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.)

>> 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?

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.

Which it is, by definition?

-1.4ish is another square root of 2

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?

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.

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.

Um, I don’t really think that’s right.

using the square root to convert area to perimeter, for example, is an application of square roots, one where the negative root is kinda useless. That’s not the definition of the operation. The definition of the operation is the solution to y = xx. And sure for many applications the negative root is useless, but you can’t argue against that both -2 -2 and 2*2 = 4.

You may not think I'm right, but that is the actual history. This operation was invented at a time where geometry was the main way mathematics was done. The square root is a much older concept than negative numbers and equations.

So, by definition, the square root of 4 is +2. x * x = 4 has two solutions, which we dub +sqrt(2) and -sqrt(2).

The negative root is also a square root, but it isn't the root denoted by the radical symbol.

But anyway, when talking about THE nth root, we're almost always talking about the principal root.

https://news.ycombinator.com/item?id=34441863

> Diophantus, in Arithmetica, c. 280 AD, was claiming that the equation 4x + 20 = 4 doesn't have any solutions/is absurd. So, the square root is more than 600 years older than negative numbers in the Hellenistic tradition.

The gap is much wider than that; double-entry bookkeeping speaks of "credit" and "debit" (defined to be exactly the same concept as positive and negative numbers) because negative numbers were still nearly unknown to Europeans in the 15th century.

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.

Sure, but doesn’t that weaken the proof? It is an arbitrary restriction on the function that is required

Barely? It proves it for the principal root, and you can trivially prove it for the negative root using the principal root's irrationality.

I can’t believe you are downvoted. In math sqrt is defined as the positive one. It seems HN readers need a refresher.

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.

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)".

Your statement assumes the educational system values consistency, which is a clearly false assumption.

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 , 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 square root.

Your evidence is consistent with my claims. Or rather: I never claimed there aren't textbooks that define it as the positive square root.

All textbooks define it that way. There is no other definition available for that symbol.

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.