[proofplan] Both identities follow by expanding the metric through its definition in terms of the norm. For translation, the translated difference $(x+a)-(y+a)$ reduces to $x-y$ by [vector space](/page/Vector%20Space) algebra. For scaling, the scaled difference $\lambda x-\lambda y$ factors as $\lambda(x-y)$, and the absolute homogeneity axiom of the norm converts this into the scalar factor $|\lambda|$. [/proofplan] [step:Expand the translated distance and cancel the common translate] Let $a,x,y\in V$. Since $d$ is defined by $d(u,v)=\|u-v\|$ for all $u,v\in V$, we have \begin{align*} d(x+a,y+a)=\|(x+a)-(y+a)\|. \end{align*} Using associativity and commutativity of vector addition, together with the definition of subtraction in the vector space $V$, \begin{align*} (x+a)-(y+a)=x+a-y-a=x-y. \end{align*} Therefore \begin{align*} d(x+a,y+a)=\|x-y\|=d(x,y). \end{align*} [guided] Fix $a,x,y\in V$. The metric $d$ is not an arbitrary metric here; it is the metric induced by the norm. Thus the first step is to replace $d$ by its defining formula: \begin{align*} d(x+a,y+a)=\|(x+a)-(y+a)\|. \end{align*} Now we simplify the vector inside the norm. Subtraction in a vector space means addition of the additive inverse, so \begin{align*} (x+a)-(y+a)=(x+a)+(-(y+a)). \end{align*} Since the additive inverse of $y+a$ is $-y-a$, and vector addition is associative and commutative, this becomes \begin{align*} (x+a)-(y+a)=x+a-y-a=x-y. \end{align*} Substituting this equality back into the norm gives \begin{align*} d(x+a,y+a)=\|x-y\|. \end{align*} Finally, applying the defining formula for $d$ again, now to the pair $(x,y)$, gives \begin{align*} \|x-y\|=d(x,y). \end{align*} Hence \begin{align*} d(x+a,y+a)=d(x,y). \end{align*} [/guided] [/step] [step:Expand the scaled distance and apply absolute homogeneity of the norm] Let $\lambda\in\mathbb{F}$ and let $x,y\in V$. By the definition of $d$, \begin{align*} d(\lambda x,\lambda y)=\|\lambda x-\lambda y\|. \end{align*} By distributivity of scalar multiplication over vector addition, \begin{align*} \lambda x-\lambda y=\lambda(x-y). \end{align*} Since $\|\cdot\|$ is a norm on the vector space $V$ over $\mathbb{F}$, it satisfies absolute homogeneity: \begin{align*} \|\lambda(x-y)\|=|\lambda|\,\|x-y\|. \end{align*} Using again that $d(x,y)=\|x-y\|$, we obtain \begin{align*} d(\lambda x,\lambda y)=|\lambda|\,d(x,y). \end{align*} [/step] [step:Conclude the two identities hold for all allowed variables] The first step proved that, for arbitrary $a,x,y\in V$, \begin{align*} d(x+a,y+a)=d(x,y). \end{align*} The second step proved that, for arbitrary $\lambda\in\mathbb{F}$ and arbitrary $x,y\in V$, \begin{align*} d(\lambda x,\lambda y)=|\lambda|\,d(x,y). \end{align*} These are exactly the [translation invariance](/theorems/4911) and homogeneity identities asserted in the theorem. [/step]