The expression in question is:

$\exists i \in \mathbb{Z} : \forall j \in \mathbb{Z} : i^2 \leq (i + j)^2$

This reads: "There exists an integer $i$ such that for all integers $j$, $i^2 \leq (i + j)^2$."

To determine if this is valid, let's analyze the inequality:

1. Expand $(i + j)^2$: $(i + j)^2 = i^2 + 2ij + j^2$

2. The inequality becomes: $i^2 \leq i^2 + 2ij + j^2$

3. Subtract $i^2$ from both sides: $0 \leq 2ij + j^2$

Now, the question is whether there exists some $i \in \mathbb{Z}$ such that for all $j \in \mathbb{Z}$, $2ij + j^2 \geq 0$.

Case 1: $i = 0$

• Substituting $i = 0$: $2(0)(j) + j^2 = j^2$ Since $j^2 \geq 0$ for all integers $j$, this inequality holds for all $j \in \mathbb{Z}$.

Case 2: $i \neq 0$

• For $i \neq 0$, we have: $2ij + j^2$ This expression can take negative values. For example, if $j = -1$ and $i = 1$, the expression becomes: $2(1)(-1) + (-1)^2 = -2 + 1 = -1$ which is negative, so the inequality does not hold for all $j$ when $i = 1$.

Conclusion:

The expression is valid for $i = 0$ but not valid for any $i \neq 0$. Therefore, the answer is:

Yes, the expression is valid, because there exists $i = 0$ that satisfies the inequality for all $j \in \mathbb{Z}$.

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Would you like more details or have any questions?

Here are 5 related questions:

1. What does the symbol $\exists$ represent in logic and set theory?
2. Can you generalize this expression for $i \in \mathbb{R}$ (real numbers)?
3. How does this inequality change if we remove the $\forall$ quantifier?
4. What happens if we reverse the inequality to $i^2 \geq (i + j)^2$?
5. How do inequalities with absolute values differ from squared terms like in this problem?

Tip: In inequalities involving squares, pay attention to when terms become negative, as they can often break the generality of the statement.