Extract 6.3
Context: This Episode is from the same tutorial as Extract 6.1 — where the tutor and the students discussed CD2.1. Now they turn to CD2.2.
The Episode:
CD2.2 was proved by the lecturer for n=2, say Andrew and Jack: he squared both sides and 'showed triangles'. The picture with triangles and the squaring, replies the tutor, is the 'genetic intuition' for understanding the statement. The principle here however is to 'develop all the properties of the real numbers from a minimum number of axioms', he continues. Andrew asks whether the squaring and the picture 'is not the right thing to do' and the tutor stresses 'it's the right way to understand it and remember it, but it's not the right way to prove it'. 'Pictures and triangles have no part' in a proof where the aim is to develop logically a property for n numbers from what is known on two numbers, he concludes. He then suggests proving the triangle inequality for n=2, namely that |x+y|£ |x|+|y|, by taking cases for x and y (fig.3). By the definition of ||, this leads to eight possibilities for x and y.
Andrew sounds hesitant about the 'need' to write explicitly the eight cases. The tutor starts writing out the table (fig.3) commenting that even though it is ' painful to write down eight possilbilities' it is 'on the other hand a back-to-basics proof'. He then claims that the eight cases can be reduced to two. Andrew suggests 'taking the absolute values off' in the cases x and y are both positive or negative and putting the appropriate sign. The tutor says he agrees. Andrew starts outlining the rest of the cases. The tutor then says he just realised that the least number of cases that could be written down is four. Then by symmetry the eight possibilities would be covered, he adds. Then Andrew has a comment on the triangle inequality.
A1: Would that inequality do? Because |x+y| is either equal to |x|+|y|, or equal to |x|-|y| or a couple more permutations, er, combinations...
T1: Well, the modulus of this is equal to the modulus of that, isn't it [he points at the various cases for |x+y|]?
A2: Yes, it's more equal to than... In fact we seem to generalise to less than or equal to when it is equal to something or in fact to something else. It can't be really in between...
T2: You're discarding a lot of information by merely writing that |x+y|£|x|+|y| but...
J1: It is true.
A3: But it's not continuous in the fact that you take a couple of values that range from equal to depending on the value of x and y and keep going down as far as you like.
T3: Well, first of all it's true. Secondly I mean what about the geometric picture?
A4: I don't know... I can't... where is the crossover... is this a geometric instance? It can't take any value...
T4: Well, what about complex numbers? Is this also true?
A5: I thought so...
T5: Indeed it's true of vectors and there the fact is in these other veins you are not throwing a lot out and this is only the special case of real numbers.
The tutor then returns to the proof: looking at the various options for x and y, they deduce that the triangle inequality is true for n=2. Now, says the tutor, they have to generalize to n numbers, by induction. He criticises Jack for starting his inductive proof from n=1 because the triangle inequality 'is a basic assertion for two. It then has some information in it. For one it doesn't'. He then writes down the inductive hypothesis for n=k and proves the statement for n=k+1 by using the proved statement for n=2 and the associative law for the real numbers. He concludes that he did this in detail ' only as a matter of style of presentation, but style matters'. Andrew then has a question.
A6: In the question there it said that the hypothesis or the statement they were talking about is for n³2 because as Jack was saying we can show for one it is true. Does this invalidate the proof?
T6: Well, it doesn't say n³2...
A7: You can start by proving that this is true for 1 and...[the tutor nods] But you would still have to show that P2 is true because that helps in the second part.
T7: Well, yes, I am thinking of P2 as being both what you are calling P2 and the triangle inequality or the basic fact that |x+y|£|x|+|y|.
Both students nod. The tutor then suggests they look at CD2.4 (Extract 6.5).