<s>[INST] Let $p(x)$ and $q(x)$ be the predicates: $p(x):x^2-3x+2=0$ and $q(x):(x-1)^{3}(x-2)(x-3)x=0$. Show: $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$. I would like to know if this is correct. Thanks in advance.
Let $p(x)$ and $q(x)$ be the predicates:
$p(x):x^2-3x+2=0$
$q(x):(x-1)^{3}(x-2)(x-3)x=0$
Show that: $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.
This is what I did:
$p(x):x^2-3x+2=0$
$$x^2-3x+2=0$$
$$(x-1)(x-2)=0$$
then,
$q(x):(x-1)^{3}(x-2)(x-3)x=0$
$$(x-1)^{3}(x-2)(x-3)x=0$$
$$(x-1)^{2}(x-1)(x-2)(x-3)x=0$$
$$(x-1)^{2}(0)(x-3)x=0$$
$$0=0$$
Then, $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.
 [/INST] I would rephrase like that:
then,
$(x-1)^{3}(x-2)(x-3)x= (x-1)^{2}(x-1)(x-2)(x-3)x=(x-1)^{2}(0)(x-3)x=0$
Then, $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.
or
To show $q(x): (x-1)^{3}(x-2)(x-3)x=0$
$L.H.S.= (x-1)^{3}(x-2)(x-3)x = (x-1)^{2}(x-1)(x-2)(x-3)x=(x-1)^{2}(0)(x-3)x = 0$
$R.H.S. = 0$
$L.H.S. = R.H.S.$
Then, $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.
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