# Do the $p$-norms in $\mathbb R^2$ have a nice geometric intuition behind them?

For $p\geq 1,$ the $p$-norm of a vector $(x,y)\in\Bbb R^2$ is the number $\|(x,y)\|_p=(|x|^p+|y|^p)^{1/p}.$

I learned this definition some time ago, but I never really understood it. Is there a useful and not too difficult to understand geometric intuition that explains it (at least for natural $p$)?

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Draw a picture of $||x||_p = 1$. – user20266 Jul 15 '12 at 12:31
@Thomas I know what the unit circles look like, more or less. I've used Wolphram Alpha to draw the pictures. But I'm not sure how it explains what the norms actually mean. – Bartek Jul 15 '12 at 12:33
I know that for $p=1$ we can think of it as a distance we have to walk to get somewere in a city with orthogonal streets. And of course I have a pretty good intuition of the $p=2$ case. – Bartek Jul 15 '12 at 12:36
Bartek: No since the points (0,0) and (2,0) are at knight-distance 2 but the points (0,0) and (1,0) are at distance 3 (instead of 1). Hence the knight-distance is not a norm distance. – Did Jul 15 '12 at 13:40
I recently found a nice article, about $π_p$, the value of $π$ in $ℓ_p$ – draks ... Jul 16 '12 at 9:45