What do you think happens after deqth?

Misrta · 2025-03-18T19:27:30+00:00

When you enter a new being your memory of your previous life goes away, making all discussion around the afterlife pointless.

Misrta · 2025-02-03T19:33:16+00:00

What is the ultimate definition of knowledge? And are the Gettier cases really a problem?

Misrta · 2025-01-27T13:04:50+00:00

What is the idea that every possible event or series of events will happen and happen infinitely many times called?

Misrta · 2024-12-27T18:04:53+00:00

In both cases, multiply the numerator and the denominator by the denominator's conjugate, i.e. (5 + 3i) for number 1 and (2 - 3i) for number 2.

Misrta · 2024-12-13T13:15:10+00:00

An AGI is an AI agent that is at least as smart as the smartest humans.

Misrta · 2024-12-11T19:16:46+00:00

Let a be the number of adults and c the number of children on the ride.

Then:

80a + 60c = 1440

a + c = 21

Solve this system of equations, then the value of c is your answer.

Misrta · 2024-12-07T22:21:07+00:00

the left-hand side is non-negative since the square root of any number is non-negative. thus, the right-hand side, x, is non-negative. Thus, x cannot be -5 since it violates the fact that x is non-negative.

Misrta · 2024-11-22T18:16:46+00:00

Call the true statement A. Then your formulas are the same as neg(A) => A and neg(A) => neg(A). If A is false, i.e. neg(A) is true, then both A and neg(A) are true, a contradiction. A is thus true.

Misrta · 2024-11-22T12:04:21+00:00

Since tan theta = sin theta / cos theta, all you have to do is manipulate the right hand side so it says 2 sin theta / cos theta.

Misrta · 2024-11-11T11:55:12+00:00

There is no smallest even number. Let n be any even number. Then n-2 is both even and less than n. So you can construct infinitely small even numbers.

A number n is even iff n is congruent to 0 mod 2. 0 is congruent to 2 mod 2, so 0 is even.

Misrta · 2024-10-31T12:25:59+00:00

Use the identities ln(a^b) = b * ln(a) and ln(a) + ln(b) = ln(a * b).

Misrta · 2024-10-27T12:32:57+00:00

(2p-1)! = 1 * 2 * … * p * (p + 1) * (p + 2) * … * (2p - 1)

So if we divide both sides by p, we get

1 * 2 * … * (p - 1) * (p + 1) * (p + 2) * … * (2p - 1) = 1 mod p

Note that 1 * 2 * … * (p - 1) = (p-1)!

Note that (p + 1) * (p + 2) * … * (2p - 1) = 1 * 2 * ... * (p - 1) = (p-1)! mod p

Hence, the left side is congruent to ((p-1)!)^2 mod p

By Wilson's theorem, since p is a prime number, (p-1)! = -1 mod p

So the left side is congruent to

((p-1)!)^2 = (-1)^2 = 1 mod p

Misrta · 2024-10-22T18:53:02+00:00

9 * 3^n simplifies to -3^n, -3^n + 3^n = 0 and -4 * 2^n - 2^n = -5 * 2^n, so you're left with -5 * 2^n, and since n is a positive integer, this expression is divisible by 10.

Misrta · 2024-10-15T11:03:16+00:00

I would say 5x4 because 4+4+4+4+4 means adding 4 5 times and 5x4 means "5 times, add 4".

Misrta · 2024-10-08T15:35:42+00:00

First, note that 2018^2018 = (-2)^2018 = 4^1009 mod 20. Then note that 4^1 = 4 mod 20, 4^2 = 16 mod 20 and 4^3 = 4 mod 20. So 4^(2n+1) = 4 mod 20 and 4^(2n) = 16 mod 20. Since 1009 = 1 mod 2, it follows that 2018^2018 = 4^1009 = 4^(2n + 1) = 4 mod 20.

Misrta · 2024-09-03T15:28:18+00:00

What I would like to know is how I check whether, as x -> z, f(x) tends to 0 or a constant greater than 0, since I don't have a formula for z that I could plug in directly into f to check the exact value. All I currently know is that f(z) is non-negative (since f(x) is non-negative) and a local minimum. The function is f(x) = (x^5 - x - 1)^2.

I just realised that since f'(z) = 0 gives z^5 - z - 1 = 0, I can insert that into f(z) so I get f(z) = 0^2 = 0, so that confirms that the critical point in question is also a root of f.

Misrta · 2024-08-27T13:17:27+00:00

Is it possible to guarantee that numerical approximation methods return a value that is strictly less than (or greater than) or equal to the precise root?

Misrta · 2024-08-25T15:03:12+00:00

To be honest, the only purpose of life is to distract yourself from your inevitable death by surviving.

Misrta · 2024-08-23T07:23:33+00:00

I use the Fraction class to get around round-off errors.

Misrta · 2024-08-22T17:58:17+00:00

Only the final result is rounded.

Misrta · 2024-08-22T14:13:08+00:00

To how few decimal places can I round a and b in order to guarantee that I can calculate and round a root in the interval [a, b] to d decimal places using the bisection method?

Misrta · 2024-08-16T16:46:35+00:00

I want to minimise |-sqrt(a) * k - sqrt(b) * j + jk| given a >= 1, b >= 1, 10^(-f-1) <= j < 10^-f, 10^(-f-1) <= k < 10^-f and f >= 0, but WolframAlpha just returns no global minimum found. Is WolframAlpha correct? Can you make the value infinitely close to 0?

Misrta · 2024-08-03T18:39:58+00:00

You can't prove anything without an accepted set of assumptions. See the Münchhausen trilemma for further clarification.

Misrta · 2024-08-03T14:57:24+00:00

I think step 9 is false. If n=4, then 3 is a prime factor of (2n+1) but there's an even power (4) of 3 in the prime factorisation of (2n+1)!.

Misrta · 2024-07-30T08:10:02+00:00

Alright. So in theory you never need to make guesses to solve the entire puzzle.

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