# Does $\|z\|=(\sqrt{|x|}+\sqrt{|y|})^{\frac{1}{2}}$ define a norm on $\mathbb R^2$

Does $\| z\|=(\sqrt{|x|}+\sqrt{|y|})^{\frac{1}{2}}$, with $z=(x,y)\in\mathbb R^2$, define a norm on $\mathbb R^2$?

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What did you try so far? do you know the definition of a norm? – Ittay Weiss Mar 12 '13 at 9:41
yes.Is exist a easy criterion for checking Triangle inequality؟ – mojipapa Mar 12 '13 at 9:44
you should check the positive homogeneity first – Quickbeam2k1 Mar 12 '13 at 9:45
I'm not sure what you mean by that, but if you know the definition of a norm then you know you need to check a few properties. Did you check those for the proposed norm you give in the question? – Ittay Weiss Mar 12 '13 at 9:45
Positive homogeneity is true.My problem is with Triangle inequality – mojipapa Mar 12 '13 at 9:47

$\|e_1\|=1$ and $$\left\|\begin{pmatrix} 2 \\ 0 \\ \end{pmatrix} \right\|= \sqrt{\sqrt{2}}\neq 2$$