I have the following claim:
“If $x$ and $y$ are real numbers and their product is irrational, then either $x$ or $y$ must be
irrational.”
I'm supposed to write this in mathematical notation. It's my first year of university and I'm not so sure how to go about this, I never did any of this in high school.
Could anyone help me write this in mathematical notation?
I'm supposed to prove this later, but I have an idea how to do that, I just need help with this part. Thank you in advance.
Best Answer
I don't understand why this seems so difficult to the other people trying to give answers. I notice the discrete-math tag.
One want to write what you want is: $$ \forall x,y \in \mathbb{R} (xy\in \mathbb{R}\setminus\mathbb{Q} \to (x\in \mathbb{R}\setminus\mathbb{Q} \lor y \in \mathbb{R}\setminus \mathbb{Q})). $$
In "English" this is saying that: for all $x$ and $y$ real numbers, if the product of $x$ and $y$ is a real number, but not a rational number (i.e. $xy$ is irrational), then $x$ is a real number, but not a rational number (i.e. $x$ is irrational) or $y$ is a real number but not a rational number (i.e. $y$ is irrational).
You could use $\implies$ instead of $\to$. Also, some might prefer fewer parentheses.
Some also write $\mathbb{R} - \mathbb{Q}$ instead of $\mathbb{R}\setminus\mathbb{Q}$. This is simply the set of real numbers minus the set of rational numbers. In general $A\setminus B$ is the set of elements in $A$ that are not in $B$. So $\mathbb{R}\setminus \mathbb{Q}$ is the set of irrational numbers.
As mentioned in a comment by @HenningMakholm one might also prefer to write $x\notin \mathbb{Q}$ instead of $x\in\mathbb{R}\setminus \mathbb{Q}$. This, however is only good because we gave the domain as $\mathbb{R}$.
As also mentioned in other comments, while we read $\forall$ as for all, the symbol doesn't just replace the words. The symbol has a precise (mathematical) meaning. Likewise, $\lor$ mean or, but it is used between the two statements $x\in\mathbb{R}\setminus \mathbb{Q}$ and $y\in\mathbb{R}\setminus \mathbb{Q}$.