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In two dimensional space, the length of a vector is $$\sqrt{x^2+y^2}$$

In three dimensional space, the length of a vector is $$\sqrt{x^2+y^2+z^2}$$

How can one prove that in n th dimensional space the length of a vector is

$$\sqrt{d_1^2+d_1^2+\cdots +d_n^2}$$

where $d_n$ represents the n dimensional axis, like $n_1=x, n_2=y, n_3=z, etc$

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    This is actually the definition of the length of a vector in $\Bbb R^n$. How do you define it if this property is what you are trying to prove? –  Mar 16 '16 at 21:57
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    The above is only one of an infinite number of possible ways to Define length of vector. WHat's your definition? – DonAntonio Mar 16 '16 at 22:00
  • Also known as the "Euclidean norm:"

    https://en.wikipedia.org/wiki/Norm_(mathematics)#Euclidean_norm

    The multidimensional definition satisfies the requirements of a norm: https://en.wikipedia.org/wiki/Norm_(mathematics)#Definition

     – Alex R. Mar 16 '16 at 22:02
  • If you assume that distance is defined so that the Pythagorean Theorem holds in two-dimensional subspaces, then there is a straightforward induction proof. – André Nicolas Mar 16 '16 at 22:12

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Using the Pythagorean theorem for higher dimensions, one can show that the length of a point $\mathrm{P}(d_1,\ldots,d_n)$ from the origin in $\mathbf{R}^n$ is precisely equal to, $$\big\lVert\vec{\mathrm{OP}}\big\lVert=\operatorname{dist}(\mathrm{O},\mathrm{P})=\sqrt{\displaystyle\sum_{k=1}^n d_k^2}.$$

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