# Quod Erat Demonstrandum

## 2011/09/19

### (x=2 or x=-2) implies x^2=-4?

Filed under: Junior Form Mathematics,NSS — johnmayhk @ 4:11 下午

Just a question asked by a student, Thomas.

$x=2$
$x=-2$

multiply the above, yield

$x^2=-4$

strange?

He puzzled that, why we need to do something like:

$x-2=0$
$x+2=0$

to obtain

$(x-2)(x+2)=0$
$\Rightarrow x^2=4$

instead of considering the product of

$x=2$ and $x=-2$ directly?

The above procedure may be found under the context of forming equation, like

“Determine a quadratic equation with roots $2$ and $-2$."

The crucial idea is, for real numbers $a$ and $b$,

($a=0$ or $b=0$) if and only if $ab=0$.

Note that, ZERO plays a crucial role to maintain the equivalence of the two statements ($a=0$ or $b=0$) and $ab=0$.

Let me make it clear, as for example,

the statement ($a=2$ or $b=3$), DOES NOT imply $ab=6$.

($a=2$ or $b=3$) is true even if ($a=2$ and $b=5$) (say), hence, in this case, $ab=(2)(5)=10$ (not 6).

Further, the statement ($a=2$ and $b=3$) implies $ab=6$ indeed. However, $ab=6$ DOES NOT imply ($a=2$ or $b=3$), because, $ab=6$ may imply other cases like ($a=6$ and $b=1$), ($a=12$ and $b=0.5$), and so on.

We see that

($a=2$ or $b=3$) DOES NOT imply $ab=6$.

or we say,

($a=2$ or $b=3$) is not equivalent to $ab=6$.

Back to the original question,

($x=2$ or $x=-2$) is not equivalent to $x^2=-4$.

however,

($x=2$ or $x=-2$) is equivalent to ($x-2=0$ or $x+2=0$) and hence equivalent to $(x-2)(x+2)=0 \Leftrightarrow x^2=4$.