# Help with proposition whether it's true or false [closed]

Is this proposition true or false?

$$\exists y \in \mathbb R \;\forall x \in \mathbb R\,(xy\neq x \rightarrow x=0)$$

And why?

I'm confused as to what exactly is being claimed.

• Yes, the proposition is true or false. :) Commented Dec 19, 2014 at 15:35
• Haha, isn't it false ? Commented Dec 19, 2014 at 15:36
• Why would it be false? (That's not a rhetorical question; it's asking for thoughts and reasons that should be in the main text of the question.) Commented Dec 20, 2014 at 2:14

It is true. $y=1\in\mathbb R$ does the job.

Note that $x\neq x\Rightarrow x=0$ is actually the same statement as $x=x\vee x=0$.

This statement is evidently true for each $x\in\mathbb R$.

• Ok, I pick y=1 and what if x is 2 what happens then? Commented Dec 19, 2014 at 15:45
• @Mr.H123: Then you're looking at $2\ne 2\Rightarrow 2=0$, which is true because the antecedent is false. Commented Dec 19, 2014 at 15:46
• what is the antecedent? Commented Dec 19, 2014 at 15:49
• $2\neq2$ is the antecedent. It is false. If $A$ denotes a false statement then every statement of the form $A\Rightarrow P$ is true. Commented Dec 19, 2014 at 15:52

Take $y = 1$ and show that $x \neq x \implies x = 0\tag{1}$

Note that you can express this in its logically equivalent contrapositive form, which might help you see that the statement is in fact true.

Take $y = 1$. Then we have $x\neq 0 \implies x = x$, which is certainly true, because $x = x$ is always true. So no matter whether $x = 0$ or $x\neq 0$, the implication must be true.

Recall: an implication $p \rightarrow q$ is true whenever $p$ is false or $q$ is true.

One way to examine whether a "there exists" statement is trus is to see whether it's true for some particular values. Suppose you say "If I pick $y = 1$, is the rest true?" You might have a little trouble answering that, so move on.

Try saying "If I pick $y = 2$, is the remainder true?" What result do you get?

The earlier answers all pull $\;y=1\;$ as a rabbit out of a hat. Let me show you a way to avoid that.

\newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\then}{\mathop{\;\Rightarrow\;}} \newcommand{\ref}[1]{\text{(#1)}} \newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}}Implicitly letting $\;x\;$ and $\;y\;$ range over $\;\mathbb R\;$ to reduce clutter, and using a slightly different notation, we calculate

$$\calc \langle \forall x :: xy \not= x \then x = 0 \rangle \calcop\equiv{logic: write \;P \then Q\; as \;\lnot P \lor Q\; -- to get rid of the \;\not=\;} \langle \forall x :: xy = x \;\lor\; x = 0 \rangle \calcop\equiv{arithmetic: divide both sides of \;xy = x\; by \;x\;, special case \;x = 0\;} \langle \forall x :: x = 0 \lor y = 1 \;\lor\; x = 0 \rangle \calcop\equiv{logic: simplify; pull part not using \;x\; out of \;\forall x\;} y = 1 \;\lor\; \langle \forall x :: x = 0 \rangle \calcop\equiv{arithmetic/logic: not all real numbers are 0; simplify} y = 1 \endcalc$$

From this, and the rule of logic that says that $\;\langle \exists y :: y = \dots\rangle\;$ is true, it immediately follows that your proposition is true.

Observe also how the above calculation doesn't just show that $\;y = 1\;$ 'does the job': the equivalences show that it is the only such real number.