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: $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.

Question

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, $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.

Answer 1

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, $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.

Answer 2

That's correct and using a good approach. As a nitpick, the proof is lacking in "filler" words, where you explicitly say that you use/assume $p(x)=0$ to do a substitution.