I have been searching for this on SO, but I just could not find something exactly similar.
What I am trying to achieve is to have a page, that is full in height and in width, but does have a fixed header. The content height needs to be the remaining area left, but that area needs to have a height of 100%, most of the html code found on SO is just using height: auto. Basically I want to do this so that I can style it: adding border etc. Here's the code so far:

&lt;html&gt;
&lt;head&gt;
    &lt;title&gt;Test&lt;/title&gt;
    &lt;style type="text/css"&gt;
    body, html {
    height:100%;
    }

    body {
    margin:0;
    padding:0;
    }

    #header{
    height:50px; 
    background:green;
    width:100%; 
    }

    #wrapper {
    background:blue;
    position:relative;
    min-height:100%;
    height:auto !important;
    height:100%;
    }

    #content {
        margin: 10px;
        background-color: black;
        height: 100%;
        width: 100%;
border: 3px dashed gray;
    }
    &lt;/style&gt;
&lt;/head&gt;
&lt;body&gt;
    &lt;div id="wrapper"&gt;
    &lt;div id="header"&gt;header&lt;/div&gt;
    &lt;div id="content"&gt;&lt;/div&gt;
    &lt;/div&gt;
&lt;/body&gt;
&lt;/html&gt;



Question: Formal proof that $\mathbb{R}^{2}\setminus (\mathbb{Q}\times \mathbb{Q}) \subset \mathbb{R}^{2}$ is connected.
Answer: Theorem. $\ \mathbb{R}^2-A\ $ is connected for any set $A\subset\mathbb{R}^2$ of cardinality less than the continuum.

Proof. Consider any two points $u,v$ in $\mathbb{R}^2-A$. There is a foliation of continuum many paths from $u$ to $v$, which are disjoint except at $u$ and $v$. For example, one could consider all the various circle fragments containing $u$ and $v$. Since only fewer than continuum many of these paths contain points from $A$, it follows that almost all of them are contained in $\mathbb{R}^2-A$, which is therefore path-connected, and even arc-connected. QED

In particular, $\mathbb{R}^2-\mathbb{Q}\times\mathbb{Q}$ is connected.