Question

Determine the truth value of the statement $\mathrm{\forall }x\mathrm{\exists }y(xy=1)$ if the domain for the variables consists of?

• A. The nonzero real numbers.
• B. The nonzero integers.
• C. The positive real numbers.

Answer 1

Answer: (A), (C)

The positive real numbers.

Identify the true value of the given statement:

The explanation for the correct answer is Option (a).

The given statement is $\mathrm{\forall }x\mathrm{\exists }y(xy=1)$. If the domain for the variables consists of the non-zero real number.

Since the given statement is $\mathrm{\forall }x\mathrm{\exists }y(xy=1)$.

Take the values as: $x=3,y=\frac{1}{3}$. So,

$\begin{array}{rcl}xy& =& 3·\frac{1}{3}\\ & \Rightarrow & xy=1\end{array}$

Thus, the statement is true because $\mathrm{\forall }x\mathrm{\exists }y(xy=1)$.

The explanation for the incorrect answer is Option (b).

the nonzero integers.

Let us take $x=3,y=-\frac{1}{3}$. So,

$\begin{array}{rcl}xy& =& 3\left(-\frac{1}{3}\right)\\ & \Rightarrow & xy=-1\left(\ne 1\right)\end{array}$

If $x=3$, then there does not exist $y$in integer such that $xy=1$.

Thus, the statement is false.

The explanation for the correct answer is Option (c).

the positive real numbers.

Let us take $x=5,y=\frac{1}{5}$. So,

$\begin{array}{rcl}xy& =& 5·\frac{1}{5}\\ & \Rightarrow & xy=1\end{array}$

Thus, the statement is true because $\mathrm{\forall }x\mathrm{\exists }y(xy=1)$.

Therefore, the correct truth value of the statement is option (a) and option (c).