Introduction to University Mathematics: Lecture 7 - Oxford Mathematics 1st Year Student Lecture

This course is taken in the first two weeks of the first year of the Oxford Mathematics degree. It introduces the concepts and ways of mathematical thinking that students need in the years ahead (if you find the writing difficult please use the captions which have been edited).

Much of the context will be familiar from high school but the way we think and write about it, and construct arguments and proofs, will be more rigorous.

In summary it is a recap and a pointer to what is to come for our students. We are showing the whole course over eight lectures. This is Lecture 7 and looks at constructing statements and proofs.

You can watch many other student lectures via our main Student Lectures playlist (also check out specific student lectures playlists):

All first and second year lectures are followed by tutorials where students meet their tutor to go through the lecture and associated problem sheet and to talk and think more about the maths. Third and fourth year lectures are followed by classes.

[knightofmathematics]

Lovely lecture, thank you! Your mentioning of the method of least criminals brings to mind an instance from my first year at Oxford (1985) when the lecturer used that very method to actually prove, to justify proof by induction. They didn't use the terminology "least criminals" then, but this was the method they used.

So, it was shown that if a statement P(n) with n a natural number is such that:
(i) P(0) is true
(ii) for any natural number k, P(k) -> P(k+1)
then P(n) is true for all n.
The proof: Suppose that there is at least one natural number n for which P(n) is not true. Consider the set A of all such numbers: A = (n | P(n) is false}. Then A has a least member; call it m. The number m cannot be 0, because of i. So m >= 1. Then m-1 is a natural number and as m-1 < m, m-1 cannot be a member of A. So P(m-1) is true. But then by ii, P(m-1+1) or P(m) is true, which is a contradiction. Hence n cannot exist and P is true for all natural numbers.

After graduation, I remember impressing a non-mathematical colleague with this proof. He complained that he did not understand why induction worked, when they taught it to him at school, so I gave him this proof and he was quite amazed by it. Although I like it very much as a proof and as an approach, I would not advocate for the use of this particular proof, as my experience with Axiomatic Theory impels me to justify induction using the Peano axioms and the construction of the natural numbers as the smallest inductive set. But that is another story.

One other thing: Another counterexample of "if g o f is injective then g need not be" is given by the couple of functions:
g: R -> R with g(x) = x^2
f: [0, +infinity) -> [0, +infinity) with f(x) = sqrt(x)
Then g(f(x)) = x and that is injective, but g is not.
Many thanks once again.

[iliyac-ioup]

All this math to become an engineer.... while you earn a lot more being a procesoperator, while doing the minimum