if $\bar{\mathbb {Q}}=\mathbb R$ then is : $\bar{\mathbb {Q}}\times\bar{\mathbb {Q}}$ = $\mathbb R^2$?

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I have read about density of $\mathbb Q$ in real number set which can be written as :$\bar{\mathbb {Q}}=\mathbb R$ , I have two question :

What does it meant :$\bar{\mathbb {Q}}$ in mathematics ?

and for my second question is :

if $\bar{\mathbb {Q}}=\mathbb R$ then is : $\bar{\mathbb {Q}}\times\bar{\mathbb {Q}}$ = $\mathbb R^2$ ?

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In your context, the notation $\overline A$ means closure of the set $A$ in some implied ambient topological space. Note that this notation is actually ambiguous because it does not say what the ambient space is. If $A\subset X$ is a subset of a topological space $X$ the closure of $A$ in $X$ is the intersection of all subsets $B\subset X$ which are (1) closed in $X$ and (2) contain $A$, i.e. $A\subset B$.

So for instance, the closure of $\mathbb Q$ in $\mathbb R$ is in fact $\mathbb R$, so one could write $\overline{\mathbb Q}=\mathbb R$. However, the closure of $\mathbb Q$ in the topological space $\mathbb Q$ is just $\mathbb Q$ itself, so in this case $\overline{\mathbb Q}=\mathbb Q$. All this to say, context is important.

Now, for the question of whether $\overline{\mathbb Q}\times \overline{\mathbb Q} = \mathbb R^2$ one has to be a little careful. We have to make sure we know what that statement even means.

One way in which this statement could be true is if we consider $\mathbb Q$ as a subsed of $\mathbb R^2$ in two different ways: once as $\mathbb Q\times \{0\}$ and once as $\{0\}\times \mathbb Q$. It is the true that if we take the closure of each of those subsets (yielding the two lines $\mathbb R\times \{0\}$ and $\{0\}\times\mathbb R$ inside $\mathbb R^2$) we can think of $\mathbb R^2$ as the cartesian product of those two lines. This however requires a lot of unmentioned identifications and so should only be done if all the identifications are clear from the context or made explicit.

Another thing one could do is to consider the subset $\mathbb Q^2$ of $\mathbb R^2$ which is dense, so we could write $\overline{\mathbb Q\times \mathbb Q}=\mathbb R^2$ which is similar but different from what you wrote.