.

Specifically to your example, perhaps you are having trouble with what's meant by closure. According to the definition, you want to prove that $S:\{(a,b)|a,b \in \mathscr R\}$ is a vector space.

You have actually done it. You have shown it is closed under addition as well as multiplication.

Let's take your example,$–10 \times (1, –7) = (–10 \times 1, –10 \times –7) = (–10, 70)$, the original vector$(1,-7) \in S$,right? Also$(-10,70) \in S$ because let $a=-10,b=70$, then you see $(-10,70)$ satisfy the requirement $a,b \in R$. Thus you have proved closure under multiplication.