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fermatslibrary's profile
Fermat's Library
Fermat's Library
Fermat's Library
@fermatslibrary

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Fermat's Library

@fermatslibrary

A platform for illuminating academic papers. We publish an annotated paper every week. Our chrome extension for arXiv: https://fermatslibrary.com/librarian 

fermatslibrary.com
Joined September 2015

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    Fermat's Library‏ @fermatslibrary 3 Nov 2019
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    Here's a proof of why the square root of any prime number is irrational.pic.twitter.com/9yuUFnTTBt

    6:18 AM - 3 Nov 2019
    • 1,300 Retweets
    • 6,337 Likes
    • zkluwutt Gabre Mvs Sanjay Dimitrios Kalemis oneninefive #JusticeforGeorgeFloyd #BlackLivesMatter samantha Ofsayt Melis
    37 replies 1,300 retweets 6,337 likes
      1. New conversation
      2. Alex! 👁️‏ @AlexHoratio_ 3 Nov 2019
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        Replying to @fermatslibrary

        Can someone explain why B^2 * P definitely has an odd number of prime factors?

        5 replies 0 retweets 14 likes
      3. This Tweet is unavailable.
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      2. Matt Hewes‏ @stellarhewes 3 Nov 2019
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        Replying to @fermatslibrary

        Serious question, is there an integer for which the square root is rational but not an integer?

        2 replies 0 retweets 7 likes
      3. Timothy Leverett‏ @zzzzBov 3 Nov 2019
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        Replying to @stellarhewes

        No because a non-integer rational number is comprised of two numbers that don't share prime factors. Squaring those numbers means the results continue to not share prime factors.

        1 reply 1 retweet 18 likes
      4. End of conversation
      1. New conversation
      2. Walter Tross  🇪🇺@ 🏠2 ✋ 👑 🦠‏ @waltertross 3 Nov 2019
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        Replying to @fermatslibrary

        isn't that easily generalisable to any integer that is not a square, i.e., whose prime factors do not all have even powers?

        2 replies 0 retweets 2 likes
      3. Matt Kelly‏ @kellmano9 3 Nov 2019
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        Replying to @waltertross @fermatslibrary

        No because if P =15 then P has an even number of factors. Of course it is irrational, but this particular proof doesn't work.

        1 reply 0 retweets 3 likes
      4. 3 more replies
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      2. Kerthorok‏ @Kerthorok 3 Nov 2019
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        Replying to @fermatslibrary

        You can make a more general argument that all integers that are not perfect nth powers have irrational nth roots by proving 2 things: 1: That the integer is between two consecutive nth powers. 2: That the nth power of any non-integer rational is a non-integer rational.

        1 reply 1 retweet 1 like
      3. Paul Hankin‏ @PaulHankin2 3 Nov 2019
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        Replying to @Kerthorok @fermatslibrary

        Or in the spirit of the given proof, given an integer x that isn't an n'th power, let p be a prime that doesn't appear a multiple of n times in the prime factorization of x. Then x * b^n = a^n, and p appears non-zero times (mod n) on the left, and zero times (mod n) on the right.

        0 replies 0 retweets 0 likes
      4. End of conversation
      1. New conversation
      2. Vivek Kumar‏ @vivekkr_kumar 3 Nov 2019
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        Replying to @fermatslibrary

        Not sure of the assumption that A^2 or B^2 will have even number of prime factors. For example, both 7 and 7^2=49, have the same number of prime factors

        1 reply 0 retweets 1 like
      3. mek‏ @erakem 4 Nov 2019
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        Replying to @vivekkr_kumar @fermatslibrary

        7*7 -> 2 factors which are prime. And yes, equal.

        0 replies 0 retweets 1 like
      4. End of conversation

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