Prove or disprove that the product of two irrational numbers is irrational.

Step 1
To prove or disprove that the product of two irrational numbers is irrational, we will consider a proof by contradiction.
Assume that $x$ and $y$ are irrational numbers, and we want to determine whether $xy$ is irrational.
We begin by assuming the opposite, that $xy$ is rational. This implies that $xy$ can be expressed as a fraction $\frac{a}{b}$, where $a$ and $b$ are integers and $b$ is not equal to 0.
$xy=\frac{a}{b}$
Step 2
Next, we can rearrange the equation to solve for $y$:
$y=\frac{a}{bx}$
Since $x$ and $y$ are assumed to be irrational, $x$ cannot be equal to 0. Therefore, we can divide both sides of the equation by $x$:
$\frac{y}{x}=\frac{a}{bx}$
Simplifying further:
$\frac{y}{x}=\frac{a}{b}$
The left-hand side of the equation, $\frac{y}{x}$, represents the division of two irrational numbers, which can also be expressed as a rational number. Let's say $\frac{y}{x}$ is equal to $\frac{c}{d}$, where $c$ and $d$ are integers and $d$ is not equal to 0.
$\frac{y}{x}=\frac{c}{d}$
Multiplying both sides of the equation by $x$:
$y=\frac{cx}{d}$
Step 3
Here, we have expressed $y$ as the division of an irrational number ($cx$) by an integer ($d$). This means that $y$ can be written as a rational number, which contradicts our initial assumption that $y$ is irrational.
Since our assumption leads to a contradiction, we can conclude that the product of two irrational numbers is indeed irrational.
Therefore, the statement is proven to be true: The product of two irrational numbers is irrational.

To prove or disprove that the product of two irrational numbers is irrational, let's consider a counterexample.
Let $x=\sqrt{2}$ and $y=\sqrt{3}$. Both $\sqrt{2}$ and $\sqrt{3}$ are known to be irrational numbers.
Now, let's calculate the product $xy$:
$xy=\sqrt{2}·\sqrt{3}=\sqrt{6}$
The value of $\sqrt{6}$ is also an irrational number. This counterexample demonstrates that the product of two irrational numbers can indeed result in an irrational number.
Therefore, we can conclude that the product of two irrational numbers is irrational, based on the counterexample of $x=\sqrt{2}$ and $y=\sqrt{3}$.

Answer:
$x=\sqrt{3}\text{ and }y=\frac{\sqrt{3}}{3}$
Explanation:
To prove or disprove that the product of two irrational numbers is irrational, let's assume we have two irrational numbers, $x$ and $y$. We need to show that their product, $xy$, is irrational.
Let's proceed with the proof by contradiction.
Assume that $x$ and $y$ are both irrational numbers, and we want to prove that $xy$ is also irrational.
We can express $x$ as $x=\sqrt{3}$ and $y$ as $y=\frac{\sqrt{3}}{3}$. (Given in the question)
Now, let's calculate their product:
$xy=\sqrt{3}·\frac{\sqrt{3}}{3}=\frac{(\sqrt{3}{)}^2}{3}=\frac{3}{3}=1$.
We have obtained $xy=1$, which is a rational number.
Since $xy$ is rational, our assumption that $x$ and $y$ are both irrational is incorrect.
Therefore, we can conclude that the product of two irrational numbers, $x=\sqrt{3}$ and $y=\frac{\sqrt{3}}{3}$, is indeed rational.
Hence, we have disproved the statement that the product of two irrational numbers is always irrational.
$\boxed{x=\sqrt{3}\text{ and }y=\frac{\sqrt{3}}{3}}$

To prove or disprove that the product of two irrational numbers is irrational, let's consider the statement:
$\mathbf{\text{Statement:}}$ The product of two irrational numbers is irrational.
$\mathbf{\text{Proof (by contradiction):}}$
Let $a$ and $b$ be two irrational numbers. We assume that their product, $ab$, is rational.
According to the definition of a rational number, a rational number can be expressed as the quotient of two integers ($\frac{p}{q}$), where $q$ is not equal to zero. Since $ab$ is rational, we can express it as $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$.
The statement to be proved is written as:
$\text{Statement: The product of two irrational numbers is irrational.}$
We use the proof by contradiction method, so we assume the opposite:
$\text{Assumption: Let }a\text{ and }b\text{ be two irrational numbers. We assume that their product, }ab,\text{ is rational.}$
Now we define a rational number:
$\text{Definition: A rational number can be expressed as the quotient of two integers (}\frac{p}{q}\text{), where }q\text{ is not equal to zero.}$
Since we assumed $ab$ to be rational, we can express it as $\frac{p}{q}$, where $p$ and $q$ are integers and $q\ne 0$.
To continue the proof, we will show that this assumption leads to a contradiction.
$\mathbf{\text{Continuation of the proof:}}$
Since $a$ and $b$ are both irrational numbers, they cannot be expressed as the quotient of two integers. Therefore, we cannot write $a$ and $b$ as $\frac{m}{n}$ and $\frac{p}{q}$, respectively, where $m,n,p,$ and $q$ are integers, and $n$ and $q$ are not equal to zero.
Now, let's consider the product $ab$. Since $a$ and $b$ are irrational, $ab$ is the product of two numbers that cannot be expressed as the quotient of two integers. Therefore, $ab$ cannot be expressed as $\frac{p}{q}$, which contradicts our assumption.
Hence, our assumption that the product of two irrational numbers is rational leads to a contradiction. Therefore, we can conclude that the product of two irrational numbers is indeed irrational.
$\mathbf{\text{Conclusion:}}$
We have proved that the product of two irrational numbers is irrational.