What's the error in my proof of the statement: The product of two irrational numbers is irrational?

Question

Statement: The product of any two irrational numbers is irrational.

Formally it can be written as: $$\Big(\forall x \forall y\Big)\,\Big(\big(x \notin \mathbb{Q} \wedge y \notin \mathbb{Q}\big) \to \big((x*y) \notin \mathbb{Q}\big)\Big)$$

Proof:

This statement is equivalent to its contrapositive, namely: $$\Big(\forall x \forall y\Big)\,\Big(\big((x*y) \in \mathbb{Q}\big) \to \big(x \in \mathbb{Q} \lor y \in \mathbb{Q}\big)\Big)$$

Because $x*y$ is rational we can write it as: $x*y = \frac{a}{b}$, where $a,b \in \mathbb{Z}$.

From this we can see that we can write x as $x = a$ and y as $y = \frac{1}{b}$. Because $a$ and $b$ are arbitrary integers, we can conclude that $x$ and $y$ will be rational numbers because they can be written as ratio of two integers.

Because we chose arbitrary $x*y$ it follows that we have proven the original statement.

However, obviously this statement is false. One counterexample would be $\sqrt{2}*\sqrt{2} = 2$

What is wrong with my proof?

Answer 1

From this we can see that we can write $x$ as $x=a$ and $y$ as $y=\frac 1b$.

No, there is no reason why you would be able to do that.

When you're trying to prove $\forall x \forall y$, the values of $x$ and $y$ will be given to you by a malevolent adversary. You don't get to choose what they are. If the adversary has chosen, for example $x=y=\sqrt 2$, you need to work with those choices, and not simply decide you'd like it better if it was $x=2, y=1$.

Answer 2

"From this we can see..." is where you have a problem.

In your "proof" $x$ and $y$ are already fixed. So while it is true that if $xy=a/b$, we might have that $x=a$ and $y=1/b$. It is not the case that $x$ must be $a$ and $y$ must be $1/b$.

For example, we could have $x=1$ and $y=a/b$ or $x=\sqrt{2}$ and $y=a/(b\sqrt{2})$.

Answer 3

You have stated that

"Because $x∗y$ is rational we can write it as:$ x∗y=a/b $, where $a,b∈Z.$

From this we can see that we can write $x$ as $x=a$ and $y$ as $y=1/b$"

You can also write $x=a\sqrt 2$ and $y=1/{b\sqrt 2}$ which are not rational.

You are assuming that a rational number has a unique representation as a fraction, which is not true as you have mentioned in your counter example.