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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?

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  • $\begingroup$ Don't use $*$ for $\cdot$, please. $\endgroup$ – Andrés E. Caicedo Nov 6 '18 at 15:31
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    $\begingroup$ "we can write $x$ as $a$ and $y$ as $\dfrac 1 b$". In this step you have already (incorrectly) assumed what you are trying to prove : that $x,y \in \mathbb Q$. $\endgroup$ – Mauro ALLEGRANZA Nov 6 '18 at 15:31
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    $\begingroup$ Consider : $2= \dfrac 2 1 = \sqrt 2 \cdot \sqrt 2$. From this does not follows that $\sqrt 2 = 2$ and $\sqrt 2 = \dfrac 1 1 = 1$. $\endgroup$ – Mauro ALLEGRANZA Nov 6 '18 at 15:32
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    $\begingroup$ You know a counterexample. One trick is walk through your proof in the specific case of your counterexample, to see where things go wrong. (And you’ll find it’s exactly where the answerers say it is.) $\endgroup$ – spaceisdarkgreen Nov 6 '18 at 15:40
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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 $x=2, y=1$.

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"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 $y=a/(b\sqrt{2})$.

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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.

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