40 ms·

What am I missing about why this couldn't also argue sqrt(4) is irrational?

thaumasiotes 4y ago
The parent comment didn't call it out, but this proof relies on the fact that √2 - 1 is less than 1. (This is necessary when we conclude that k(√2 - 1) is less than k.) The same is not true of √4 - 1. If you look at my more explicit proof sidethread, the problem occurs in line L1, where instead of going from "√2 < 2" to "√2 - 1 < 1", you'd go from "√4 < 4" to "√4 - 1 < 3" and then you'd be stuck. (My proof won't even work for √3, but it seems safe to say that the problem with √3 is correctable and the problem with √4 is not.)
- phkahler 4y ago
  >> The parent comment didn't call it out, but this proof relies on the fact that √2 - 1 is less than 1. Doesn't it also rely on √2 being positive?
  - estomagordo 4y ago
    Which it is, by definition?
    - thehappypm 4y ago
      -1.4ish is another square root of 2
      - johnnny 4y ago
        I'll repeat myself: it's very obvious in context that they meant the positive solution of xˆ2 = 2. My definition of the square root is as follows: the square root of a positive real number x is the positive number, noted √x, such that (√x) ^ 2 = x. To make this a useable definition, we need to prove that equation has a solution (using the fact the function t -> t^2 is zero for t=0, diverges to +inf when t -> +inf, and is continuous between the two) and that solution is unique (using the fact the same function is strictly increasing). Do you have any other definition of √x?
        thehappypm 4y ago
        The most satisfying answer I have for the nature of the square root is to consider complex numbers. For an arbitrary complex number, a + bi, we can plot this as a vector on a two axis scale (x axis real and y axis imaginary). We can also convert any complex number to the form z = r e^(i θ). In other words, draw the complex number vector as an angle and a magnitude, in polar form. So any number can be drawn as a vector onto the complex plane. Multiplying two vectors together causes the angles to add and the magnitude to multiply — creating a rotated and extended (or shrunken) new vector as the product. So what’s the square root of a complex number z? It’s the vector a that, when multiplied by itself, winds up with z. When z has no imaginary component, it lies flat on the x axis; its magnitude is z and the angle is 0. What’s its square root? There are two. |a|e^(i 0) and |a|e^(i 180). 180 degrees multiplied together rotates the vector back to 0. And the e^(i 180), in radians, is e^(i pi), or negative one. So putting it together: there are only two solutions, with the same magnitude, and different signs.
        mannerheim 4y ago
        The radical symbol always refers to a single-valued function by convention. x^(1/n) is used for multi-valued functions in complex analysis, not the 'n-radical' symbol, which would usually refer to the principal root.
        thehappypm 4y ago
        Sure, but doesn’t that weaken the proof? It is an arbitrary restriction on the function that is required
        mannerheim 4y ago
        Barely? It proves it for the principal root, and you can trivially prove it for the negative root using the principal root's irrationality.
        simiones 4y ago
        The initial definition of a square root of a number X (from where the name actually comes from) is "the length of the sides of a square whose area is X". There are some straightedge and compass constructions for this even in Euclid's Elements. That's why the square root is always a positive number; bringing in complex numbers only serves to confuse the issue, and is anachronistic.
    - zeroonetwothree 4y ago
      I can’t believe you are downvoted. In math sqrt is defined as the positive one. It seems HN readers need a refresher.
      - BeetleB 4y ago
        Most people are taught that there are two values for the square root. It was only in later math courses (e.g. precalculus) that they reserved the square root symbol exclusively for the positive square root.
        thaumasiotes 4y ago
        That is not what people are taught. If that were true, the numerator in the quadratic equation would be "-b + √(b² - 4ac)", not "-b ± √(b² - 4ac)".
        BeetleB 4y ago
        Your statement assumes the educational system values consistency, which is a clearly false assumption.
        thaumasiotes 4y ago
        Textbooks aren't that bad. Here's what Pearson's Algebra 1 says when introducing the novel concept of the radical sign: > The radical symbol √ indicates a nonnegative square root, also called a principal square root. I quoted that one because it's available on libgen ( http://library.lol/main/BE522034F8D3680B248162CD2C35D455 http://library.lol/main/BE522034F8D3680B248162CD2C35D455 , page 16), but obviously every basic algebra textbook covers this. The symbol is never discussed -- or used -- in any other terms than that it uniquely refers to a positive (or zero) square root.
        BeetleB 4y ago
        Your evidence is consistent with my claims. Or rather: I never claimed there aren't textbooks that define it as the positive square root.
  - vlovich123 4y ago
    I’m very rusty but I think this proves the positive roots are irrational. If the positive roots are irrational then the negative ones would be too.
- mannykannot 4y ago
  If I am not mistaken, the proof in the article works for √3, and your lemma 1 can be modified by noting that when x > 1, √x > 1 and also 1 < 3 < 4 so 1 < √3 < 2, 0 < √3 - 1 < 1, and k' = k(√3 - 1) < k.

