Why is $x=2 \implies (x-2)(x-3)=0$ false?

Question

Let $P(x)$ be the equation $x=2$ and $Q(x)$ be the equation $(x-2)(x-3)=0$

By definition of implication I see that $P(x)$ implies $Q(x)$...

As I see it, any premise that is false can give any consequence. With $x=2$ both sides of the arrow are true.. so implication is true.

It is clear that $Q(x)$ implies $P(x)$ and it is also clear that $P(x)$ and $Q(x)$ are not equivalent. Since $Q(x)$ implies $P(x)$ and $P(x)$ and $Q(x)$ are not equivalent it follows that $P(x)$ cannot imply $P(x)$.

Where am I wrong?

EDIT: I knew that $P(x)\Leftrightarrow Q(x)$ was not valid. And I thought wrongly that $P(x)\Leftarrow Q(x)$ is valid which should, if it was valid, imply that $P(x)\nRightarrow Q(x)$. But testing $P(x)\Rightarrow Q(x)$ by the truth table showed $P(x)\Rightarrow Q(x)$ is valid which confused me. I was simply wrongly thinking $P(x)\Leftarrow Q(x)$ which is the correct answer. I have also posted a soultion below, which typo is corrected.

Answer 1

On your comment to Jack Bauer's answer you write

"it is supposed to be false since x=3 is a solution... but by the definition of implication I see it true... what is wrong?"

The problem is when you say "It is supposed to be false since $x=3$ is a solution." The fact that $x=3$ is a solution does not make "$P\implies Q$" false; it makes "$Q\implies P$" false. There is absolutely no problem with saying "$P$ implies $Q$, but is not equivalent to $Q$."

Answer 2

$$(x-2)(x-3)=0 \iff x = 2 \ \, \text{or} \, \ x = 3$$

If $x=2$, then the hypothesis of the RHS is satisfied and hence we have the LHS.

I don't see why that is supposed to be false.

Answer 3

$$(2-2)(2-3)=0(-1)=0$$

In what universe is that false?

Answer 4

Perhaps you mean the converse $Q(x) \Rightarrow P(x)$...that is false. If $(x-2)(x-3) =0$, then $x=2$ or $x=3$, so it is not necessarily the case that $x=2$.

Answer 5

I think the problem is in how to understand logical expressions and how to use them. If you look at the truth table of $P\Rightarrow Q$, you have: \begin{array}{ccc} P&Q&(P\Rightarrow Q)\\\hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T \end{array} Therefore you are correct in saying that if $P$ is false, then $P\Rightarrow Q$ is true, no matter what $Q$ is. This doesn't mean, however that $Q$ must be true.

The way to use the logical statements is: \begin{align} P \text{ is true}&&\color{blue}{\text{and}}&&(P\Rightarrow Q)\text{ is true} \end{align} imply that \begin{equation} Q\text{ is true}. \end{equation}

Let's apply this to $Q\Rightarrow P$. In order to infer that $P$ is true, you need \begin{align} Q \text{ is true}&&\color{blue}{\text{and}}&&(Q\Rightarrow P)\text{ is true}. \end{align} Here we run into a problem. As was mentioned in other answers and comments, if $Q$ is true, this means that $x=2$ or $x=3$. So $Q\Rightarrow P$ is false. We can't infer that $P$ is true.

Answer 6

In the logic of statement, $P \Rightarrow Q$ is only wrong when $P$ is wrong, $Q$ is true. You can read here for more details https://en.wikipedia.org/wiki/Material_conditional. (mainly on the Truth table section)

In your question, it is a statement depend on variable $x$, that means the characteristic of true or wrong of $P(x) \Rightarrow Q(x)$ depend on the value of $x$.

Go on more specific, $x=2 \Rightarrow (x-2)(x-3)=0$, we consider $x \in \mathbb{R}$ (of course you can deal with the case $x \in \mathbb{C}$)

$\bullet x=2$, we get $2=2 \Rightarrow (2-2)(2-3)=0$, or $P(x), Q(x)$ are true, so the statement is true

$\bullet x \neq 2$, we conclude that $P(x)$ is always wrong, so by the Truth table, the statement is true Conlusion: $P(x) \Rightarrow Q(x)$ is always true

So, apply the same method, you can answer why we can't use the argument $Q(x) \Rightarrow P(x)$, because when $x=3$, $Q(x)$ is wrong, $P(x)$ is true, so the statement $Q(x) \Rightarrow P(x)$ is wrong, of course it is true for $x \neq 3$ but if you want to use this, you must restrict to condition $x\neq 3$.

A little further: this question remind me of the reasonable transformations in solving an equation. That is $P(x) =0 \Rightarrow Q(x)=0$ if and only if the solution set of $P(x)$ is a subset of the solution set of $Q(x)$, that means $P(x) \Rightarrow Q(x)$ is alwayse true if and only if the solution set of $P(x)$ is a subset of the solution set of $Q(x)$ (you can use the method above to prove that). When the solution set of $P(x)$ is exactly the same as the solution set of $Q(x)$, we can write $P(x)\Leftrightarrow Q(x)$. That make sense because when you solve $Q(x)$, there are some solution of $Q(x)$ that not the solution of $P(x)$ and you have to check again (this section is one of the most section that high school student often forget)

Answer 7

Usually in logic, we talk about truth value for sentences, which are formulas without free variables. Here you have a free variable $x$. However, for any particular choice of $x$ the statement is true.

Answer 8

For the case $P(x)\Rightarrow Q(x)$:

If we choose $x=2$ then we got true on both sides of the arrow. If we choose $x\neq 2$ then P(x) is false and Q(x) can be false (if $x\neq 3$) or true (if $x=3$). The implication $P(x)\Rightarrow Q(x)$ is therefore valid.

Following explanation will explain why I was wrong. This is for the case $P(x)\Leftarrow Q(x)$:

If we choose $x=2$ then we got true on both sides of the arrow. If we choose $x=3$ then we got Q(x) is true and P(x) is false. Since this scenario is possible, it means the implication $P(x)\Leftarrow Q(x)$ is not valid, i.e. we have $P(x)\nLeftarrow Q(x)$.

A bit more intuitive and more self explaining similar example is when we consider $x=3 \Rightarrow x^2=9$ which is fully valid. What if we change the direction of the arrow in the implication? Changing direction of the arrow, gives us $x=3 \nLeftarrow x^2=9$ since $x=-3$ is true for $x^2=9$ but not for $x=3$.

Answer 9

There is a tacit "$\forall x$" here that should be discussed. That is, if $P(x)$ is the proposition "$x=2$" and $Q(x)$ is the proposition $(x-2)(x-3)=0$, the full-fledged logical statements whose true/false status is in question are

$$\forall x(P(x)\implies Q(x))\quad\text{and}\quad \forall x(Q(x)\implies P(x))$$

The first of these is true, the second is false. The reason the second is false is simple: $Q(3)$ is true, but $P(3)$ is false, so the implication $Q(x)\implies P(x)$ is not true for all $x$. (Note, the implication is true for all $x$ other than $3$, mostly because the proposition $Q(x)$ is false for most values of $x$.)

It's perhaps worth adding why the statement $\forall x(P(x)\implies Q(x))$ is true. The propositions $P(2)$ and $Q(2)$ are both true, so the implication $P(2)\implies Q(2)$ is true. For all $x$ other than $2$, the proposition $P(x)$ is false, which makes any implication beginning "$P(x)\implies\ldots$" automatically true. Hence the implication $P(x)\implies Q(x)$ is true for all $x$.