To prove the statement:

Given $X$ is an inner product space and $x, y \in X$, if $x \perp y$ (i.e., $x$ is orthogonal to $y$), then prove that:

$\|x + y\|^2 = \|x\|^2 + \|y\|^2.$

Proof:

1. Start with the definition of the norm in an inner product space: The squared norm of a vector $z$ is given by: $\|z\|^2 = \langle z, z \rangle,$ where $\langle \cdot, \cdot \rangle$ denotes the inner product.

2. Apply this to $\|x + y\|^2$: $\|x + y\|^2 = \langle x + y, x + y \rangle.$

3. Expand using the linearity of the inner product: $\|x + y\|^2 = \langle x, x \rangle + \langle x, y \rangle + \langle y, x \rangle + \langle y, y \rangle.$

4. Use the property of orthogonality ($x \perp y$): By definition of orthogonality, $\langle x, y \rangle = 0$ and $\langle y, x \rangle = 0$. Thus: $\|x + y\|^2 = \langle x, x \rangle + \langle y, y \rangle.$

5. Rewrite the terms as squared norms: $\|x + y\|^2 = \|x\|^2 + \|y\|^2.$

Thus, the given statement is proven.

Would you like further clarification or details about any of the steps?

Here are five related questions to deepen understanding:

1. What properties of an inner product space are used in this proof?
2. How does this proof relate to the Pythagorean theorem in Euclidean geometry?
3. Can this result be generalized to complex inner product spaces? If so, what changes?
4. What happens if $x$ and $y$ are not orthogonal? How does the formula change?
5. Can you provide examples of inner product spaces where this property holds?

Tip: The orthogonality property in inner product spaces is fundamental for concepts like projections, least squares, and Fourier analysis.