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intsu11midtermsecondtrys

intsu11midtermsecondtrys - Introductory Number Theory...

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Introductory Number Theory. Midterm Exam-Take two–Solutions Note: Every question either appeared on the study guide or is a particular case of a question appearing there. In your proofs, there should be no non sequiturs 1 , irrelevancies, etc. Speculative statements, numerical examples play a role in a classroom presentation, are essential when trying to figure something out, but have no place in a proof you write in an exam or homework for a grade. Finally, if I find anything hard to read, I will assume it is wrong. In other words, write clearly. 1. Prove: There exists an infinity of prime numbers. 2. Prove that the product of any three consecutive numbers is divisible by 6. Solution. This is a fairly obvious result, not much to prove. One can use that because gcd (2 , 3) = 1, a number is divisible by 6 if (and only if) it is divisible by 2 and by 3 (a reason nobody gave 2 ) and of any two consecutive numbers one has to be even, of three consecutive ones, one is divisible by 3. One can justify this by saying that if the numbers are n, n + 1 , n + 2, we can write n = 3 k + r , where r = 0 , or 2. If r = 0, then n is divisible by 3, if r = 1, then n + 2 = 3( k + 1) is divisible by 3, and if r = 2, then n + 1 = 3( k + 1) is divisible by 3. Proofs using Fermat’s little theorem (killing flies with a sledgehammer!) had one general serious flaw. Fermat’s little theorem does NOT say that a p - 1 1 (mod p ) if p is prime; it says this holds if p is prime AND p ̸ | a . However, it is in fact always true that a p a (mod p ), because if p ̸ | a , then it follows from Fermat’s theorem multiplying both sides by a ; if p | a , then a p 0 a (mod p ). So the people who proved that 3 must divide the product of three consecutive numbers by using that n 3 1 (mod 3) used a correct result, though either without proof or with an incorrect proof. But when they then wrote n 1 (mod 2) to somehow conclude from this that n 3 n (mod 2), they committed several mistakes.
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