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 $\|x + y\| = \|x - y\|.$
• 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 $\|x + y\| = \|x - y\|,$ we note that \begin{align*} \|x + y\| &= (x + y, x + y)^{1/2} \\ &= [(x + y, x) + (y, x + y)]^{1/2} \\ &= [(x, x) + (x, y) + (y, x) + (y, y)]^{1/2} \\ &= [(x, x) + 2(x, y) + (y, y)]^{1/2} \end{align*} and \begin{align*} \|x - y\| &= (x - y, x - y)^{1/2} \\ &= [(x - y, x) - (y, x - y)]^{1/2} \\ &= [(x, x) - (x, y) - (y, x) + (y, y)]^{1/2} \\ &= [(x, x) - 2(x, y) + (y, y)]^{1/2} \end{align*} But if $(x, y) = 0$ we can see that $\|x + y\| = \|x - y\|.$ Now to prove that $\|x + y\| = \|x - y\|$ implies $(x, y) = 0,$ we refer to the equations for $\|x + y\|$ and $\|x - y\|.$ If $\|x + y\| = \|x - y\|$ then we have the following: \begin{align*} [(x, x) + 2(x, y) + (y, y)]^{1/2} &= [(x, x) - 2(x, y) + (y, y)]^{1/2} \end{align*} Squaring and simplifying both sides, we get \begin{align*} (x, y) = -(x, y) \end{align*} which implies that $(x, y)$ must be zero. As such, we have shown that $(x, y) = 0$ if and only if $\|x + y\| = \|x - y\|.$ This completes the proof.