To all the lovely people commenting on my previous post (highest number of comments on any post in 4 years, thank you): You are assuming that that Hayfever, cold, and suck are all in the Ring of Integers. There may well be Rings where the equation 2x = x^2 has many non-obvious solutions – I’m ill so can’t really face thinking about some examples (or maybe there are only ever two solutions in any non-trivial ring – there’s an exercise for the interested).
I guess what might have more clearly expressed what I was thinking was that the function f : {hayfever, cold, plus enough other stuff to make an additive group} -> N (natural numbers representing units in some metric on the space of suckitude) is quadratic rather than linear. But, you know, I was ill and grumpy so tried to say that in a more satirical way (maths satire, just what the world needs). But again I’m not exactly firing on all cylinders so please be gentle.
Anyway I’m going to start thinking up some more random algebraic posts and start cashing in the Adwords revenue.
x=0 will be a solution for any ring, by the definition of the multiplicative identity.
It’s certainly possible to construct rings in which this is the only solution, Z_2 being an almost-trivial example (or Z_2*Z_2, etc).
A ring Z_n will have (at least?) three solutions if n is a multiple of two, and n is greater than 4. So for example, in Z_8, 0 and 2 are solutions as before, but also 2*4 = 8 = 0 = 4*4.
You could generate rings with more solutions by looking at product rings formed by various numbers of suitable Z_n rings “crossed” together.
An interesting question though is whether there are any infinite rings with anything other than two solutions…
— Tristan Jun 16, 11:40 pm #
Gah, I meant additive identity of course, not multiplicative.
— Tristan Jun 16, 11:58 pm #
Well, the equation 2x=x^2 is polynomial, and can be factored as x(x-2)=0, where we see that 0 and 2 are solutions (if they’re distinct…).
If you’re working in an integral ring, those are the only ones. If not, then you can have more solutions. As an example, 4 is a solution too in the ring of integers modulo 8, and 7 in the ring of integers modulo 35…
— Snark Jun 17, 06:14 am #
2x = x^2 has two real solutions: 0 and 2. If you are in the ring of functions from R to R (with the pointwise addition and multiplication), then any function whose image is included in {0, 2} will fit. So there are infinitely many solutions, including the function always equal to 0, the function always equal to 2, and for instance the function f(x) = 0 iff E(x) is even, 2 iff odd, where E(x) is the integral part of x.
If you impose further that the function be continuous, then the only solutions remaining are the constant functions 0 and 2.
Does this example suit you ?
— _FrnchFrgg_ Jun 17, 10:00 am #