# Another counting numbers proof

• Mar 10th 2009, 08:04 PM
noles2188
Another counting numbers proof
1) For each counting number n let Sn := "Sigma(sub n) = (n(n+1))/2.
Prove that if n is a counting number, then Sn is true.
(The definition of counting number that we are using is meant to be a number in the minimal induction set. By the minimal induction set is meant an induction set no proper subset of which is an induction set.)
• Mar 11th 2009, 12:21 AM
Counting number proof
Hello noles2188
Quote:

Originally Posted by noles2188
1) For each counting number n let Sn := "Sigma(sub n) = (n(n+1))/2.
Prove that if n is a counting number, then Sn is true.
(The definition of counting number that we are using is meant to be a number in the minimal induction set. By the minimal induction set is meant an induction set no proper subset of which is an induction set.)

I'm not sure whether this proof conforms to your definition of a counting number (which I haven't seen before), but the standard 'induction' proof of this is:
Let $P(n)$ be the propositional function: $S_n =\sum_{i=1}^ni =\tfrac{1}{2}n(n+1)$.

Then $P(n) \Rightarrow S_n+(n+1) = \tfrac{1}{2}n(n+1) +(n+1)$

$= (n+1)(\tfrac{1}{2}n +1)$

$=\tfrac{1}{2}(n+1)(n+2)$

$\Rightarrow P(n+1)$

Now $P(1)$ is: $S_1 = 1 = \tfrac{1}{2}\cdot 1\cdot 2$, which is true.

So by induction, $P(n)$ is true for all $n \in \mathbb{N}$