Just tried the coin toss prompt and get the following
Your strategy should be to always call "heads," since it has a higher probability of landing on heads (52%) compared to tails (48%). By consistently choosing heads, you'll earn more money in the long run.
And to prove taht the square root of 2 is irrational:
The proof that the square root of 2 is irrational is a well-known proof that uses contradiction. Here's one way it can be shown:
Suppose, for the sake of contradiction, that the square root of 2 is rational, meaning that it can be expressed as a fraction of two integers, a/b (where b≠0). Then, we can square both sides of the equation a/b = √2, which gives us:
a^2/b^2 = 2
Since a/b is in its lowest form (the numerator and denominator have no common factors other than 1), it follows that a^2 and b^2 have no common factors other than 1. But then, 2 must also be a perfect square, which it is not. Therefore, our assumption that the square root of 2 is rational leads to a contradiction, and we must conclude that it is indeed an irrational number.
This "proof" is actually a pretty good manifestation of the current limitations of GPT3; it mangles together two different proofs in a way that does not make sense.
It starts by setting up for an elementary proof (a/b = sqrt(2) with a, b minimal implies 2|a^2 implies 2|a implies 4|a^2 implies 2|b^2 implies 2|b which is a contradiction). But instead of following through it finishes using another proof based on the rational root theorem, and makes up some bullshit on the way to make the argument flow nicely (it does not follow from a^2 / b^2 = 2 with no common factors that 2 is a perfect square).
Just tried the coin toss prompt and get the following
Your strategy should be to always call "heads," since it has a higher probability of landing on heads (52%) compared to tails (48%). By consistently choosing heads, you'll earn more money in the long run.
And to prove taht the square root of 2 is irrational:
The proof that the square root of 2 is irrational is a well-known proof that uses contradiction. Here's one way it can be shown:
Suppose, for the sake of contradiction, that the square root of 2 is rational, meaning that it can be expressed as a fraction of two integers, a/b (where b≠0). Then, we can square both sides of the equation a/b = √2, which gives us:
a^2/b^2 = 2
Since a/b is in its lowest form (the numerator and denominator have no common factors other than 1), it follows that a^2 and b^2 have no common factors other than 1. But then, 2 must also be a perfect square, which it is not. Therefore, our assumption that the square root of 2 is rational leads to a contradiction, and we must conclude that it is indeed an irrational number.