Mathematical Immaturity

1.13 Exercises

Prove that each of the statements in Exercises 3 through 7 is valid for all elements $x$ and $y$ in a real Euclidean space.

• $(x, y) = 0$ if and only if $\|x + y\|^2 = \|x\|^2 + \|y\|^2.$
• Recall from Section 11 that for an element $x$ in a Euclidean space $V$ with inner product $(x, x),$ the nonnegative number $\|x\| = (x, x)^{1/2}$ is called the norm of $x.$
• 
  Proof.$\quad$ To show that $(x, y) = 0$ implies that $\|x + y\|^2 = \|x\|^2 + \|y\|^2,$ we note that \begin{align*} \|x + y\|^2 &= (x + y, x + y) \\ &= (x + y, x) + (y, x + y) \\ &= (x, x) + (x, y) + (y, x) + (y, y) \\ &= (x, x) + 2(x, y) + (y, y) \end{align*} But if $(x, y) = 0$ then $\|x + y\|^2 = (x, x) + (y, y) = \|x\|^2 + \|y\|^2.$
  
  To prove that $\|x + y\|^2 = \|x\|^2 + \|y\|^2 $ implies $(x, y) = 0,$ we note that the equation $\|x + y\|^2 = \|x\|^2 + \|y\|^2$ can be rewritten as \begin{align*} (x, x) + 2(x, y) + (y, y) &= (x, x) + (y, y) \end{align*} Subtracting $(x, x)$ and $(y, y)$ from both sides, we find that $(x, y) = 0.$ Hence, $(x, y) = 0$ if and only if $\|x + y\|^2 = \|x\|^2 + \|y\|^2.$ This completes the proof.