ZURAB JANELIDZE'S BLOG

So I wanted to check how well will ChatGPT-4 perform with the theorem-proving (in a variant of incidence geometry) that I give to second-years at the start of Foundations of Abstract Mathematics I... it takes a few weeks for the students to master the new, axiomatic approach to mathematics -- and precision of proofs that comes with it. For ChatGPT-4 it was a breeze at first, but then it started to mess up. It was, however, after some effort, able to correct its mistakes once they were pointed out. Below is the record of our conversation.  I will state some axioms and then ask you to prove theorems based on them. Please do not use any of your previous knowledge in the proofs, just the axioms. Sure, please state the axioms and the theorems you would like me to prove. Axiom 1. A point may only lie on a line or a plane, and a line may only lie on a plane. If a point lies on a line which lies on a plane, then the point lies in the same plane. Thank you for providing the axiom. Plea...

Use mathematical induction to show that n^2-n is always divisible by 2. To prove that n^2 - n is always divisible by 2 using mathematical induction, we'll follow these steps: Base case (n=1): Show that the statement is true for the lowest value of n. Inductive step: Assume that the statement is true for n=k (k is an arbitrary positive integer), and show that it is also true for n=k+1. Base case (n=1): When n=1, the expression becomes: 1^2 - 1 = 0 Since 0 is divisible by 2, the statement holds for n=1. Inductive step: Assume that the statement is true for n=k, i.e., k^2 - k is divisible by 2. We can write: k^2 - k = 2m, where m is an integer. Now, we need to prove that the statement is true for n=k+1. So, let's consider the expression for (k+1)^2 - (k+1): (k+1)^2 - (k+1) = k^2 + 2k + 1 - k - 1 = k^2 + k Now, we want to show that k^2 + k is divisible by 2. To do this, we can rewrite the expression using the assumption that k^2 - k is divisible by 2: k^2 + k = (k^2 - k) + 2k = 2m ...

  Model: GPT-4 18 March 2022 Prove that 2^n is bigger than n, for every natural number n. To prove that 2^n is greater than n for every natural number n, we can use mathematical induction. Mathematical induction is a technique that involves proving a base case and then proving that if the statement is true for some natural number k, it must also be true for k+1. Base case: Let n = 1. Then, 2^1 = 2, and 2 > 1. So the statement holds true for n = 1. Inductive step: Assume the statement is true for some natural number k, i.e., 2^k > k. Now, we need to prove the statement for k+1, i.e., 2^(k+1) > k+1. Consider 2^(k+1). We can rewrite it as 2^k * 2. Since we assumed 2^k > k, and 2^k * 2 > k * 2, we can say that 2^(k+1) > 2k. Now, we need to show that 2k > k+1 for k ≥ 1. Note that k ≥ 1 implies that k > 0, so 2k > k. Therefore, 2k > k+1 for k ≥ 1. Putting it all together, we have 2^(k+1) > 2k > k+1. Thus, the statement holds true for k+1. By the princi...