Explaining maths

In response to a request on Piazza, I gave a detailed proof that the uniform really is a norm, including some comments and warnings on possible pitfalls along the way. Here (essentially) is what I said. Let’s have a look at $\|\cdot\|_\infty$ on the bounded, real-valued functions on $[0,1]$ (but you can also work with bounded functions on any non-empty set you like, and the same proof will work). We define $\ell_\infty([0,1]) = \{f: f \textrm{ is a bounded, real-valued function on }[0,1]\,\}\,,$ and this is the set/space we will look at. So we set $Y=\ell_\infty([0,1])$. Then you can check for yourselves that $Y$ really is a vector space over $\mathbb{R}$ (using the usual "pointwise" operations for functions). Note that the zero element in this vector space is the constant function $0$ (from $[0,1]$ to $\mathbb{R}$), which I'll denote by $\mathbf{0}$ here for clarity. (We also used $Z$ for this function in one of the exercises.) I'll first give...

One of our first-year students asked (on Piazza) whether converse and negation were the same thing. One of my colleagues explained the differences in terms of propositional logic. I added some comments afterward to see if some specific examples might help. I don’t know whether this helped or not! Here is what I said. Dr Feinstein adds: Sometimes specific examples can help here. So consider the following statement, which we will call statement $P$:   ($P$)                          $6$ is divisible by $3$ Then the negation of $P$ is the statement $\neg P$:   ($\neg P$)                    $6$ is not divisible by $3$ Notice here that $P$ is true and $\neg P$ is false.  Consider instead the statement $Q$:   ($Q$)                            $3$ i...