1. First, state the problem: Prove that $\sqrt{7} - t + \sqrt{7} + t \in \mathbb{N}$ where $t$ is a variable. 2. Simplify the expression inside the set membership: $$\sqrt{7} - t + \sqrt{7} + t = (\sqrt{7} + \sqrt{7}) + (-t + t) = 2\sqrt{7} + 0 = 2\sqrt{7}$$ 3. Now the expression reduces to $2\sqrt{7}$. We need to check if $2\sqrt{7} \in \mathbb{N}$ (the set of natural numbers). 4. Note that $\sqrt{7}$ is an irrational number, since 7 is not a perfect square. 5. Multiplying an irrational number $\sqrt{7}$ by 2 (a rational number) remains irrational. 6. Since $2\sqrt{7}$ is irrational, it cannot be a natural number. 7. Therefore, the initial expression $\sqrt{7} - t + \sqrt{7} + t$ simplifies to an irrational number and is not an element of the natural numbers $\mathbb{N}$. Hence, the original statement is false for any real $t$.