
#19
Feb1714, 08:50 PM

Sci Advisor
PF Gold
P: 9,185

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.




#22
Feb1814, 11:22 PM

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 



#23
Feb1814, 11:32 PM

P: 1,623





#24
Feb1914, 04:31 PM

P: 81

I have something which is a little bit simpler to handle for a novice: Suppose k is a positive integer. Then k = a_{n}10^{n} + a_{n1}10^{n1} + ... + a_{3}10^{3} + a_{2}10^{2} + a_{1}10^{1} + a_{0}, for some nonnegative integers n, a_{n}, a_{n1}, ..., a_{3}, a_{2}, a_{1}, a_{0}. (Note that the a_{i}'s are the digits of k.) Dividing the positive integer k by 9 yields the following equation: k/9 = (a_{n}10^{n} + a_{n1}10^{n1} + ... + a_{3}10^{3} + a_{2}10^{2} + a_{1}10^{1} + a_{0})/9. The above equation can be rewritten as follows: k/9 = [a_{n}(10^{n} 1 +1) + a_{n1}(10^{n1} 1 +1) + ... + a_{3}(10^{3} 1 +1) + a_{2}(10^{2} 1 +1) + a_{1}(10^{1} 1 +1) + a_{0}]/9. It follows that k/9 = [a_{n}(10^{n} 1) + a_{n1}(10^{n1} 1) + ... + a_{3}(10^{3} 1) + a_{2}(10^{2} 1) + a_{1}(10^{1} 1)]/9 + [ a_{n}+ a_{n1} + ... + a_{3} + a_{2} + a_{1} + a_{0}]/9. Clearly, the first term on the right handside of the above equation is an integer since a_{n}(10^{n} 1) + a_{n1}(10^{n1} 1) + ... + a_{3}(10^{3} 1) + a_{2}(10^{2} 1) + a_{1}(10^{1} 1) is divisible by 9. (Note how each (10^{i} 1) contains only 9's as digits, so each (10^{i} 1) is divisible by 9.) Now suppose that k is not divisible by 9. Then k/9 is not an integer. It follows that [ a_{n}+ a_{n1} + ... + a_{3} + a_{2} + a_{1}]/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 [ a_{n}+ a_{n1} + ... + a_{3} + a_{2} + a_{1}]/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. 



#25
Feb2114, 08:00 PM

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.




#26
Feb2214, 12:17 AM

P: 81

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 proofwriting) because most if not all of your future courses might be heavily proofbased. 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 proofwriting (or if you will have already mastered proofwriting 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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