11–20 of 94 posts
The new proof rephrased: Let K be the set of positive integers k such that k √2 is an integer. Suppose K is non-empty, denote its minimum as k. Consider the positive integer k' = k √2 - k = k (√2 - 1). Then k' < k and k' is also in K since k' √2 = 2 k - k √2, contradicting non-emptyness of K.
The new proof rephrased: Let K be the set of positive integers k such that k √2 is an integer. Suppose K is non-empty, denote its minimum as k. Consider the positive integer k' = k √2 - k = k (√2 - 1). Then k' < k and k' is also in K since k' √2 = 2 k - k √2, contradicting non-emptyness of K.
... which contradicts the hypothesis that k is the minimum. Therefore, K is empty.
The new proof rephrased: Let K be the set of positive integers k such that k √2 is an integer. Suppose K is non-empty, denote its minimum as k. Consider the positive integer k' = k √2 - k = k (√2 - 1). Then k' < k and k' is also in K since k' √2 = 2 k - k √2, contradicting non-emptyness of K.
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.
The new proof rephrased: Let K be the set of positive integers k such that k √2 is an integer. Suppose K is non-empty, denote its minimum as k. Consider the positive integer k' = k √2 - k = k (√2 - 1). Then k' < k and k' is also in K since k' √2 = 2 k - k √2, contradicting non-emptyness of K.
The new proof rephrased: Let K be the set of positive integers k such that k √2 is an integer. Suppose K is non-empty, denote its minimum as k. Consider the positive integer k' = k √2 - k = k (√2 - 1). Then k' < k and k' is also in K since k' √2 = 2 k - k √2, contradicting non-emptyness of K.
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'
L1. When x > 1, √x 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.
The new proof rephrased: Let K be the set of positive integers k such that k √2 is an integer. Suppose K is non-empty, denote its minimum as k. Consider the positive integer k' = k √2 - k = k (√2 - 1). Then k' < k and k' is also in K since k' √2 = 2 k - k √2, contradicting non-emptyness of K.
The new proof rephrased: Let K be the set of positive integers k such that k √2 is an integer. Suppose K is non-empty, denote its minimum as k. Consider the positive integer k' = k √2 - k = k (√2 - 1). Then k' < k and k' is also in K since k' √2 = 2 k - k √2, contradicting non-emptyness of K.
What am I missing about why this couldn't also argue sqrt(4) is irrational?
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.)
For integers p, q with q not 0, suppose
2 = (p/q)^2
Then
2 q^2 = p^2
and the left size has an odd number of factors of 2 and the right side has an even number of factors of 2, a contradiction.
Therefore such p, q do not exist, and 2 does not have a rational square root.
Done.
The new proof rephrased: Let K be the set of positive integers k such that k √2 is an integer. Suppose K is non-empty, denote its minimum as k. Consider the positive integer k' = k √2 - k = k (√2 - 1). Then k' < k and k' is also in K since k' √2 = 2 k - k √2, contradicting non-emptyness of K.
This is similar to the proof of the fundamental theorem of arithmetic (from which the irrationality of the square root of 2 trivially follows): https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithme...
Let's have a simple proof so that we don't have to argue about more complicate issues: For integers p, q with q not 0, suppose 2 = (p/q)^2 Then 2 q^2 = p^2 and the left size has an odd number of factors of 2 and the right side has an even number of factors of 2, a contradiction. Therefore such p, q do not exist, and 2 does not have a rational square root. Done.
[1] https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithme...