Proving that a larger integer can have a smaller square order

Key concept: Integer inequalities and squaring rules behave differently when negative numbers are involved. The comparison between $x$ and $y$ must be checked against the sign of each value before concluding anything about $x^2$ and $y^2$ [1].

Step by step

Core idea

The statement "if $x,y\in \mathbb{Z}$ and $x>y$ then $x^2>y^2$" is not true in general. The reason is that squaring does not preserve order for all integers. It preserves order only under extra conditions, such as both numbers being nonnegative and one being larger than the other.

A quick counterexample is enough to disprove the claim. Take $x=-1$ and $y=-2$. Then $x>y$ because $-1>-2$, but $x^2=1$ and $y^2=4$, so $x^2<y^2$. That single example shows the implication fails.

Why the logic breaks

The expression $x^2-y^2$ factors as $(x-y)(x+y)$. Since $x-y>0$ when $x>y$, the sign of $x^2-y^2$ depends entirely on $x+y$.

That means there are three possible situations:

• if $x+y>0$, then $x^2-y^2>0$ and so $x^2>y^2$;
• if $x+y=0$, then $x^2=y^2$;
• if $x+y<0$, then $x^2-y^2<0$ and so $x^2<y^2$.

So the original statement is missing an important condition. The order of integers alone is not enough to guarantee the order of their squares [2].

A correct version of the statement

A true related statement is: if $x$ and $y$ are nonnegative integers and $x>y$, then $x^2>y^2$.

Here the proof is straightforward. Because $x>y\ge 0$, both factors in $x^2-y^2=(x-y)(x+y)$ are positive. Hence $x^2-y^2>0$, which gives $x^2>y^2$.

This corrected version is often what exam questions intend when they ask about comparing squares. In the original form, however, one counterexample is enough to reject the claim.

Common mistake to avoid

A frequent error is to assume that multiplying both sides of an inequality by the same expression always preserves the direction. Squaring is not a simple linear multiplication step; it can change order when negative values are present. The sign of the numbers matters as much as the inequality itself.

Pitfall alert

The trap in this inequality is assuming that squaring is order-preserving for every integer. That fails as soon as negative numbers enter the picture. A student may see $x>y$ and immediately write $x^2>y^2$, but that shortcut is only safe when both numbers are nonnegative. The quickest way to test the statement is to choose a negative pair such as $x=-1$ and $y=-2$, where the original inequality is true but the squared inequality reverses. Another subtle mistake is trying to prove the claim by expanding $x^2-y^2$ without checking the sign of $x+y$. Since $(x-y)$ is positive, the entire sign depends on $x+y$, and that can be negative, zero, or positive. Examiners often award full marks for a clean counterexample or for stating the corrected condition precisely.

Try different conditions

If the question were changed to "If $x,y\in \mathbb{Z}$ and $x>y\ge 0$, then $x^2>y^2$," the claim becomes true. In that version, both numbers are nonnegative, so squaring preserves the order. The proof uses the factorization $x^2-y^2=(x-y)(x+y)$. Because $x>y\ge 0$, we have $x-y>0$ and also $x+y>0$, so their product is positive. Therefore $x^2-y^2>0$, which implies $x^2>y^2$. If the condition were instead changed to $x>y$ and $x+y>0$, the statement would also be true, even if one number is negative. That variant shows exactly which extra assumption is needed to make the inequality valid.

Further reading

counterexample for inequalities, squaring preserves order, factorization of difference of squares