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  "body": "\n## Proof\n\nWe expand the square of the norm with the rules of the inner product.\n\n1. Let $V$ be an inner product space over $\\mathbb{K} = \\R$ or $\\C$, with [[inner product|0VGSYBWHV2]] $\\ip{\\cdot}{\\cdot}$. Let $x, y \\in V$ with $\\ip{x}{y} = 0$, which means that $x$ and $y$ are [[orthogonal|0TS7VZJBWD]].\n2. For all $u, z, w \\in V$, conjugate symmetry and linearity in the first argument give $\\ip{u}{z + w} = \\overline{\\ip{z}{u} + \\ip{w}{u}} = \\ip{u}{z} + \\ip{u}{w}$.\n3. By the definition of the [[induced norm|0KG4D8AP6G]], $\\norm{x + y}^2 = \\ip{x + y}{x + y}$.\n4. Linearity in the first argument and step 2 give $\\ip{x + y}{x + y} = \\ip{x}{x} + \\ip{x}{y} + \\ip{y}{x} + \\ip{y}{y}$.\n5. By conjugate symmetry, $\\ip{y}{x} = \\overline{\\ip{x}{y}} = \\overline{0} = 0$. Also $\\ip{x}{y} = 0$ by step 1.\n6. So $\\norm{x + y}^2 = \\ip{x}{x} + \\ip{y}{y} = \\norm{x}^2 + \\norm{y}^2$.\n\n## Idea\n\nThe square of the norm of a sum expands like the square of a sum of numbers. There are two cross terms, $\\ip{x}{y}$ and $\\ip{y}{x}$. Orthogonality makes both cross terms zero, and only the two squares remain.\n",
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