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 \(Q\):   (\(Q\))                            \(3\) is divisible by \(6\) and its negation   (\(\neg Q\))                    \(3\) is not divisible by \(6\) This time \(Q\) is false and \(\neg Q\) is true. This is no coincidence, because for any (co