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This is something I've been struggling to understand since the past few days. Let's take an example:

Prove/Disprove: $(x+1)^2 = (x+1)^3$ for all real values of $x$.

Proof:

Let us assume the opposite, i.e., $(x+1)^2 \neq (x+1)^3$.

Now, we can split this into two inequalities:

$(x+1)^2 > (x+1)^3$ or $(x+1)^2 < (x+1)^3$

Multiply both sides by $0$ in both inequalities:

$0 > 0$ or $0 < 0$

which are both false, therefore ($x+1)^2 = (x+1)^3$ for all real values of $x$, which is b

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    A nitpick: $(x + 1)^2$ is an expression, not an equation. An expression is a bit of notation that has a value; an equation is a statement that two things are equal.2011-08-30

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While you can multiply both sides of an equation by the same thing and it remains an equality, $x = y \quad \Rightarrow \quad ax = ay,$ the same is not true for inequalities, that is, $x < y \quad \nRightarrow \quad ax < ay.$ What happens depends on the sign of the thing you're multiplying by. For example, if $x < y$, then $2x < 2y,$ but $-2x > -2y$ (think about it), and, coming to the point of the question, $0x = 0y$ (obviously).

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    Ah, I see. Thank you very much.2011-08-30