Talk:High School Mathematics Extensions/Primes/Solutions

In the Division and Inverses section exercise 3 I don't know how to explain properly that 5n can never be divided by 3 without using the gcd. I hope somebody else can find a good explanation for this.Mmartin


 * Actually it can be done. 1/3 = 3-1 = 4, as 3 * 4 = 12 = 1. Xiaodai 00:13, 22 Aug 2004 (UTC)


 * But shouldn't it be 3-1 instead of 1\3. Because If I just calculate the value the outcome is 190904339063484856909638312798434267580915957296383567154407501220703125/3 which can't be reduced modulo because it's not a whole number. Mmartin 08:04, 22 Aug 2004 (UTC)


 * I think you may have been confused. as you know 1/3 does not exist mod n, only 0,1,2,3...,n-1 does, 1/3 It is used to denote 3-1 (the inverse of 3). We can't really do that ("190904339063484856909638312798434267580915957296383567154407501220703125/3") because we are in mod modular arithmetic and when we make the jump to real number arithmetic, we can't just carry the 1/3 across. Xiaodai 11:58, 22 Aug 2004 (UTC)


 * I understand that but you said in the text "From now on, we will use x-1 to denote the inverse of x if it exists." And I am still used to the fact that you can't really do division in modular arithmetic which was taught to me in a cryptography class. Although the inverse of a number exists as long as gcd(number,modulo)=1. Mmartin 13:54, 22 Aug 2004 (UTC)