0
$\begingroup$

How can we prove that x = 2 or 3 step-by-step:

$$(2-x)(3-x) = 0$$

I know the answer, but how do we get it without plugging in 2 and 3 directly?

Thank you!

$\endgroup$
3
  • $\begingroup$ Not very clear... For sure, it is enough to substitute $2$ (or $3$) in place of $x$ and compute to prove that : "if $x=2$ or $x=3$, then $(2-x)(3-x)=0$ is true". $\endgroup$ Commented May 30, 2018 at 8:09
  • $\begingroup$ Maybe you want to show that : "if $ (2−x)(3−x)=0$ holds, then necessarily : either $x=2$ or $x=3$" .... $\endgroup$ Commented May 30, 2018 at 8:10
  • 1
    $\begingroup$ If so, by contradiction: assume an $a$ such that $a \ne 2$ and $a \ne 3$ and show that $(2-a)(3-a) \ne 0$. $\endgroup$ Commented May 30, 2018 at 8:11

2 Answers 2

4
$\begingroup$

A product in a ring with no divisors of zero can only be zero if one of the factors is zero and thus $$(2-x)(3-x)=0$$ implies $2-x=0$ or $3-x=0$.

$\endgroup$
2
  • $\begingroup$ I see. That's much clearer now. Thanks! $\endgroup$
    – Zilong Li
    Commented May 30, 2018 at 8:12
  • $\begingroup$ You´re welcome! $\endgroup$ Commented May 30, 2018 at 8:12
1
$\begingroup$

$$ab=0\implies a=0\lor b=0$$ because $$a\ne0\land b\ne0\implies ab\ne0.$$

Now

$$2-x=0\lor 3-x=0\implies 2=x\lor 3=x,$$ by adding $x$ to the equalities.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .