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Euclid's Proof that √2 is Irrational

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Re: Euclid's Proof that √2 is Irrational

#41

Post by ginkoleaf »

Smaug123 wrote: post #7

1) 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…

This distinction is only made by a small number of mostly constructivists. It is not common usage, and most working mathematicians will have no idea what you're talking about.

Re: Euclid's Proof that √2 is Irrational

#43

Post by thaumasiotes »

wazdra wrote: post #29
Smaug123 wrote: post #25

Earlier quoted context omitted.

As I said, according to the three sources above, which are the first sources I clicked on which didn't seem like blogspam, the phrase "proof by contradiction" is a term of art which means "uses the law of excluded middle to conclude the truth of a statement given a proof that its negation is false". It may be unfortunate that the mathematical world has standardised on the phrase "proof by contradiction" for this, but…

> but it has standardised on that phrase! To me, the only formal distinction you can make between the two lies in the use of the excluded middle. However, this distinction has not standardised in mathematics, as many mathematicians simply do not care for intuitionistic logic. Such a mathematician could see the above proof as: I want to show ¬P by contradiction. Therefore I assume ¬(¬P) which is just P to me (the unin…

> To me, the only formal distinction you can make between the two lies in the use of the excluded middle.

It's much dumber than that, since he's invoking the law of the excluded middle to use contradiction at all.

Re: Euclid's Proof that √2 is Irrational

#45

Post by cyberax »

kevinventullo wrote: post #32

For those who are interested in connections to more advanced mathematics, there is a sense in which √2 is still an integer, even though it is irrational. Specifically there is the notion of “algebraic integers”, which are the set of all complex numbers expressible as the root of a monic polynomial: x^n + a_{n-1}x^(n-1) + … + a_1x + a_0. Here each a_i is a usual integer in ℤ, and monic refers to the leading coefficien…

Perhaps a translation issue?

I've always referred to that set as "algebraic numbers" ( https://en.wikipedia.org/wiki/Algebraic_number ). Since they are equipotent with integers, you _can_ call them that, but it's misleading.

Re: Euclid's Proof that √2 is Irrational

#46

Post by LegionMammal978 »

cyberax wrote: post #45
kevinventullo wrote: post #32

For those who are interested in connections to more advanced mathematics, there is a sense in which √2 is still an integer, even though it is irrational. Specifically there is the notion of “algebraic integers”, which are the set of all complex numbers expressible as the root of a monic polynomial: x^n + a_{n-1}x^(n-1) + … + a_1x + a_0. Here each a_i is a usual integer in ℤ, and monic refers to the leading coefficien…

Perhaps a translation issue? I've always referred to that set as "algebraic numbers" ( https://en.wikipedia.org/wiki/Algebraic_number ). Since they are equipotent with integers, you _can_ call them that, but it's misleading.

In general, won't some algebraic numbers' minimal polynomials have a leading coefficient greater than 1, when written with integer coefficients?

Re: Euclid's Proof that √2 is Irrational

#47

Post by thaumasiotes »

Smaug123 wrote: post #7

1) 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…

> 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).

That isn't actually a critical part of the proof; you can just assume that your initial two integers are relatively prime and then derive a contradiction directly.

Re: Euclid's Proof that √2 is Irrational

#48

Post by adrian_b »

cyberax wrote: post #45
kevinventullo wrote: post #32

For those who are interested in connections to more advanced mathematics, there is a sense in which √2 is still an integer, even though it is irrational. Specifically there is the notion of “algebraic integers”, which are the set of all complex numbers expressible as the root of a monic polynomial: x^n + a_{n-1}x^(n-1) + … + a_1x + a_0. Here each a_i is a usual integer in ℤ, and monic refers to the leading coefficien…

Perhaps a translation issue? I've always referred to that set as "algebraic numbers" ( https://en.wikipedia.org/wiki/Algebraic_number ). Since they are equipotent with integers, you _can_ call them that, but it's misleading.

As already mentioned by another poster, algebraic numbers are more general than algebraic integers, because the leading coefficient of the polynomial does not have to be one, similarly to the difference between rational numbers and integer numbers, where for the former the denominator does not have to be one, like for the latter.

Re: Euclid's Proof that √2 is Irrational

#49

Post by bmacho »

cyberax wrote: post #45
kevinventullo wrote: post #32

For those who are interested in connections to more advanced mathematics, there is a sense in which √2 is still an integer, even though it is irrational. Specifically there is the notion of “algebraic integers”, which are the set of all complex numbers expressible as the root of a monic polynomial: x^n + a_{n-1}x^(n-1) + … + a_1x + a_0. Here each a_i is a usual integer in ℤ, and monic refers to the leading coefficien…

Perhaps a translation issue? I've always referred to that set as "algebraic numbers" ( https://en.wikipedia.org/wiki/Algebraic_number ). Since they are equipotent with integers, you _can_ call them that, but it's misleading.

No, algebraic integers are a different set than algebraic numbers. (A subset.)

Algebraic integers are much cooler, since there is a number theory on them: https://en.wikipedia.org/wiki/Algebraic_integer . (And also because the most basic facts about it, like that it forms a ring, are not trivial to prove, that's a good sign for a concept to be cool and useful.)

These two number sets are more or less in a relationship like regular integers (with a number theory), and rational numbers. In fact A = O/Z where A denotes the set of algebraic numbers, O denotes the set of algebraic integers, and Z denotes the set of integers.

Re: Euclid's Proof that √2 is Irrational

#50

Post by jl6 »

kevinventullo wrote: post #32

For those who are interested in connections to more advanced mathematics, there is a sense in which √2 is still an integer, even though it is irrational. Specifically there is the notion of “algebraic integers”, which are the set of all complex numbers expressible as the root of a monic polynomial: x^n + a_{n-1}x^(n-1) + … + a_1x + a_0. Here each a_i is a usual integer in ℤ, and monic refers to the leading coefficien…

Yes, but did God make the algebraic integers? Because this looks suspiciously like the work of man.
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