# After calculus first pure math course, having a hard time.

by chimath35
Tags: calculus, math, pure, time
 Sci Advisor PF Gold P: 9,377 My long standing position is number theory is extremely useful in calculus. I had the same problem in my undergrad days. It caused me ... difficulty.
P: 87
 Quote by crownedbishop Proof by contradiction sounds like a great way to attack some of the problems you listed. Unless you're a constructivist. In number theory, induction is a very important tool too. For the first one, why can't a divide b if a>b?
beacause it would be less than one, not a positive integer
 P: 87 So do you think if I take discrete math concurrently, it will help?
 P: 87 Also if you are bored, here is a number theory problem: An integer is divisible by 9 if and only if the sum of its digits is divisible by 9 Proof by induction? I am stuck ac=b 9c=b ex. 9c=81 so c is an int. thus divisible let b= d+dn so 9c=d+dn then 9c=d+dn+dn+9
P: 1,622
 An integer is divisible by 9 if and only if the sum of its digits is divisible by 9
Write n = ∑ak10k and notice that n is divisible by 9 if and only if n ≡ 0 (mod 9). So this means n is divisible by 9 if and only if ∑ak ≡ 0 (mod 9) and the result follows.
P: 81
 Quote by chimath35 Also if you are bored, here is a number theory problem: An integer is divisible by 9 if and only if the sum of its digits is divisible by 9 Proof by induction? I am stuck ac=b 9c=b ex. 9c=81 so c is an int. thus divisible let b= d+dn so 9c=d+dn then 9c=d+dn+dn+9
 Quote by jgens Write n = ∑ak10k and notice that n is divisible by 9 if and only if n ≡ 0 (mod 9). So this means n is divisible by 9 if and only if ∑ak ≡ 0 (mod 9) and the result follows.
Hi jgens, I am pretty sure that your sketch of the proof is very elegant and efficient, but I feel "mod" is an unnecessary artifice for somebody trying to understand (and like) this type of pure math course.

I have something which is a little bit simpler to handle for a novice:

Suppose k is a positive integer. Then k = an10n + an-110n-1 + ... + a3103 + a2102 + a1101 + a0, for some non-negative integers n, an, an-1, ..., a3, a2, a1, a0.
(Note that the ai's are the digits of k.)

Dividing the positive integer k by 9 yields the following equation:
k/9 = (an10n + an-110n-1 + ... + a3103 + a2102 + a1101 + a0)/9.

The above equation can be re-written as follows:
k/9 = [an(10n -1 +1) + an-1(10n-1 -1 +1) + ... + a3(103 -1 +1) + a2(102 -1 +1) + a1(101 -1 +1) + a0]/9.

It follows that
k/9 = [an(10n -1) + an-1(10n-1 -1) + ... + a3(103 -1) + a2(102 -1) + a1(101 -1)]/9 + [ an+ an-1 + ... + a3 + a2 + a1 + a0]/9.

Clearly, the first term on the right hand-side of the above equation is an integer since an(10n -1) + an-1(10n-1 -1) + ... + a3(103 -1) + a2(102 -1) + a1(101 -1) is divisible by 9.
(Note how each (10i -1) contains only 9's as digits, so each (10i -1) is divisible by 9.)

Now suppose that k is not divisible by 9. Then k/9 is not an integer. It follows that [ an+ an-1 + ... + a3 + a2 + a1]/9 is not an integer, so the sum of the digits of k is not divisible by 9. Thus, if the sum of the digits of k is divisible by 9, k is divisible by 9.

Next suppose the sum of the digits of k is not divisible by 9. Then [ an+ an-1 + ... + a3 + a2 + a1]/9 is not an integer. It follows that k/9 is not an integer, and hence k is not divisible by 9. Thus, if k is divisible by 9, the sum of its digits is divisible by 9.

Therefore, we conclude that integer k is divisible by 9 if and only if the sum of its digits is divisible by 9. This completes the proof.
 P: 87 Ya I think I may just change the class from credit to audit; considering I have not taken an intro to proofs class, and the fact that my gpa is almost a 4. My uni. does not offer proof intro this sem. I think this way I will still build up my mathematical thinking abstractly and with proofs without the consequence of possibly tanking my gpa. My professor also just states theorems and then proves them. I am not sure if that is typical or not? The way that I learn the best I think is seeing the theorem and then doing examples of similar problems. Not just stating a theorem and then taking almost the entire lecture time to be complete and prove the theorem; I don't think him being complete and proving the theorems helps with homework or test problems.
P: 81
 Quote by chimath35 The way that I learn the best I think is seeing the theorem and then doing examples of similar problems. Not just stating a theorem and then taking almost the entire lecture time to be complete and prove the theorem; I don't think him being complete and proving the theorems helps with homework or test problems.
I would advise you to be careful and not to adopt such an attitude. Think about it for a second: you posted some of your homework problems, and evidently, all of them involved proofs. Given that you don't have much experience with proof-writing, I believe that it would be wiser to carefully follow the proofs (and especially, the proof strategies that he uses) of your professor in the lectures, and then try to apply those similar proof strategies on your homework problems.

Concerning your decision to audit the course instead of taking it for a credit, I think that was a smart move on your part. However, you should still try to make the most out of this course (especially about proof-writing) because most if not all of your future courses might be heavily proof-based.

Another important thing: luck seems to be smiling at you, as you'll have this summer before your next semester at college. It would be in your interest to pick up a book on proof-writing (or if you will have already mastered proof-writing by summer, pick up another pure math text like introductory real analysis or abstract algebra), and work through it. Trust me, that's the best way to really "get" proofs if you are not a "natural" (it worked out pretty well for me).

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