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.

• $(x + y, x - y) = 0$ if and only if $\|x\| = \|y\|.$
• 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.$\quad$ First, suppose that $(x + y, x - y) = 0.$ This means: \begin{align*} (x + y, x - y) &= (x + y, x) - (x + y, y) \\ &= (x, x) + (x, y) - (y, x) - (y, y) \\ &= (x, x) - (y, y) \\ &= 0 \end{align*} Adding $(y, y)$ to both sides, we get $(x, x) = (y, y).$ Taking the square root of both sides, we find that $\|x\| = \|y\|.$
  
  Now, suppose we have $\|x\| = \|y\|.$ This means that $(x, x) = (y, y),$ or that $(x, x) - (y, y) = 0.$ We can rewrite this as \begin{align*} (x, x) - (y, y) &= (x, x) + (x, y) - (y, x) - (y, y) \\ &= (x, x) + (x, y) - [(y, x) + (y, y)] \\ &= (x, x + y) - (y, x + y) \\ &= (x + y, x) + (x + y, -y) \\ &= (x + y, x - y) \\ &= 0 \end{align*} Thus, we have shown that $(x + y, x - y)=0$ if and only if $\|x\| = \|y\|.$ This completes the proof.