You nailed it. This is exactly what I want to do!
Can you suggest some good resources and a sort of rough path for me out of your experience? I know that it may not be possible for you to do so and it is quite okay if you can't do the latter.
Thank you for replying.
On a more serious note: Other people have recommended very good books that will teach you problem solving so I would like to give you a feeling of what "real" math is. In high school and in most University degrees you only do applied math => how to use math to solve this or that problem. The more math you know, the more tools you have to solve your problem. For a given problem: could I solve it with a partial differential equation? with a Bayesian approach? etc. You can see how the more math you know, the wider your search for a solution can be.
However, most people never get to do "pure" math => to create your own new math. When you read a book or attend to a class you only get to see the finished product. When you see a finished building, they have removed all the scaffolding, the cranes, and all the other tools that were used to build it. If you spend some time creating your own new math you will get a very deep understanding of what math is and it will widen your learning abilities.
Of course I don't expect a high school student to create something truly new, this is just a learning exercise. I just want to provide a 'feeling', I won't be formal or rigorous.
Definition 1: An object of class TIAM is composed of an integer and a set. The set can be empty or contain letters from the English alphabet with no duplicates.
Definition 2: Operation "+" This operation will take two TIAM objects and produce a new TIAM object. The integer part of the new object will be the sum of the cardinalities of the sets. 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.
Example 1: [1, {a,b,c}] + [2, {c, d}] = [cardinal({a,b,c}) + cardinal({c,d}), {a,b,c}U{c,d}>>2] = [3+2,{a,b,c,d}>>2] = [5, c,d,e,f]
Example 2: [0,{z}] + [1, {}] = [cardinal({z}) + cardinal({}), {z}U{}>>1] = [1+0,{z}>>1] = [1,{a}]
* TIAM objects are closed under +. (See http://en.wikipedia.org/wiki/Closure_%28mathematics%29). Do you understand why? Can you prove it?
* Is the operation + conmutative or non conmutative? (See http://en.wikipedia.org/wiki/Commutativity) Why? Can you prove that it is / it isn't?
* Can you find a neutral element? (See http://en.wikipedia.org/wiki/Identity_element) I mean, can you find a TIAM object (I will call it TIAM-ZERO) so that TIAM-A + TIAM-ZERO = TIAM-A for any given value of TIAM-A?
Etc.
This is just an example, try to build your own stuff and reason about it. Ask yourself questions about it and answer them. Once you get used to this kind of thinking, you will learn faster and deeper.
#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, {}]
5) Is question 4 different than question 3 and why or why not?
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.
When I think on this, let's say TIAM-A is
[1, {a}], and you add [0, {}]
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
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 think you are shifting a little bit too much.
I interpreted "the second TIAM object" as "the new TIAM object", as opposed to the second of the original objects. Thank you for the clarification :)
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?