# Is $(x+1)^2 = (x+1)^3$ for any $x$?

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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The multiplying by zero right there is your problem... note that this trickery figures in fake proofs of $1=2$ and ilk. – J. M. Aug 30 '11 at 6:23
You have to be careful when multiplying inequalities by things. If $a < b$, then $2a < 2b$, but $0a = 0b$, and $-2a > -2b$... – Rahul Aug 30 '11 at 6:26
First, the opposite of "$(x+1)^2=(x+1)^3$ for all real values of $x$" is that "there exists some $x$ such that $(x+1)^2\neq(x+1)^3$", which is more subtle than what you say. Second, I do not understand why the right hand side of your inequalities change from $(x+1)^3$ to $(x+2)^2$. Third, multiplying my zero on both sides of an inequality does not preserve the inequality. – alex.jordan Aug 30 '11 at 6:27
$0*x = 0*y \Leftarrow x=y$ but NOT the other way around. – Per Alexandersson Aug 30 '11 at 6:38
What's wrong with just taking x=1 and observing that 4 is not equal to 8, thereby disproving the original statement? – yatima2975 Aug 30 '11 at 8:04

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).