# Recent content by Martin Rattigan

1. ### B Equation with modulus

As I said in parentheses. And yes, I want the division performed in the integers mod 2. I assumed from OP's inclusion of "mod 2" in his equation that that was exactly the problem he wanted solved. The expression ##\frac{200}{15x}## is undefined in the ring of integers for any integral ##x##...
2. ### B Equation with modulus

The problem with the operator you describe is that its not very useful mathematically if it's applied to the real numbers, because it doesn't necessarily respect arithmetic in the reals. E.g. <p id="a"> <script> function m2(exp){document.getElementById("a").innerText += (exp)%2+"\n"}...
3. ### B Equation with modulus

But really the whole usefulness of arithmetic modulo n is that the canonical isomorphism from \mathbb{Z} to \mathbb{Z}/n\mathbb{Z} allows you to ignore the ks. There is only one value\mod 2 that satisfies OP's equation as stated in my original post.
4. ### B Equation with modulus

But if you run <p id="a"> <script> function m2(exp){document.getElementById("a").innerText += (exp)%2+"\n"} m2(2) m2(3.141) m2(-1) m2(-5.5) m2(200/(15*1)) </script> in e.g. Chrome, javascript gives: 0 1.141 -1 -1.5 1.333333333333334 You can't rely on different computer programs to give the...
5. ### B Equation with modulus

The solution is x=1\mod 2. If x=0\mod 2 then \frac{200}{15x} is undefined\mod 2 (so not a solution according to Russell's theory of definite descriptions). This assumes you are looking for solutions in the field of integers \mod 2, so there are but two possible solutions. The equation reduces...
6. ### B Are parabolas ellipses?

The answer is almost. Look at the focus/directrix definition of conics http://mathworld.wolfram.com/ConicSection.html. Parabolas are defined as those conics with eccentricity 1. Anything below and it's defined to be an ellipse. Anything above and it's defined to be a hyperbola. It's easy to...
7. ### I When is D_{n} abelian? What's wrong with the proof?

##e,\ (ab),\ (cd)## and ##(ab)(cd)## gives an Abelian group that contains two-cycles from more than two elements and that are disjoint.
8. ### I Confused about Dedekind cuts

That it still wrong on two counts. The real numbers are the cuts, which are pairs of sets. None of them appear in A or B for any cut. The cut is a rational real number if it is the image of some ##q## in the set ##\mathbb{Q}## of rationals from which the reals are constructed under the mapping...
9. ### I Dedekind Cuts

stevendaryl wrote: The real associated with the pair ##L,R## is the unique number ##r## that is greater than or equal to every element of ##L## and less than or equal to every element of ##R## Since Dedekind cuts are used to construct the reals I think it would be better to say that the real...
10. ### I Proving an inverse of a groupoid is unique

Stephen Tashi wrote: "If we take the definition of monoid and remove the requirement that it be associative then we create a definition of a new algebraic structure." I propose the the name oneoid after J. Milton Hayes.
11. ### I Proving an inverse of a groupoid is unique

If you mean a two sided identity how could it fail to be unique? 1a1b=???.
12. ### I Monte Hall Problem confusion?

Take the goat. Cheaper to run.
13. ### I Partitions of Euclidean space, cubic lattice, convex sets

No it's not homework (bit old for that). I was doing some number theory revision a couple of years ago and the question occurred to me after reading Minkowski's theorem on convex regions symmetric about a lattice point (Hardy and Wright 3.9/3.10), having noticed that Theorem 38, used in the...
14. ### B Can someone solve this exponential equation for me?

I was actually playing Devil's advocate to point out that there was something missing from the original logic. I did prove that \lim_{n \to \omega}(\sqrt{2} \uparrow\uparrow n)=2 without establishing the convergence range you give. The solution x=-\sqrt{2} also needs to be discounted.
15. ### B Can someone solve this exponential equation for me?

The "trick" mentioned by Dr. Tom does in this instance give a correct solution, but it's not a proof. As mentioned by Peter Winkler in his book "Mathematical Puzzles" one has only to consider the similar equation x^{x^{x^{.^{.^.}}}}=4 Using the same logic gives x^4=4, i.e. x=\sqrt{2}. If the...