matt grime

Let * denote the group operation. Commutative/Abelian means x*y=y*x for all x and y. Or if you like when you see an expression of the form: x*y*z*w*p*q (which is well defined because of transitivity) then you can rearrange the terms to get it into a better form. Because the most common abelian group is the integers under addition, and there's something nice about abelian groups (we understand them completely) we often make the leap of writing + for * in abelian groups (personally i think this is stupid, but tradition dictates...) It also means that we need to write 0 for the identity and that we write 2a for a+a. So the first question asks you to show that the set of all elements of order 2 plus the identity is a (sub)group. Associativity of the operation comes for free since it is inherited from the larger group. The identity is there too since 0+0=0, and inverses are there as well since every element is its own inverse. Now, at no point have I used the abelian nature of the group. You must show closure (that the composition of two elements of order two has order 2, or is the identity) and that requires the commutativity - it isn't true for non-abelian groups. As for the second one. In an abelian group, if x and y have orders n and m resp, then show that the order of x+y (or x*y) is the least common multiple of n and m.

Let * denote the group operation. Commutative/Abelian means x*y=y*x for all x and y. Or if you like when you see an expression of the form: x*y*z*w*p*q (which is well defined because of transitivity) then you can rearrange the terms to get it into a better form. Because the most common abelian group is the integers under addition, and there's something nice about abelian groups (we understand them completely) we often make the leap of writing + for * in abelian groups (personally i think this is stupid, but tradition dictates...) It also means that we need to write 0 for the identity and that we write 2a for a+a. So the first question asks you to show that the set of all elements of order 2 plus the identity is a (sub)group. Associativity of the operation comes for free since it is inherited from the larger group. The identity is there too since 0+0=0, and inverses are there as well since every element is its own inverse. Now, at no point have I used the abelian nature of the group. You must show closure (that the composition of two elements of order two has order 2, or is the identity) and that requires the commutativity - it isn't true for non-abelian groups. As for the second one. In an abelian group, if x and y have orders n and m resp, then show that the order of x+y (or x*y) is the least common multiple of n and m.

It's not about the difficulty, it's a query about utility. Writing out group tables doesn't really help you do anything that you're not better off doing without them you see. If you have to use tables, have you heard of the latin square principle?

It's not about the difficulty, it's a query about utility. Writing out group tables doesn't really help you do anything that you're not better off doing without them you see. If you have to use tables, have you heard of the latin square principle?

not useful, nor indeed correct since not every element in a field or ring is obtained by adding 1 to itself (only the subring of integers at most is obtained). yes.

You may not add elements to an element in the ideal like that and expect it to work. An ideal is closed only under adding elements *within* the ideal. It is however closed under multiplication by *ANY* element. As you correctly point out, if the ideal is not 0, then it contains some element x, and more importantly every element in a field has a *MULTIPLICATIVE* inverse. Just by adding -r (which you've called f) above you're not even staying in the ideal, nor are you using the fact that the underlying ring is also a field. So your proof is wrong (and in fact would show that no ring had ideals). As for the second question. If R is not a field, then there is some element without a multiplicative inverse. Look at the ideal it generates.

Of course it's not useless (no more so than any other unit). It is analytically (ie in terms of calculus) and geometrically the correct measure of angle. It simply and directly relates angle to arc length and sector area. Degrees are the convenient one - their sole benefit is to divide the circle in to units such that 2,3,4,5,6,8,9,10,12 divide the whole - it serves no other use than this numerical one.

No, a step function is trivially a regulated function.

The convenience is that if the argument is in radians then differentiating sin gives cos, etc. If it were degrees then the derivative of sin would be something like (2pi/360)cos

Yep.

What does Euclid tell us? That given a and p coprime, there are integers x and y with ax+py=1 can you now see how to express b as "a" times something plus something congruent to zero mod p? Note that if all we require is that a is not zero then p must be a prime, but it suffices just for a and p to be coprime.

Firstly, this isn't true unless p is a prime, if it is a prime then say so. The answer as always follows from the hint: all you know (if p is a prime), is that a and p are coprime. Hence all you can do is use the single result at your disposal, ie Euclid's Algorithm. Let us know where that gets you

I imagine the first question means "have the same integral part", not "are the same integer", and there are an infinite number of answers, indeed any number greater than 10 is a solution. 2. perhaps can be solved by rearrangement (eg, severian, jsut because i don't know a or b doesn't mean I can't write down a/b given that 3a=4b).

And what's the full set doron? the set of all sets? so what about russell's paradox? is it because we are restricted by or 0 xor 1 thinking?

It's a finite subset sof the complex numbers closed under multiplication and possessing inverses and identity, hence it is an abelian group. They really asked you to write out the group table? What level course is this?

the possible orders of elements of H are indeed 1,2,5,10, so a^3 has possible orders 1,2,5,10 and hence a has possible order..?

What are the elements of G/H and what are their orders is anything of order 4? What is the order of K? hence the order of G/K (there are only two abelian groups of order 4).

a^3 is an element of H, an element of H has what possible orders?

So, once more, by putting conditions on set theory that no one else does you make ZF inconsistent? That is nonsense since consistency is a purely an internal attribute.

and what's the definition of non-artificial, and hence artificial?

Firstly, this is your interpretation of what symmetry means (with respect to sets and natural numbers), and if it contradicts the usual model of ZF then it is inconsistent with ZF. This would make ZF inconsistent and would be important if you could demonstrate how to deduce your interpretation of symmetry from the axioms of ZF or in some claimed model of ZF. Can you do that?

None of that shows ZF or Cantor to be inconsistent (you understand what that word means?) It shows that if *you* assign meanings to things that directly contradict those of ZF that you get an inconsistent system. That is neither important nor news. You have said there is no absolute truth in mathematics (and there isn't), however if you are insisting that something *must* be always true, which is contradictory behaviour.

the probability of what is 1/7? here's how to use the pigeon hole principle in the example of people and friends. pick one person. of the other five he must either know or not know at least 3 of them by the pigeon hole principle. assume he knows three of them, if any of two those three know each other then we have three common friends, if not then all three of them must be mutual strangers and we are done. (the other option is completely analogous.)

Why should S(n-1) equal S(n)? one is the sum of the first n integers, the other the sum of the first n-1 integers. S(n-1)+n=S(n), fortunately