Base case 1 and 2Q) Prove that if n=m^3-m, where m is an integer, then n is a multiple of 6.
assume n=k
Show k+1
Note that by hypothesis
and because either or is even.
QED
I'm trying to get to grips with mathematical proofs. I think I almost have this, but am missing something.
Q) Prove that if n=m^3-m, where m is an integer, then n is a multiple of 6.
A) A proof by contradiction:
Assume the inverse of the statement, that if n=m^3-m then n is not a multiple of 6.
This implies that n = 6k +{1,2,3,4,5} where k is an integer. In other words, n has a remainder of between 1 and 5 if you try to divide it by 6.
Now I need to show that this implies a contradiction with the original statement (n=m^3-m) for every case, {1,2,3,4,5}.
Case 1: n = 6k+1
By the orignal statement, this is equal to m^3-m. But m^3-m is always even, and 6k+1 is always odd. This is therefore obviously false. The same logic applies to cases 3 and 5.
Case 2: n = 6k+2
Again, by the original statement this is equal to m^3-m. But how do I show that this is false. This time both sides are even?
This is where I get into trouble. I am a bit confused on how to prove that 6k+2=m^3-m leads to an obvious contradiction. I can see that you can disprove it by counter example: If k=1 then n=8, but m^3-m can only take values {0,0,6,24,...}. So in general, the statement is false. In fact, it is only true when the number we add on to 6k is 6, otherwise we always 'miss' the values that m^3-m can take. So I think I can see it in words, but how do I formulate it into a proper proof?
thank you, but although I see that you've written a recurrence relation for n, and tried to show that every term is a multiple of 6, does that really qualify as a proof in the strict mathematical sense? I understood that you must always assume the opposite of the given statement, then show this leads to a obviously false statement, such as 0=1. Otherwise you are just deriving implied relationships, not actually proving anything.
Note that:
That is, the product of 3 consecutive natural numbers. Now, given 2 consecutive numbers one of them must be divisble by 2, and given three consecutive natural numbers one must be divisible by 3. Thus the product is divisble by 6
Or you can remember that