>textbook meant for highschoolers and early undergrads>stuck on the first pagehow long is it supposed to take you to read these things? I've been stuck on the first page of the first proof for an hour and it makes me want to die. I know what the symbols mean on their own but together it's like a bunch of garbled up noise. I've never combed through a textbook on my own before.
>>15464983Post what it is that is confusing you, and 50 of the brightest mathematicians here will solve it for you. Then you rinse and repeat. It never gets easy, you never reach the summit. Embrace the struggle.
>>15465094Elementary Number Theory by Jones & Jones. I don't really care about number theory that much but I have slightly over a month to cram as much of a subject as I can into my head for a summer research project. I need to study something new because my only other math classes are calc i and ii, which I did well in, but everybody seems to insist that you can't do any interesting projects or research with just that knowledge base.
>>15465111Right, but where exactly do you get stuck? Post the very first part that blocks you from proceeding. Remember, you're anonymous so it doesn't matter if your ego finds it embarrassing.
>>15465132for proving If a and b are integers with b > 0, then there is a unique pair of integers q and r such that a = qb + r and 0 =< r < b.
>>15465144oh this is the part right before it
Have you tried doing an example with specific numbers?
>>15465144Well in plain English, that's just the division algorithm. If I want to perform 14/3, then I have 4 remainder 2. Or, 14 = 4*3 + 2. They're demonstrating that the remainder cannot be equal to or larger than the divisor (3 in this case). Of course this makes sense, if you divide by 3, then the remainder can only be 0, 1, or 2.
>>15465144Every non-empty subset of the natural integers has a smallest element. This property is constantly used to prove things about integers. For example, a very widely known proof of "p(0) and forall n: nat, p(n) implies p(n+1)" for some predicate n uses the well-ordering (meaning every non empty subset has a smallest element) of N (sketch: assume p(0) and forall n: nat, p(n) imp p(n+1) is true then let n: nat, consider the set of all integers for which the predicate p is not true. It is a subset of nat, consequently it has a smallest element m, by definition of m, p(m-1) needs to be true, and by the hypothesis since you have p(m-1) you also have p(m) true - contradiction). The key idea here to take away is to reduce the problem set into a NON EMPTY subset of the Natural integers and it will immediately give you a smallest element to work with. Try it yourself
>>15464983>>15465144Don't learn math like that just do actual examples and memorize the process, lolI taught myself Diff Eq by reading a dry boring ass textbook but filtered out the examples provided, and managed to pass the course with an A
