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...