When comparing integer squares, greater input need not mean greater square

Expert intro: Integer inequalities with squares depend on sign, not just order, so the step from x > y to x^2 > y^2 needs careful checking.

Detailed walkthrough

Core idea: order and squaring are not the same

The implication x > y ⇒ x^2 > y^2 is not true for all integers. Squaring preserves order only when both numbers are nonnegative, because the function $f(t)=t^2$ increases on $[0, fty)$ but decreases on $(- fty,0]$ toward 0. If one number is negative and the other is less negative, the larger number can still have the smaller square.

For example, take $x=-1$ and $y=-2$. Then $x>y$ because $-1>-2$, but $x^2=1 \quad \text{and} \quad y^2=4,$ so $x^2>y^2$ is false.

Why the statement fails

The inequality $x-y>0$ only tells you that $x$ is to the right of $y$ on the number line. It does not tell you how far each number is from 0. Since squaring measures distance from 0, a number closer to 0 can have a smaller square even if it is larger.

A second counterexample is $x=2$, $y=-3$. Here $2>-3$, but $2^2=4$ and $(-3)^2=9$, so again the conclusion fails. These examples show that the claim cannot be proved as written.

What would make it true

The statement does become true under extra conditions. If $x\ge y\ge 0$, then squaring preserves the inequality and $x^2\ge y^2$. If both are negative, the direction can reverse in surprising ways, so you must be explicit about the sign range before squaring.

A useful algebraic rewrite is $x^2-y^2=(x-y)(x+y).$ Since $x-y>0$, the sign of $x^2-y^2$ depends entirely on $x+y$. If $x+y>0$, then $x^2>y^2$; if $x+y<0$, then $x^2<y^2$. That is the real condition hidden behind the statement.

Common pitfall

The most common mistake is assuming that any increasing input gives an increasing square. That is only safe after checking the numbers are nonnegative. Another pitfall is trying to multiply or square an inequality without tracking whether the operation preserves order. Here, the sign of the numbers matters more than the order itself.

So the correct conclusion is: the statement is false in general, but it is true on a restricted domain such as nonnegative integers.

💡 Pitfall guide

The trap here is treating squaring like a universally order-preserving operation. For integers, that is only safe when both numbers are known to be nonnegative. A student often sees $x>y$ and immediately writes $x^2>y^2$, but the counterexample $x=-1$, $y=-2$ breaks it instantly. Another subtle issue is that $x-y>0$ does not control the sign of $x+y$. Since $x^2-y^2=(x-y)(x+y)$, the second factor decides the outcome. If you forget to examine whether $x+y$ is positive, negative, or zero, you can reach the wrong sign conclusion even when the original inequality is correct.

🔄 Real-world variant

If the statement were changed to if $x,y\in\mathbb{Z}$ and $x>y\ge 0$, then $x^2>y^2$, it would be true. For instance, with $x=5$ and $y=3$, we get $5>3$ and also $25>9$. The reason is that on the nonnegative integers, squaring is strictly increasing. A slightly different variant, if $x>y$ and $x+y>0$, then $x^2>y^2$, is also true because $x^2-y^2=(x-y)(x+y)$ becomes the product of two positive factors. If instead the variant were if $x>y$ and both are negative, then $x^2>y^2$, that would generally be false, as $-1>-2$ but $1<4$.

🔍 Related terms

difference of squares, order of integers, squaring inequalities