I am sorry if I make a fool out of myself, and I am really sorry if they are some mistakes over here (it's 4 AM and I need to sleep), but I'll try to do this.
#TIAM objects are closed under + as it is a union of two sets and we are not including any element outside that set. So basically I get a bigger set that includes all the elements of the previous set sans repetition. Now, since I am including all the unique elements of the previous set this means that no new symbol is over there.
However, the last operation (the shift one) isn't one with closure.
#It is commutative as it is the addition of the two sets with an algebraic operation. So it doesn't matter what order in which I add the stuff together. The results always remain the same.
4) The identity element is the same as that in question 3.
5) It's the same question as you've asked me to come up with something that when it's put through a machine that computes the '+' operation with another thing it comes out the same on the other side. In this operator it's limited to this {0, []}. I don't know how I know this, and I don't know if I can prove this but I think that this is the answer.
that's [card({a})+card({}), ({a}U{})>>1], or [1,{b}]
so that can't be the neutral element.
I think that there must not be an identity element, because the >> operator forces a change under any circumstance which is dependent upon TIAM-A's contents.
Similarly (I haven't thought about this enough yet), I suspect that the operation is non-commutative, because unions can change the number of elements (deletion of duplicates), and the shift is involved (which changes elements, as opposed to just combining them). So let's say you shift and then union, you could get a different result than if you union and shift.
I'm gonna see if we can get a concrete example here...
at somewhat random:
TIAM-A (now referred to as A): [0, {a,b,f}]
TIAM-B (now referred to as B): [1, {b,c}]
TIAM-C (now referred to as C): [2, {d,e}]
A+B+C=T1, A+C+B=T2, if T1 != T2, not commutative (pardon me as I riff on notation for ease of writing)
A+B = [0, {a,b,f}] + [1, {b,c}]
A+B = [3+2, (abf U bc)>>3+2]
A+B = [5, (abcf)>>5]
A+B = [5, fghk]
A+B+C = [5,fghk] + [2, de]
A+B+C = [4+2, (fghk U de)>>4+2]
A+B+C = [6, (defghk)>>6]
A+B+C = [6, jklmnq] = T1
then:
A+C = [0, {a,b,f}] + [2, {d,e}]
A+C = [3+2, (abf U de)>>3+2]
A+C = [5, abdef>>5]
A+C = [5, fgijk]
A+C+B = [5, fgijk] + [1, {b,c}]
A+C+B = [5+2, (fgijk U bc)>>5+2]
A+C+B = [7, (bcfgijk)>>7]
A+C+B = [7, ijmnpqr] = T2
T1 = [6, jklmnq], does not equal T2 = [7, ijmnpqr]
hence, non-commutative - but I could have misunderstood something entirely here
Hi, yes, I made it non-conmutative to make it a little bit more difficult, I think you don't get to see a non conmutative algebra in high school, it was a way to take her out of her comfort area.
However you made a sligth mistake with the shifting part:
"The set part of the new object will be the union of the two sets with all the letters shifted as many places as specified by the integer part of the second TIAM object."
I should have done it on paper and worked out it out more extensively. Is there anything else conceptually wrong in my reply?
Also, is there any way at all I can contact you? I would love to learn more from you. If possible can you please drop me a line at yesthisisananonymousid at gmail?
#TIAM objects are closed under + as it is a union of two sets and we are not including any element outside that set. So basically I get a bigger set that includes all the elements of the previous set sans repetition. Now, since I am including all the unique elements of the previous set this means that no new symbol is over there.
However, the last operation (the shift one) isn't one with closure.
#It is commutative as it is the addition of the two sets with an algebraic operation. So it doesn't matter what order in which I add the stuff together. The results always remain the same.
#Yeah,
[0, {}]