Science:Math Exam Resources/Courses/MATH312/December 2008/Question 06 (c)
{{#incat:MER QGQ flag|{{#incat:MER QGH flag|{{#incat:MER QGS flag|}}}}}}
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q1 (e) • Q1 (f) • Q2 (a) i • Q2 (a) ii • Q2 (a) iii • Q2 (a) iv • Q2 (b) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q5 (a) • Q5 (b) • Q5 (c) • Q6 (a) • Q6 (b) • Q6 (c) • Q6 (d) •
Question 06 (c) |
|---|
|
The purpose of this problem is to prove the following tThe purpose of this problem is to prove the following theorem. Theorem 1. For all positive integers we have Let be positive integers. For , the theorem holds trivially, so we assume from now on that and write its prime-power factorization as for different primes and positive integer exponents and some positive integer k. Let and focus on the prime power in the prime-power factorization of m. (c) Prove that |
|
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? |
|
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! |
Hint |
|---|
|
Split this problem up into the two cases as mentioned in parts (a) and (b). |
|
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
|
Solution |
|---|
|
We proceed in cases. Case 1: . In this case, part (a) states that . Since , the element above has an inverse showing that . Multiplying both sides by yields . which is what we wanted to show in this case. Case 2: . In part (b), we showed that
Since
We also have that
and hence that . |
{{#incat:MER CT flag||
}}
