Chatbot math/Gemini/24.03/Surreal numbers
- See also Surreal number/ChatBot assistance
Summary
[edit | edit source]Gemini might not understand surreal numbers. On two occasions, it gave me what I strongly suspect are wrong answers.
Another possibility is that there are too many ways of looking at topics like this. The topic "surreal numbers" is a genuine mathematical subject. But it's primary purpose seems to be entertainment. Perhaps different authors created different versions of surreal numbers.
I had trouble understanding Wikipedia's Surreal numbers. So I skimmed a couple of articles until I could sort of guess what they were, asking Gemini whenever I had a question. It turns out that Gemini is quite versed on the topic, although for some reason, it struggled with the word "birthday" when I asked how many birthdays it takes to construct 13/31 from 0. Gemini's response was:
- Unfortunately, the concept of "birthdays" doesn't directly translate to constructing surreal numbers. Birthdays typically refer to increments of 1 year, which represents a specific type of positive number within the surreal number system.
Gemini went on to give me an example that confused me. When I rephrased the question, Gemini "apologized" about the birthday misunderstanding and gave me a cogent answer. However, upon reflection, I have serious doubts as to whether Gemini gave me a correct answer.
I didn't record the ensuing conversation, but was satisfied with Gemini's answers, and eventually uncovered what might have been the misconception that was blocking my effort to compssrehend the Wikipedia article. The emphasis on might have been is essential because it is entirely possible for a chatbot to create a misconception that leaves the human believing they understand something when they don't. Chatbot tutors come with a risk!
[edit | edit source]This is the question that either cleared things up (or falsely appeared to clear things up) in my mind:
Question: Is it possible to construct a surreal tree starting with 0, where the first number greater than 0 is pi?
Gemini's answer
[edit | edit source]Yes, it is possible to construct a surreal tree starting with 0 where the first number greater than 0 is pi.
Surreal trees are defined with two branches for each node: a left branch containing numbers less than the node's value and a right branch containing numbers greater than the node's value. Here's how we can construct such a tree:
# Start with a root node containing the value 0. # Since pi is greater than 0, assign it to the right child of the root node. # Complete the tree by assigning negative infinity to the left child of the root node. This ensures any number less than 0 is "smaller" than 0 in the surreal tree.
Here's a breakdown of the tree structure:
0
/ \
-∞ π (or any other number greater than 0)
This structure adheres to the surreal tree definition where the left branch contains a value less than the root (negative infinity in this case) and the right branch contains a value greater than the root (pi).
Comments by the human being coached by Gimini
[edit | edit source]I don't know if it the typesetting skills of the chatbot, or the flexibility of the WMF visual editor, but the fact that this tree was correctly rendered in wikitext is quite welcome.--Guy vandegrift (discuss • contribs) 13:55, 3 April 2024 (UTC)
Gemini's response
[edit | edit source]Here's how to prove 1 + 1 = 2 using surreal numbers in Conway notation:
- Representing the numbers:
- 1: As mentioned earlier, 1 is { 0 | }.
- Another 1: Since we're adding 1 to itself, we can again use { 0 | } to represent another 1.
- Adding surreal numbers:
Addition of surreal numbers involves considering elements from both left and right sets.
- Left set (result): We need to consider all elements that are less than both individual ones (1 and 1). Since there are no elements less than 1 (empty left set), the resulting left set is also empty.
- Right set (result): Here, things get interesting. We need to consider elements greater than both individual ones (both { 0 | }). In this case, any number greater than 0 (which is included in the right set of both 1s) will also be greater than the sum (2). So, the right set remains empty to maintain the well-formedness of the surreal number (no element in the right set can be less than or equal to an element in the left set).
- Putting it together:
Combining the empty left and right sets, we get the result of adding 1 and 1: { | }.
- Equivalence to 2:
This notation ({ | }) represents the surreal number 2 in Conway notation. Here's why:
- There are no elements smaller than 2 (empty left set).
- Any number greater than 0 (including 1) is also greater than 2 due to the empty right set.
Therefore, 1 + 1 (represented as { | }) is equivalent to 2 in the surreal number system.
Infinity plus one
[edit | edit source]The following conversation occurred on 6/11/2024. I believe Gemini also got it wrong here.
In surreal numbers, is \omega +1 greater than \omega? If so, what is the difference?
[edit | edit source]Show drafts
No, in surreal numbers, omega + 1 (ω + 1) is not greater than omega (ω). Here's why:
- Omega (ω): This represents the first infinite ordinal number. Imagine an infinitely long sequence starting from 1, going on forever (1, 2, 3, ...). Omega (ω) refers to this concept of infinity.
- Adding with Surreal Numbers: Unlike regular addition, surreal number addition relies on comparing sets. A surreal number is written in the form {a | b}, where elements to the left of the vertical bar (a) are greater than those on the right (b). A number {a | b} is considered greater than another {c | d} if the first element in set b (on the right) is smaller than the first element in set a (on the left).
- Why ω + 1 isn't greater:
- Omega plus one (ω + 1) would be written as {ω | 1}.
- In this case, we have infinitely many positive elements (ω) on the left, representing omega's infiniteness.
- Adding 1 only introduces a single finite element (1) on the right.
The crux lies in how surreal numbers handle infinity:
* Adding a finite quantity (like 1) to an infinite set (ω) doesn't change the overall infiniteness in the surreal number system.
* The resulting set {ω | 1} still contains infinitely many positive elements, making it essentially the same size as ω itself.
Therefore, ω + 1 doesn't represent a new kind of infinity larger than ω. It remains infinite and falls under the same category as omega within surreal numbers.