Euclid's Proof that √2 is Irrational
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Euclid's Proof that √2 is Irrational
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Re: Euclid's Proof that √2 is Irrational
#2I thought this was basically the same proof that got someone killed in the Pythagorean cult. https://www.scientificamerican.com/article/how-a-secret-soci... shows it was a different proof in a similar vein, though. Fun times.
Re: Euclid's Proof that √2 is Irrational
#3> Likewise if a number is even and is a square of an integer, then its square root must be even.
The proof would be more compelling if this was proven instead of being taken as an obvious fact.
Re: Euclid's Proof that √2 is Irrational
#4> Likewise if a number is even and is a square of an integer, then its square root must be even. The proof would be more compelling if this was proven instead of being taken as an obvious fact.
It is really trivial though.
Re: Euclid's Proof that √2 is Irrational
#5> Likewise if a number is even and is a square of an integer, then its square root must be even. The proof would be more compelling if this was proven instead of being taken as an obvious fact.
Even x Even results Even
Even x Odd irrelevant if squaring
Odd x Odd results Odd
Re: Euclid's Proof that √2 is Irrational
#6These days I much prefer a different proof of this fact, attributed to Conway: https://www.youtube.com/watch?v=wNOtOPjaLZs -- I'm actually about half-way through teaching it to my kids right now!
Re: Euclid's Proof that √2 is Irrational
#71) it's not a proof by contradiction, it's a proof of a negation :grump:
2) I am not a fan of this phrasing "we can't simplify forever". Why can't we? It's obvious if you phrase it in the usual way as "the denominator is strictly smaller than it was before", but the "simplify" operation is kind of complex! They don't even mention "decreasing" until the very final Note box where they say offhand that actually it's an infinite descent (which is a critical part of the proof they've otherwise handwaved).
Re: Euclid's Proof that √2 is Irrational
#8> Likewise if a number is even and is a square of an integer, then its square root must be even. The proof would be more compelling if this was proven instead of being taken as an obvious fact.
Let n = 2r, and n = xx for some integers r and x, because n is even and n is a square. So xx = 2r.
Because of the fundamental theorem of arithmetic, we know that x must be representable as the product of a unique string of prime numbers.
Because 2 is prime, then since xx = 2r, there must be a 2 in the string of primes for xx.
But since 2 is prime, it must be in x as well, because a prime cannot come out of nowhere. In other words, if there is a given prime P in xx, there must be at least two P in xx, because there was at least one in x, and the number of each one got doubled in xx.
Therefore xx = 2r = 2*2*y = 4y for some integer y.
Therefore n = 4y and sqrt(n) = sqrt(4y) = sqrt(4)sqrt(y) = 2sqrt(y) which is an even number.
Therefore sqrt(n) is even.
Re: Euclid's Proof that √2 is Irrational
#9> Likewise if a number is even and is a square of an integer, then its square root must be even. The proof would be more compelling if this was proven instead of being taken as an obvious fact.
Let n = 2r, and n = xx for some integers r and x, because n is even and n is a square. So xx = 2r. Because of the fundamental theorem of arithmetic, we know that x must be representable as the product of a unique string of prime numbers. Because 2 is prime, then since xx = 2r, there must be a 2 in the string of primes for xx. But since 2 is prime, it must be in x as well, because a prime cannot come out of nowhere. I…
FTA is massive overkill. For every number n, either n can be expressed as 2k for some k, or 2k+1 for some k, but not both (proof: by induction); in particular the square root can too. If the square root is (2k+1), then the square is 4k^2 + 4k + 1 = 2(2k^2+2k) + 1, which is by definition odd, not even as we supposed.
Re: Euclid's Proof that √2 is Irrational
#10Earlier quoted context omitted.
Let n = 2r, and n = xx for some integers r and x, because n is even and n is a square. So xx = 2r. Because of the fundamental theorem of arithmetic, we know that x must be representable as the product of a unique string of prime numbers. Because 2 is prime, then since xx = 2r, there must be a 2 in the string of primes for xx. But since 2 is prime, it must be in x as well, because a prime cannot come out of nowhere. I…
FTA is massive overkill. For every number n, either n can be expressed as 2k for some k, or 2k+1 for some k, but not both (proof: by induction); in particular the square root can too. If the square root is (2k+1), then the square is 4k^2 + 4k + 1 = 2(2k^2+2k) + 1, which is by definition odd, not even as we supposed.
True, but the FTA proof is just really intuitive for me and I like it.