For any rational number p/q (with q > 0) there exists a smallest positive natural number k such that k * p/q is natural: Clearly q works, so we check the finitely many numbers 1, ..., q and pick the smallest.
Suppose the square root of 2 is rational. Let k be this smallest number for \sqrt 2, and proceed with the rest of the proof to find 0 < k' < k that also works.
This is why it also works for √3, but doesn’t work for √5 or 1.5
k' = √n k - a k (instead of √2 k - 1 k)
= k(√n - a) (so 0 < k' < k)
√n k' = n k - a k √n (is integral; k' ∈ K)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.)
Doesn't it also rely on √2 being positive?
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?
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.
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.
So, by definition, the square root of 4 is +2. x * x = 4 has two solutions, which we dub +sqrt(2) and -sqrt(2).
But anyway, when talking about THE nth root, we're almost always talking about the principal root.
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.
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.
> 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.
1. Let K be the set of positive integers {n: n√2 ∈ ℤ}
2. Supposing K is non-empty, denote its minimum as k.
3. Consider k' = k√2 - k = k(√2 - 1)
Lemma 1: k' < k
L1. When x > 1, √x < x. Therefore, √2 < 2, √2 - 1 < 1, and k' = k(√2 - 1) < k.
Lemma 2: k' is a positive integer
L2a: By hypothesis, k√2 ∈ ℤ.
L2b: By hypothesis, k is a positive integer.
L2c: The difference of any two integers is itself an integer, so k' ∈ ℤ.
L2d: When x > 1, √x > 1. Therefore, √2 > 1, k√2 > k, and so k' = k√2 - k > 0.
4. k'√2 = (k√2 - k)√2 = 2k - k√2 ∈ ℤ. Therefore, k' ∈ K.
5. This contradicts the assumption that k is the minimum element of K. Thus K must be empty.
To make this proof work, we needed some general information about the behavior of the square root function over the entire real domain. (Or, some special knowledge about the numerical value of √2, but from an elegance perspective that's even worse.) This is something you can easily get away with, but it's somewhat annoying. The proof based on "assume this rational number is represented in its simplest form, such that gcd(n,d) = 1" seems a little more self-contained.
That won't work; in that case, k' will not be a positive number (k/2 - k is negative whenever k is positive), so you'll fail to produce a contradiction.
k * (3/2 - 1)(3/2) = k'(9/4) - k(3/2)
You'd only get an integer if 4 divides k', which doesn't happen, because
k = 2
k' = k(3/2 -1) = k/2 = 1p^2=2q^2
with p and q mutually prime, which is impossible.
Instead of irreducible fractions with coprime numbers, he is doing the “dual” using the minimum possible value of the denominator.
Suppose K is non-empty, denote its minimum as k.
Consider the positive integer k' = k 3 - k = k (3 - 1).
Then k' < k (?!) and k' is also in K since k' 3 = 2 k - k 3, contradicting non-emptyness of K.
--
I guess my questions are why k' < k and why k'√2 is positive?
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?
x^2 = a
where a is a positive real number has two solutions. An equation can have multiple solutions (even infinitely many solutions). Sqrt(x) is a function and as such it can have only one output for a given input. Functions don’t have solutions. By definition, sqrt(a) is the nonnegative solution of the equation x^2=a. Here, a can be zero.
Let K be the set of positive integers k such that k * sqrt(2) is an integer.
Suppose K is non-empty, denote its minimum as k.
Consider the positive integer k' = k * sqrt(2) - k = k * (sqrt(2) - 1).
Then k' < k and k' is also in K since k' * sqrt(2) = k * 2 - k * sqrt(2).