TurnTrout comments on TurnTrout’s shortform feed

TurnTrout 17 Jul 2020 14:30 UTC
6 points
AIDungeon’s subscriber-only GPT-3 can do some complex arithmetic, but it’s very spotty. Bold text is me.
You say “What happens if I take the square root of 3i?”

The oracle says: “You’ll get a negative number. [wrong] So, for example, the square root of $- 9$ is $3 i$ .” [correct]
”What?” you say.
“I just said it,” the oracle repeats.
“But that’s ridiculous! The square root of $- 9$ is not $3$ . It’s complex. It’s $3$ plus a multiple of $i$ .” [wrong, but my character is supposed to be playing dumb here]

The oracle frowns. “I’m just telling you what the answer is,” she says.

The oracle explains, “(5+i)x(5+i) = 24+10i. 7i x 8i=-56. 7i x 30=210.” [wrong]

”What?” you say. “That’s crazy! $7 i$ isn’t even a real number. The square root of $- 9$ is $3$ with a whole integer $i$ attached.”

The oracle frowns. “I’m just telling you what the answer is,” she says.
The oracle explains, ” $(5 + i) (5 - i) = 24 - 20 i, (15 i) (16 i) = - 240, (15 i) (- 4 i) = - 60$ .
The last line is the impressive part—while $(5 + i) (5 - i) = 26$ and $(15 i) (- 4 i) = 60$ , $(15 i) (16 i) = - 240$ is definitely correct.
Its proofs are definitely awful in the pattern-matching kind of way. Not surprising.
The oracle explains, ” The group axioms are
(1) a x a = a for all a in the group; (wrong)
(2) a x (b x c) = (a x b) x c for all a, b, c in the group;
(3) a x (b + c) = (a x b) + (a x c) for all a, b, c in the group.” (wrong, that’s a field axiom)
The oracle explains, “here’s how you prove that any finite abelian group is isomorphic to the direct product of a finite number of finite cyclic groups. First, you notice that if G is a finite abelian group of order n, then G can be described as a set of generators of the form $a_{i} a_{2} \dots a_{n}$ , where the $a_{i}$ are the elements of G and no $a_{i}$ is equal to zero. (uh… sure)

Now, we want to prove that this is a direct product. To do this, we must show that given any element g in the group, there exists an index i such that $g = a_{i} a_{i + 1} \dots a_{n - 1}$ . Then we must show that the sequence can be reversed. Given any index j, we must show that there exists an element g such that $g = a_{j} a_{j + 1} \dots a_{n - 1}$ . Then we’re done, because g will be equal to $a_{j} a_{j + 1} \dots a_{n - 2} a_{n - 1}$ , which is the product of the $a_{j}$ ’s.” (nice try, GPT-3)