The parent comment didn't call it out, but this proof relies on the fact that √2 - 1 is less than 1. (This is necessary when we conclude that k(√2 - 1) is less than k.) The same is not true of √4 - 1. If you look at my more explicit proof sidethread, the problem occurs in line L1, where instead of going from "√2 < 2" to "√2 - 1 < 1", you'd go from "√4 < 4" to "√4 - 1 < 3" and then you'd be stuck. (My proof won't even work for √3, but it seems safe to say that the problem with √3 is correctable and the problem with √4 is not.)

phkahler 4y ago
>> The parent comment didn't call it out, but this proof relies on the fact that √2 - 1 is less than 1. Doesn't it also rely on √2 being positive?
- estomagordo 4y ago
  Which it is, by definition?
  - thehappypm 4y ago
    -1.4ish is another square root of 2
    - johnnny 4y ago
      I'll repeat myself: it's very obvious in context that they meant the positive solution of xˆ2 = 2. My definition of the square root is as follows: the square root of a positive real number x is the positive number, noted √x, such that (√x) ^ 2 = x. To make this a useable definition, we need to prove that equation has a solution (using the fact the function t -> t^2 is zero for t=0, diverges to +inf when t -> +inf, and is continuous between the two) and that solution is unique (using the fact the same function is strictly increasing). Do you have any other definition of √x?
      - thehappypm 4y ago
        The most satisfying answer I have for the nature of the square root is to consider complex numbers. For an arbitrary complex number, a + bi, we can plot this as a vector on a two axis scale (x axis real and y axis imaginary). We can also convert any complex number to the form z = r e^(i θ). In other words, draw the complex number vector as an angle and a magnitude, in polar form. So any number can be drawn as a vector onto the complex plane. Multiplying two vectors together causes the angles to add and the magnitude to multiply — creating a rotated and extended (or shrunken) new vector as the product. So what’s the square root of a complex number z? It’s the vector a that, when multiplied by itself, winds up with z. When z has no imaginary component, it lies flat on the x axis; its magnitude is z and the angle is 0. What’s its square root? There are two. |a|e^(i 0) and |a|e^(i 180). 180 degrees multiplied together rotates the vector back to 0. And the e^(i 180), in radians, is e^(i pi), or negative one. So putting it together: there are only two solutions, with the same magnitude, and different signs.
        mannerheim 4y ago
        The radical symbol always refers to a single-valued function by convention. x^(1/n) is used for multi-valued functions in complex analysis, not the 'n-radical' symbol, which would usually refer to the principal root.
        thehappypm 4y ago
        Sure, but doesn’t that weaken the proof? It is an arbitrary restriction on the function that is required
        mannerheim 4y ago
        Barely? It proves it for the principal root, and you can trivially prove it for the negative root using the principal root's irrationality.
        simiones 4y ago
        The initial definition of a square root of a number X (from where the name actually comes from) is "the length of the sides of a square whose area is X". There are some straightedge and compass constructions for this even in Euclid's Elements. That's why the square root is always a positive number; bringing in complex numbers only serves to confuse the issue, and is anachronistic.
  - zeroonetwothree 4y ago
    I can’t believe you are downvoted. In math sqrt is defined as the positive one. It seems HN readers need a refresher.
    - BeetleB 4y ago
      Most people are taught that there are two values for the square root. It was only in later math courses (e.g. precalculus) that they reserved the square root symbol exclusively for the positive square root.
      - thaumasiotes 4y ago
        That is not what people are taught. If that were true, the numerator in the quadratic equation would be "-b + √(b² - 4ac)", not "-b ± √(b² - 4ac)".
        BeetleB 4y ago
        Your statement assumes the educational system values consistency, which is a clearly false assumption.
        thaumasiotes 4y ago
        Textbooks aren't that bad. Here's what Pearson's Algebra 1 says when introducing the novel concept of the radical sign: > The radical symbol √ indicates a nonnegative square root, also called a principal square root. I quoted that one because it's available on libgen ( http://library.lol/main/BE522034F8D3680B248162CD2C35D455 http://library.lol/main/BE522034F8D3680B248162CD2C35D455 , page 16), but obviously every basic algebra textbook covers this. The symbol is never discussed -- or used -- in any other terms than that it uniquely refers to a positive (or zero) square root.
        BeetleB 4y ago
        Your evidence is consistent with my claims. Or rather: I never claimed there aren't textbooks that define it as the positive square root.
- vlovich123 4y ago
  I’m very rusty but I think this proves the positive roots are irrational. If the positive roots are irrational then the negative ones would be too.
mannykannot 4y ago
If I am not mistaken, the proof in the article works for √3, and your lemma 1 can be modified by noting that when x > 1, √x > 1 and also 1 < 3 < 4 so 1 < √3 < 2, 0 < √3 - 1 < 1, and k' = k(√3 - 1) < k.