What Is a Counterexample to ∀x∃y (x = 1/y)?

A counterexample to a statement of the form “for every $x$, there is a $y$” must be one allowed value of $x$ for which no allowed value of $y$ makes the equation true. The answer depends on the domains of the variables, so those should be stated first.

For the usual interpretation $x\in\mathbb{R}$ and $y\in\mathbb{R}\setminus\{0\}$, the counterexample is $x=0$. There is no nonzero real number whose reciprocal is zero, so the existential claim fails for this $x$. Therefore, the full statement is false.

Why $x=0$ works

The statement is

$$\forall x\,\exists y\;(x=1/y).$$

To disprove it, you do not need to test every $x$. You need just one value of $x$ that makes the inner claim false for every permitted $y$. Set $x=0$. For any permitted $y$, we have $y\ne 0$, and

$$\frac{1}{y}\ne 0.$$

Consequently, $0=1/y$ is false for every permitted $y$. This establishes the counterexample.

The restriction $y\ne 0$ is necessary because $1/0$ is undefined. In a formal statement, write the domains explicitly, for example:

$$\forall x\in\mathbb{R}\;\exists y\in\mathbb{R}\setminus\{0\}\;(x=1/y).$$

How the quantifiers determine a counterexample

The negation of “for every $x$, there exists a $y$” is “there exists an $x$ such that for every $y$, the claim fails.” For this equation, the negation is

$$\exists x\in\mathbb{R}\;\forall y\in\mathbb{R}\setminus\{0\}\;(x\ne 1/y).$$

Choosing $x=0$ satisfies that negation. The order matters: a counterexample is a single fixed $x$ that defeats all possible $y$ values. It is not enough to show that one particular choice of $y$ fails.

A check using ordinary values

For a nonzero real number $x$, the equation can be solved by taking its reciprocal:

$$y=\frac{1}{x}.$$

For instance, if $x=4$, take $y=1/4$; then $1/y=4$. If $x=-2$, take $y=-1/2$; then $1/y=-2$. These examples show that every nonzero real $x$ works. The only real value that fails is $x=0$, which is enough to disprove a universal statement.

In fact, if the domain for $x$ is restricted to the nonzero reals and $y$ may be any nonzero real, the statement becomes true: for each allowed $x$, choose $y=1/x$. This illustrates why the variable domains are part of the mathematical claim, not a minor detail.

If the variables are integers

If both variables are integers and $y=0$ is excluded, $x=0$ is still a counterexample: the reciprocal of a nonzero integer is never zero. Other values can fail too. For example, $x=2$ would require $y=1/2$, which is not an integer.

If instead $x$ ranges over positive integers and $y$ over nonzero integers, choose $x=2$ as the counterexample. The only positive integer that can be written as the reciprocal of a nonzero integer is $1$, obtained with $y=1$. On the other hand, if both variables range over nonzero reals, the statement is true, as shown above.

Common mistakes

• Forgetting the domain: The truth can change when the allowed values of $x$ or $y$ change.
• Trying $y=0$: The expression $1/y$ is undefined there, so $y=0$ is not a possible witness.
• Checking only one witness: Showing that a particular $y$ does not work does not refute “there exists a $y$.” You must show that every permitted $y$ fails for the chosen $x.
• Confusing a counterexample with a failing pair: To refute the quantified statement, give an $x$ for which no suitable $y$ exists. Under the real-number interpretation, $x=0$ does exactly that.