6 ms·

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

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.

tromp 4y ago
It is simple if you can assume unique factorization of positive integers [1]. [1] https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithme...
- hgsgm 4y ago
  Which is proven in a similar way as OP's proof.
  - ColinWright 4y ago
    ... but genuinely requires a lot more work to do carefully and properly. People underestimate how much work it is to create a non-handwavey proof of the FTA. Tim Gowers has written about this (although I don't have the reference to hand, and I'm on a train with a dodgy connection).
- graycat 4y ago
  > It is simple if you can assume unique factorization of positive integers [1]. Since the fundamental theorem of arithmetic is a classic and fundamental pillar of math, in writing math is it not appropriate to say "if you can assume".
jameshart 4y ago
I don’t think you can make assumptions about how factors combine on multiplication (ie that multiplying by two adds one factor of two) in a context where integers can have rational noninteger roots. If we can have x=(a/b)^2 and y=(b/c)^2, then xy=(a/c)^2 so any factors of b^2 have somehow dropped out, rather than being added in to the factors of the product. Obviously in the end this all ends up contradicting the fundamental theorem of arithmetic, but it’s not clear this doesn’t wind up being circular.
raldi 4y ago
> 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. Can you explain this step further?
- graycat 4y ago
  If I had explained it further, then the proof would have been longer! Right, we need the Fundamental Theorem of Arithmetic, that each integer factors uniquely into a product of prime numbers; right, that statement leaves out a little about the order of the factors. Well, 2 is prime number. So however many factors of 2 q has, q^2 has twice that many and, thus, an even number of factors of 2. Same argument for p^2; it also has an even number of factors of 2. But then 2 q^2 has an odd number of factors of 2. Uh, more generally, we can keep arithmetic simple or make it complicated. E.g., multiplication of integers is associative so that (2 x 3) x 4 = 2 x (3 x 4) The intermediate results are different, but the final results are the same. So there is a little something there (matrix multiplication is also associative, with more there -- right, easily function composition is associative so regard a matrix as a function and matrix multiplication as function composition). We can keep the Fundamental Theorem of Arithmetic simple, e.g., ask a 4rd grader, or make it complicated: To make arithmetic complicated, we can take a college course in abstract algebra that proves all this. Then for still more complicated we can go through axiomatic set theory before the abstract algebra course. Then in the set theory course we can get to undecidable issues. I went through all that. Instead, what I wanted was differential geometry as background for general relativity, partial differential equations for the various applications, but I'd put up with exterior algebra, of course the inverse and implicit function theorems (proved in Fleming's book and an exercise in Rudin's), fluid flow, Maxwell's equations, Schroedinger's equation, Fourier theory, Lebesgue integration, Leibniz's rule (differentiation under the integral sign; there's a simple proof if assume enough continuity and a more general proof from Lebesgue theory), how come, with Lagrange, action is stationary, on and on. I was willing to take arithmetic as okay!
  - raldi 4y ago
    Sounds like the crux is that 2(x^2) = y^2 can never be true for any integers x and y because any square of an integer must have an even number of 2's in its prime factorialization, and any doubled square of an integer must have an odd number.

tromp 4y ago

It is simple if you can assume unique factorization of positive integers [1]. [1] https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic https://en.wikipedia.org/wiki/Fundamental_theorem_of_arithme...

hgsgm 4y ago
Which is proven in a similar way as OP's proof.
- ColinWright 4y ago
  ... but genuinely requires a lot more work to do carefully and properly. People underestimate how much work it is to create a non-handwavey proof of the FTA. Tim Gowers has written about this (although I don't have the reference to hand, and I'm on a train with a dodgy connection).
graycat 4y ago
> It is simple if you can assume unique factorization of positive integers [1]. Since the fundamental theorem of arithmetic is a classic and fundamental pillar of math, in writing math is it not appropriate to say "if you can assume".

jameshart 4y ago

I don’t think you can make assumptions about how factors combine on multiplication (ie that multiplying by two adds one factor of two) in a context where integers can have rational noninteger roots. If we can have x=(a/b)^2 and y=(b/c)^2, then xy=(a/c)^2 so any factors of b^2 have somehow dropped out, rather than being added in to the factors of the product. Obviously in the end this all ends up contradicting the fundamental theorem of arithmetic, but it’s not clear this doesn’t wind up being circular.

raldi 4y ago

> 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. Can you explain this step further?

graycat 4y ago
If I had explained it further, then the proof would have been longer! Right, we need the Fundamental Theorem of Arithmetic, that each integer factors uniquely into a product of prime numbers; right, that statement leaves out a little about the order of the factors. Well, 2 is prime number. So however many factors of 2 q has, q^2 has twice that many and, thus, an even number of factors of 2. Same argument for p^2; it also has an even number of factors of 2. But then 2 q^2 has an odd number of factors of 2. Uh, more generally, we can keep arithmetic simple or make it complicated. E.g., multiplication of integers is associative so that (2 x 3) x 4 = 2 x (3 x 4) The intermediate results are different, but the final results are the same. So there is a little something there (matrix multiplication is also associative, with more there -- right, easily function composition is associative so regard a matrix as a function and matrix multiplication as function composition). We can keep the Fundamental Theorem of Arithmetic simple, e.g., ask a 4rd grader, or make it complicated: To make arithmetic complicated, we can take a college course in abstract algebra that proves all this. Then for still more complicated we can go through axiomatic set theory before the abstract algebra course. Then in the set theory course we can get to undecidable issues. I went through all that. Instead, what I wanted was differential geometry as background for general relativity, partial differential equations for the various applications, but I'd put up with exterior algebra, of course the inverse and implicit function theorems (proved in Fleming's book and an exercise in Rudin's), fluid flow, Maxwell's equations, Schroedinger's equation, Fourier theory, Lebesgue integration, Leibniz's rule (differentiation under the integral sign; there's a simple proof if assume enough continuity and a more general proof from Lebesgue theory), how come, with Lagrange, action is stationary, on and on. I was willing to take arithmetic as okay!
- raldi 4y ago
  Sounds like the crux is that 2(x^2) = y^2 can never be true for any integers x and y because any square of an integer must have an even number of 2's in its prime factorialization, and any doubled square of an integer must have an odd number.