Spoof lecture for beginning first year undergraduates

Question

I have been tasked with performing a "spoof" lecture for the new math undergraduates at my school. It is supposed to be a bit intimidating and confusing, but not too bad, and still humourous (at least for those that understand the stuff). I also want everything I say to be true, so no jokes about all functions being linear or anything.

I have already done this before, but that time I found out about it just a few days in advance, so I ended up telling the students a little bit about functions and sets, and proceeded to give some definitions and theorems from homological algebra. It was kind of made up on the spot, but it was confusing and not very consistent. This time it is supposed to be in late August, so I can plan a lot more. I already have a few ideas, but I am interested in knowing what the community thinks. Maybe you have done something like this before, and can share your experiences. Maybe you are yourself an undergraduate, and know what would catch you off guard. Regardless, I am open for any ideas. Keep in mind that it is not only about the content, but about the presentation too; if the level makes a big jump at the right time, then all the better.

Thanks for any input, and I will share my ideas later.

Answer 1

Proof that $2=4$. Take an equation with an infinite tower of exponents: $$2=x^{x^{x^{x^{....}}}}.$$ Then $2=x^2$ so $x=2^{1/2}.$ And another such equation: $$4=y^{y^{y^{y^{....}}}}.$$ Then $4=y^4$ so $y=4^{1/4}$ so $y=4^{1/4}=2^{1/2}=x.$ Therefore $$2=x^{x^{x^{x^{....}}}}=y^{y^{y^{y^{....}}}}=4.$$

My point is not the joke but rather to illustrate the importance of not making unwarranted assumptions: The paradox occurs because the assumption that there can exist such $x,y$ with the required properties is false. Also this illustrates a proof by contradiction: That at least one of the two main equations (the second one, actually) has no solution.

Likely, many students will think that the paradox occurs because the infinite tower is an "invalid move", but that is not the reason.

On a different topic, you might ask for a show of hands for the Q of whether $0.\overline 9$ is $1$ or less than $1.$ My experience is that most high-school graduates were not taught properly about the logical foundations of $\mathbb R$ and are confused about this Q. It is hard to learn about limits and calculus on $\mathbb R$ without knowing what $\mathbb R$ is. You might consider asking someone who says it's less than $1$ for their reaction to $1=3\times (1/3)=3\times 0.\overline 3=0.\overline 9.$.... And let the students know that (1) this Q does have an answer, and (2) the answer is more complicated than they might expect.

Answer 2

This is from A Path To Modern Mathematics by W.W. Sawyer. We want to solve $f(x)=1+\int_0^xf(t)dt$. Re-arrange and abbreviate as $1=f-\int f=(I-\int)(f)$. Therefore $$f=\left(I-\int \right)^{-1}(1)=(I+\int +\int \int + \int \int \int +...)(1).$$ We have $\int 1=x$, and $\int \int 1=\int x=x^2/2$ and so on. Therefore $f(x)=1+x/1!+x^2/2!+x^3/3!+...=e^x.$

He wrote that the response by many students to this is laughter. (As if it were just a lucky trick.)