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Prove that $d(\vec{x} ,\vec {y})=0$ iff $\vec{x}=\vec{y}$

Note:$d(\vec{x} ,\vec {y})= ||\vec{y}-\vec{x}||$ and $\vec{y},\vec{x}\in \mathbb{R}^n$

My attempt:

Backwards direction (<=)

Supose $\vec{x}=\vec{y}$. Then $$||\vec{y}-\vec{x}||= \sqrt{(y_1-x_1)^2+...(y_n-x_n)^2}$$

$$||\vec{y}-\vec{x}||= \sqrt{(y_1-y_1)^2+...(y_n-y_n)^2}$$

Clearly, $||\vec{y}-\vec{x}||=0$.

Forward direction (=>)

Suppose $||\vec{y}-\vec {x}||=0$. Then $$||\vec{y}-\vec{x}||= \sqrt{(y_1-x_1)^2+...(y_n-x_n)^2}=0$$ $$=(\sqrt{(y_1-x_1)^2+...(y_n-x_n)^2})^2=0^2$$ $$=(y_1-x_1)^2+...(y_n-x_n)^2=0$$ by series expansion,

$$(y_1-x_1)^2+...(y_n-x_n)^2=-2+=0$$

  • 1
    If a sum of squares is zero, then all the squares have to be zero themselves, right? – 2017-01-20
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    are u saying that $=0$ and $=0$? – 2017-01-20
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    I am saying that $(y_1 - x_1)^2 = \ldots = (y_n-x_n)^2 = 0$, which means that $y_1 = x_1, y_2 = x_2 \ldots y_n = x_n$ – 2017-01-20

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For the second part, all you need is to show that for any $x\in\mathbb{R}^n$, $\|x\|^2=0$ implies $x=0$. But if $$ \sum_{i=1}^n x_i^2=0 $$ one must have $x_i=0$ for all $i$.