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##...
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"}...
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.
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...
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...
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...
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...
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...
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.
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...
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.
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...
If the Euclidean plane is partitioned into convex sets each of area A in such a way that each contains exactly one vertex of a unit square lattice and this vertex is in its interior, is it true that A must be at least 1/2?
If not what is the greatest lower bound for A?
The analogous greatest...
This assumes that a shuffle is choosing one of the 354! orderings at random (analogous to shuffling a pack of cards) when mfb's comments are obviously the way to do it. The remarks about calculating the probability of not happening apply when the songs are chosen at random from the whole set for...
The easiest way to be convinced is to assume that the game becomes so successful they turn it into a spectacular and have 1,000,000 boxes instead of 3, one of which contains a million dollars and the other 999,999 bananas. After the contestant makes his choice the host opens 999,998 doors to...
To avoid spreading confusion it might be worth pointing out that this is not what the well ordering theorem states. It is equivalent to the weaker statement that every set can be totally ordered.
The sequences g1,g2,...,gp (if any) where g1g2...gp=e and it isn't the case that g1=g2=...=gp fall into sets of p that are cyclic permutations of each other, because if g1g2...gp=e then g2g3...gpg1=e etc.
If we remove these from A, since |A| = pk that leaves some multiple m=rp of p sequences...
Then again it doesn't seem unlikely. Has it been proved?
The non-abelian groups of orders 2n will probably arrange it by themselves (e.g. there are over ten times as many non-abelian groups of order 1024 as there are other groups up to and including 1024) so this may not be too difficult to prove.
There are as many cyclic groups (\aleph_0) as finite groups. By, "almost all finite groups are non-abelian", do you mean \lim_{n\rightarrow \infty}a_n/g_n=0, where a_n is the number of abelian groups of order \leq n and g_n is the number of groups of order \leq n? (This doesn't seem likely.)
If you want to move the whole parabola you can amend it as follows, but you may be better off writing something more general if you find you need any further extensions.
<script>
function interpol(p1,p2,p3,parm){
for(i=1;i<=3;i++)eval('x'+i+'='+'p'+i+'[0];y'+i+'='+'p'+i+'[1];')...
You pretty much explained it yourself in the second paragraph.
If S is a set of cabbages then \mathcal{P}(S)=\{O:O\subset S\} is a topology on S. It is unnecessary to define a distance function d_S(b,c) between cabbages b,c\in S in order to define the topology \mathcal{P}(S). If no such...