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

My attempt:

$$d(\vec{x},\vec{y})=||\vec{y}-\vec{x}||$$ $$=\sqrt{(y_1-x_1)^2+...(y_n-x_n)^2}$$ Squarring then taking the square root gives us $$=|(y_1-x_1)|+...|(y_n-x_n)|= |(x_1-y_1)|+...|(x_n-y_n)|$$ Now, $$\sqrt{(|(x_1-y_1)|)^2+...(|(x_n-y_n)|)^2}= \sqrt{(x_1-y_1)^2+...(x_n-y_n)^2}$$ $$=||\vec{x}-\vec{y}||=d(\vec{y},\vec{x})$$

Is my proof correct?

  • 2
    Or just use directly the fact that $(a-b)^2 = (b-a)^2$ without squaring and retaking a square root. – 2017-01-25
  • 3
    Also, $\sqrt{a^2+b^2}\neq a+b$ – 2017-01-25

0 Answers 0