I fed in the latex question file to ChatGPT, although instead of the image of the axioms P1-P12 I referred it to the axioms in Spivak’s book. It came up with the following, which I would grade as 80/100.
The mean for the midterm was 64 (by section: 59 for one section and 72 for the other). A smoothed density graph of the grades is as follows:
You can also find solutions at this link.
A reminder about the HW and grading:
Finally, here are some requests from the graders! Following these instructions will help you avoid unneccessarily losing points.
Although I did this in class, I thought it might be useful to write out a proof here. There are many variations, but it should look something quite similar to this.
Claim: For \(n = 0,1,2,\ldots\) a non-negative integer, the winner of the sticks game is as follows:
Proof: We give a proof by induction starting from \(n = 0\). Let \(P(n)\) mean that the three listed statements above are true for \(n\).
Base Case: Suppose that \(n=0\).
These three statements prove \(P(0)\) is true, which is the base case.
Inductive Step: We shall assume \(P(k)\) and show \(P(k+1)\).
Thus assuming \(P(k)\) we have shown \(P(k+1)\) is true, and so by induction we are done!
Remarks:
Added: A real life application of the sticks game (or a simple variant): Try to solve it in real time
Let’s discuss a number of classic mistakes one can make when writing down a proof.
THE WRONG ORDER: We start with proofs where the argument goes in the wrong order.
Claim: For all numbers \(a\) and \(b\), there is an inequality \(a^2 + b^2 \ge 2ab.\)
Let us also recall that we have already proven that, for any number \(c\), we have \(c^2 \ge 0\) (this is discussed in Spivak).
Bad proof: We have:
\[\begin{aligned} a^2 + b^2 & \ge 2 a b \\
a^2 + b^2 – 2 a b & \ge 0 \\
(a – b)^2 & \ge 0,\\
\end{aligned}
\]
and all squares are positive, so the inequality holds □
The problem: This argument starts with the claim that is to be proved, and then makes (correct!) deductions to end up with a true statement. But that is the reverse of what we want to do. If you start with a true statement and then make correct deductions you end up with something that is true. But you can start with a false statement, make correct deductions, and end up with something that is true. For example, if you start with \(1 = 2\) and you multiply both sides by \(0\) you get \(0 = 0\) which is, true, even though \(1 = 2\) is false! If you find yourself arguing this way, the fix is to to try to “turn your argument upside down”, like this:
The fix: All squares are positive, and so if \(c = a-b\) then \(c^2 \ge 0\). Hence
\[\begin{aligned} (a-b)^2 & \ge 0 \\
a^2 + b^2 – 2 a b & \ge 0 \\
a^2 + b^2 & \ge 2 a b,\\
\end{aligned}
\]
so the inequality holds □
This is almost the same word for word!
NOT ENOUGH WORDS OR DETAILS
Claim: Show that
\[ \frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}}
+ \frac{1}{\sqrt{3} + \sqrt{4}} = 1.\]
Bad Proof:
\[ \begin{aligned}
\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}}
+ \frac{1}{\sqrt{3} + \sqrt{4}} & = \sqrt{2} – \sqrt{1} + \sqrt{3} – \sqrt{2} + \sqrt{4} – \sqrt{3} \\
& = 1. \end{aligned} \]
□
This is not wrong exactly, but it might really be confusing if you don’t know what’s going on. Here’s an improvement:
The fix: Remember that \((b+a)(b-a)=b^2-a^2\). Now we note that
\[ \begin{aligned}
\frac{1}{\sqrt{a+1} + \sqrt{a}} & = \frac{1}{\sqrt{a+1} + \sqrt{a}} \cdot
\frac{\sqrt{a+1} – \sqrt{a}}{\sqrt{a+1} – \sqrt{a}} \ \text{where we multiplied a factor $A/A = 1$} \\
& = \frac{\sqrt{a+1} – \sqrt{a}}{(\sqrt{a+1} + \sqrt{a})(\sqrt{a+1} – \sqrt{a})}\\
& = \frac{\sqrt{a+1} – \sqrt{a}}{(\sqrt{a+1})^2 – (\sqrt{a})^2} \\
& = \frac{\sqrt{a+1} – \sqrt{a}}{(a+1) – a} \\
& = \sqrt{a+1} – \sqrt{a}. \end{aligned}\]
Hence using this formula on each term:
\[ \begin{aligned}
\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}}
+ \frac{1}{\sqrt{3} + \sqrt{4}} & = (\sqrt{2} – \sqrt{1}) + (\sqrt{3} – \sqrt{2}) + (\sqrt{4} – \sqrt{3}) \\
& = 1.\end{aligned}\]
□
We shall return to more examples of bad proofs later!
This will be the class blog for 16100 in Autumn 2025, and will be an occasional supplement to the regular course lectures. (I started blogging about courses during the pandemic, but students find the blog useful so have continued to use it.)
You should really think about this course as several courses in one. What are those courses?
Syllabus: You can find a syllabus here which has some information about grading for the course.
