Official Math Thread
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Official Math Thread
I doubt there'll be much traction here but who knows.
Mathematics is the art of thinking and problem solving. The numbers don't matter, mathematics is the beauty that emerges from creativity in creating structure and finding pattern.
anyways, I installed LaTeX on here, see the thread in the News/Info section for more about formatting mathematics.
Proof that there are infinitely many primes, by contradiction.
Assume that there are finintely many primes. Call them \(p_1, p_2,...,p_k\) and we shall assume they are sorted smallest to largest. Then consider the integer \(N\) that is one more than the product of these primes. That is,
$$N = p_1\cdot p_2 \cdots p_k +1$$
By matter of fact, this integer \(N\) must be divisible by some prime number. If it was itself prime, we arrive at a contradiction, as \(N\) is certainly larger than \(p_k\), which was presumed to be the largest prime number in our collection. If it is divisible by a prime, it must be divisible by one of the \(p_1,p_2,...,p_k\). However, \(N\), upon division by any of these presumed complete collection of primes, the remainder is always 1.
Thus, in any case, we have a contradiction.
Therefore the assumption that there are finitely many primes was an incorrect assumption, and so there must be infinitely many prime numbers. \(\square\)
Mathematics is the art of thinking and problem solving. The numbers don't matter, mathematics is the beauty that emerges from creativity in creating structure and finding pattern.
anyways, I installed LaTeX on here, see the thread in the News/Info section for more about formatting mathematics.
Proof that there are infinitely many primes, by contradiction.
Assume that there are finintely many primes. Call them \(p_1, p_2,...,p_k\) and we shall assume they are sorted smallest to largest. Then consider the integer \(N\) that is one more than the product of these primes. That is,
$$N = p_1\cdot p_2 \cdots p_k +1$$
By matter of fact, this integer \(N\) must be divisible by some prime number. If it was itself prime, we arrive at a contradiction, as \(N\) is certainly larger than \(p_k\), which was presumed to be the largest prime number in our collection. If it is divisible by a prime, it must be divisible by one of the \(p_1,p_2,...,p_k\). However, \(N\), upon division by any of these presumed complete collection of primes, the remainder is always 1.
Thus, in any case, we have a contradiction.
Therefore the assumption that there are finitely many primes was an incorrect assumption, and so there must be infinitely many prime numbers. \(\square\)
wickedsmaht- Posts : 295
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Re: Official Math Thread
https://aperiodical.com/2024/07/the-big-internet-math-off-2024-quarter-final-1/
Saw this a while back. I never knew this about the Euclidean algorithm and continued fractions. Pretty cool.
Saw this a while back. I never knew this about the Euclidean algorithm and continued fractions. Pretty cool.
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wincel- Posts : 165
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Re: Official Math Thread
^ Interesting. I posted an answer on math.se and some guy said "obviously its the same as dividing"
Not exactly the same but similar idea, and it is cool
Not exactly the same but similar idea, and it is cool
wickedsmaht- Posts : 295
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wincel likes this post
Re: Official Math Thread
wickedsmaht wrote:^ Interesting. I posted an answer on math.se and some guy said "obviously its the same as dividing"
Not exactly the same but similar idea, and it is cool
Yea in retrospect, it is obvious, but I had never learned or thought of it this way.
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