I am a bit confused on what this question is asking me to prove:

Prove $$ \exists z\forall x\in\mathbb{R}^{+}[\exists y(y - x = y/x)\leftrightarrow x \neq z] $$

Am I asked to prove that there exists a z where the bi-conditional statement is true? Or is z a given and I should just prove the statement in the brackets for some z?

I put that thought aside and showed that the statement in brackets is true when z = 1, but I just want to be sure that I did what the question I was asked.

  • $\begingroup$ Well, stupid gold badge superpowers ... hmm, hey guys, this exercise was previously discussed as math.stackexchange.com/questions/1167153; perhaps the question ought to be closed as a duplicate. $\endgroup$ – hmakholm left over Monica May 18 '15 at 0:54
  • $\begingroup$ My bad! I didn't realize the question was previously asked. $\endgroup$ – Stefan G. May 18 '15 at 1:08
  • $\begingroup$ x @Stefan: I don't think you could reasonably be expected to know; I only found the earlier question because I remembered answering it and could search for some specific phrases from it. $\endgroup$ – hmakholm left over Monica May 18 '15 at 1:11

It is conventional that quantifiers bind from left to right. Therefore, to prove this statement, you must first choose a $z$. Then, your adversary picks an $x$, and regardless of this choice, you must now prove that the biconditional holds.

To do so you must prove that under certain circumstances $y$ exists, and under other circumstances it does not.


This question is from Velleman's How to Prove it? (1994). There is no secret:

You are asked to prove that there exists a $z \in \mathbb{R}^+$ that, for all $z \in \mathbb{R}^+$, the biconditional statement


is true. Then we just recall what Velleman says about proving quantified sentences:

To prove a goal of the form $∀x P(x)$:

Let x stand for an arbitrary object and prove P(x). The letter x must be a new variable in the proof. If x is already being used in the proof to stand for something, then you must choose an unused variable, say y, to stand for the arbitrary object, and prove $P(y)$.


To prove a goal of the form $∃x P(x)$:

Try to find a value of x for which you think $P(x)$ will be true. Then start your proof with “Let x = (the value you decided on)” and proceed to prove $P(x)$ for this value of x. Once again, x should be a new variable.

This means that, for the outer universal quantifier, x must be an arbitrary (new) variable and, for the out-most existential quantifier you must chose a value $z=a$ for it. You should prove the universal quantifier before, and the existential next.

Indeed, the structure of the main sentence you have to prove is of a biconditional, which means you proof should have more or less the following structure:

  1. Let $z=a \in \mathbb{R}^+$ (whatever value you need for it).

  2. Let $x \in \mathbb{R}^+$

    • Proof of $\Rightarrow$
    • Proof of $\Leftarrow$
  3. Proof of $\forall$

  4. Proof of $\exists$


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