Changes

<math>1 + 2 + 3 + ... + n + (n+1)= \frac{n(n+1)}{2} + (n+1)\quad|</math>simplify the right side<br>

<math>1 + 2 + 3 + ... + n + (n+1)= \frac{n(n+1)}{2} + (n+1)\quad|</math>simplify the right side<br>

<math>\Leftrightarrow</math><br>

<math>\Leftrightarrow</math><br>

<math>1 + 2 + 3 + ... + n + (n+1)=\frac{n(n+1)}{2} + \frac{2(n+1)}{2}=\frac{n(n+1)+2(n+1)}{2}=\frac{(n+1)(n+2)}{2}</math>

<math>1 + 2 + 3 + ... + n + (n+1)=\frac{n(n+1)}{2} + \frac{2(n+1)}{2}=\frac{n(n+1)+2(n+1)}{2}=\frac{(n+1)(n+2)}{2}</math></li></ol>

 

Now, we're finished: the hypothesis A holds for all the Natural Numbers.

Now, we're finished: the hypothesis A holds for all the Natural Numbers.

== Transfinite Induction ==

== Transfinite Induction ==

Assume hypothesis A is true for a finite number n or an infinite [[cardinal]] k. Prove it must be true for n+1 (respectively, k+1).

Assume hypothesis A is true for a finite number n or an infinite [[cardinal]] k. Prove it must be true for n+1 (respectively, k+1).

[[Category:Mathematics]]

[[Category:Mathematics]]