Context: This Episode consists of two extracts from the two tutorials mentioned in Extracts 6.1 and 6.4. It is about CD2.4. Kelle's Extract refers to CD2.4i and ii and Jack' s and Andrew's to CD2.4ii.

According to the tutor, Kelle has not proved 'what they want'. What they want, says the tutor, is to prove CD2.4i for 'as small epsilon as you can get' via CD2.3 and the Approximation Lemma. This is equivalent to proving that qⁿ®0, she adds. She then asks him whether sequences have been mentioned in the lectures. Kelle says hesitantly yes and, when the tutor asks him how he will prove that qⁿ®0, he responds reluctantly, referring to sequences, that he doesn't 'think they are meant to read anything more'. The tutor then returns to the terminology of the question and asks what is it they have to prove.

K1: We have to prove that there exists an n,... that there exists an epsilon...

T1:... not that there exists an epsilon, no...

K2:... there exists an n in...

T2. The tutor shows then on paper that what they want to prove is that starting with any 'kind of slice', any epsilon, all qⁿ will 'eventually lie in between' -e and +e. So, how will he write that? she asks.

K3: Er,... you can find an element such that if m equals N this will be less than epsilon, eh for m>N qⁿ will be less than...

T3: Yes! We are getting there. Now what is this epsilon, is it some particular epsilon or... what?

K4: It can be any... You can choose..You can find an epsilon such that...

K5: Because it might be greater than that before that, beyond that ne tries to be smaller for all the rest of the numbers.

T5: Why?

K6: Because if it's converging they should have been higher...

The tutor says she is not pleased that Kelle is using the convergence of qⁿ in his justification because this is what they are trying to prove. She repeats the statement and stresses that it is important to see that because q^n_e<e and the sequence is decreasing qⁿ<e will be true "n>n_e. 'It's so obvious that it goes to nought' remarks Kelle. She says that she agrees but they 'still have to prove it'.

The tutor then turns to CD2.4ii: if h>1 then what about 1/h? Kelle replies it will be less than 1. The tutor asks him to use part i in order to show that hn will then be arbitrarily large. Kelle suggests 'inverting', she nods and writes down the symmetrically inverted statement.

The tutor comments on the students' drafts: CD2.4i was fine in their drafts but CD2.4ii was a 'disaster'. The students look surprised and the tutor invites Andrew to the b/b to 'defend his writing'. He starts by defining 'the group of all powers hⁿ'. The tutor stops him and asks what he means by a 'group'. Andrew then corrects 'group' to 'set of all powers hⁿ'. So S={hⁿ, nÎN} and by contradiction Andrew wants to prove that 'we can find some m which is an upper bound of hⁿ, in which case we can show that we can't have... it's got to be less than this specific value...'. The tutor says he is confused.

A1: If we can find some m for which that's true, this is an upper bound of S, hⁿ cannot take any value, cannot be large enough... so I suppose that an upper bound for S exists and then we show it can't exist.

T1: OK, so we're going to do it with contradiction.

A2: OK. So... no... Suppose for some m in N... No. Suppose m is an upper bound of S and m is a natural number...

The tutor repeats what Andrew has said so far (assume that there exists mÎN an upper bound for S, that is hⁿ<m"nÎN and reach a contradiction) and says that Andrew needs to stress that the existence of m is an assumption and that he is trying to 'demolish it'. Andrew says he is sorry and writes the above on the b/b (see fig.5).

A3: Now if n is an element of N, then n+1... has... (n+1)ⁿ is in S... because n+1 is a natural number...

A4: But eta can take any value, can't we let it be...?

Jack laughs and the tutor stresses that h was given from the beginning and he cannot change it. 'Then I guess I resign' says Andrew and asks whether in any case the idea of contradiction is 'along the right lines'. The tutor asks him to justify his belief by showing some strategy. 'We really want to find an element of S which is greater than m. Thus m is not an upper bound of S. That's the strategy'. Then Jack has an idea: since we are given eta and we assume the existence of m we might reach contradiction by trying to prove that h^m is the greatest number in S. The tutor says he disagrees: even if we assume that m is an upper bound for S, h^m is not necessarily the largest number in S, he stresses. Andrew suggests proving that 'you can actually get as close as you want to h^m but you can't actually get there'. The tutor reminds them that whenever they have an idea they should be ready to confront his criticisms and suggests two ways for the proof. The first in based on the ArchPr and on using CD2.4i and on defining q as 1/h. In his presentation the tutor repeats his comment from Extract 6.1 on defining unknown things by means of known things. So q is defined as 1/h, he stresses. He also corrects Andrew's use of £ instead of < in the definition of q.

In his second suggestion, the tutor says that h>1 implies that z:=h-1>0 (Jack says that too). The tutor then asks what can be done with hⁿ. The students are silent. Then Jack suggests expanding it. The tutor says that he does not disagree but that he had something else in mind: he reminds them they are trying to prove that hⁿ is getting arbitrarily large. Andrew suggests factorization. The tutor then decides to employ Jack's suggestion. Jack dictates the binomial expansion for (1+z)ⁿ and the tutor says it is strictly greater than 1+nz, since as Andrew remarks, all terms are positive. But 1+nz gets apparently as large as n can be which completes the proof. The tutor reminds them that the inequality (1+z)n>1+nz was proved last week and so they can use it without employing 'anything as sophisticated as the binomial theorem'. This sparks off the following discussion:

A5: Well, if you do something like that would you have to say right in the beginning. Fine, I'll prove this by induction, then by this inequality that was question...

T2: Well, if you've done it before can't you refer back to it? We always go back to theorems and...

A6: Yes, it depends... but in the exams you don't know that we've done it.

T3: I know you've done it! As long as you're clear. You see your job is to produce proofs which you know really are proofs and which are clear to the person you are trying to communicate with. That he can understand these proofs. That's what you have to do.

J1: When we came here they said that we have to wipe out all knowledge of math, out of our mind... and now we start assuming things that we learned since we've been here.

T4: That's the idea, yes. Two weeks ago you knew absolutely nothing. Last week you proved by induction this inequality. So this week...

A7: So we can have the data... all kinds of uses...

The tutor says he agrees: if they have any doubts, in this case the proof via the ArchPr is a 'more direct' one.

Return to Section 6(v).

Return to Appendices for Chapter 6.