Induction to demonstrate n-1 is a natural number

  • Thread starter Thread starter chaotixmonjuish
  • Start date Start date
  • Tags Tags
    Induction Natural
chaotixmonjuish
Messages
284
Reaction score
0
The question is as follows: Using an induction argument if n > 1 is a natural number then n-1 is a natural number.

P(n)=n-1 such that n-1 is a natural number

Following the steps:

Base case: n=2, P(2)=1 which is a natural number.

We fix a natural number n and assume that P(n)=n-1 is true.

So P(n+1)=(n+1)-1=n. We choose n to be a natural number, therefore this is true.

Is this proof complete? It seems rather...light.
 
Physics news on Phys.org
Seems pretty solid to me ;-)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top