Transposing vacuously true propositions: if $x^2 < 0$, then $x = 0$.

Question

According to vacuous truth, the proposition $(1)$ is true.

$$\forall x\in\mathbb{R}~(x^2<0\implies x=0)\tag{1}$$

Now, if I transform the proposition $(1)$ using transposition, I arrive at $(2)$. Here I encounter an issue, as without transforming back, $(2)$ is not as easy to prove as $(1)$.

$$\forall x\in\mathbb{R}~(x\neq0\implies x^2\geq0)\tag{2}$$

How do I prove proposition $(2)$ in the case where $x\neq 0$? How do I interpret the meaning of this new proposition $(2)$?

Answer 1

Although the comments pretty much answer your question, I'll try to be slightly more pedantic in this answer. Before that, note that $(2)$ is indeed true since $(1)$ is and taking contrapositives doesn't change the truth value of a statement (this is easy to see by contructing a truth table, for example).

You are asking to prove that the statement $$\forall x\in\mathbb{R}~(x\neq0\implies x^2\geq0)\tag{2}$$ is true. I present here two proofs, one direct and the other not so much. The important point to note (and where I think the source of confusion resides) is that showing that $(2)$ is true is equivalent to showing that one of the following statements is true:

• $\forall x\in\mathbb{R}~(x\neq0\implies x^2>0)$.
• $\forall x\in\mathbb{R}~(x\neq0\implies x^2=0)$.

Why is this the case? This is due to our interpretation of the symbol $\geq$ in $\mathbb R$; if $x, y \in \mathbb R$, $x \geq y$ means that either $x > y$ or $x = y$; therefore, if you assume that $x \neq 0$ and you prove that $x^2 >0$, then $(2)$ follows.

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Proof 1: Let $x \in \mathbb R$ and suppose that $x \neq 0$. We have to show that $x^2 \geq 0$, i.e. we have to show that either $x^2>0$ or $x^2=0$; if any of these hold, the truth of statement $(2)$ follows. Since $x \neq 0$ we have two cases:

1. $x>0$. Then $x^2>0$ since $x$ is a positive real number.
2. $x<0$. Then there exists $z>0$ such that $x = -z$. Then $x^2 = (-z)^2 = (-1)^2z^2 = z^2>0$, where the last inequality follows since $z$ is a positive real number.

In either case $x^2>0$, so if $x$ is a real number distinct from zero, $x^2 >0$. Therefore statement $(2)$ is true.

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Proof 2: Assume for contradiction that $(2)$ is not true, i.e. assume that the statement $$\neg\left(\forall x\in\mathbb{R}~(x\neq0\implies x^2\geq0)\right)\tag{$\dagger$}$$ is true. Using the fact that $p \to q$ is logically equivalent to $\neg p \vee q$ and "moving the negation inside the expression", we have that $(\dagger)$ is logically equivalent to $$\exists x \in \mathbb R (x \neq 0 \wedge x^2<0) \tag{$\dagger\dagger$}.$$ However, note that $(\dagger\dagger)$ directly contradicts statement $(1)$ in your question, since any $x \in \mathbb R$ such that $x^2 < 0$ must be $0$. Therefore our assumption was false and hence $(2)$ is true.