CFJ's The One I Cannot Kill [Game Ended]
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
copypasta from wikipedia:
Proof by unique factorization
An alternative proof uses the same approach with the fundamental theorem of arithmetic which says every integer greater than 1 has a unique factorization into powers of primes.
Assume that √2 is a rational number. Then there are integers a and b such that a is coprime to b and √2 = a/b. In other words, √2 can be written as an irreducible fraction.
The value of b cannot be 1 as there is no integer a the square of which is 2.
There must be a prime p which divides b and which does not divide a, otherwise the fraction would not be irreducible.
The square of a can be factored as the product of the primes into which a is factored but with each power doubled.
Therefore, by unique factorization the prime p which divides b, and also its square, cannot divide the square of a.
Therefore, the square of an irreducible fraction cannot be reduced to an integer.
Therefore, √2 cannot be a rational number.
This proof can be generalized to show that if an integer is not an exact kth power of another integer then its kth root is irrational. For a proof of the same result which does not rely on the fundamental theorem of arithmetic, see: quadratic irrational.time will end-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
lifeIn post 49, Elements wrote:one iota of what?
An interesting choice. You may be town I guess.In post 50, Elements wrote:pinktime will end-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
-
-
TemporalLich Grand Scheme
- Grand Scheme
- Grand Scheme
- Posts: 5815
- Joined: January 30, 2019
- Location: A Lost Timeline
Copyright © MafiaScum. All rights reserved.