- #1

mumatics

- 2

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I'm a high school student. I gave a proof for the following theorem, but I was told by some professors that this is trivial and using natural induction twice for the rationals will do the same thing. What do you think? Is it just redundant?

Theorem:

Let P(r) be a statement about r, then if :

1) P(1) is true and,

2) [tex]\forall[/tex] m,n [tex]\in[/tex] [tex]N[/tex] , m[tex]\geq[/tex]n ; P([tex]\frac{m}{n}[/tex])[tex]\rightarrow[/tex] P([tex]\frac{m+1}{n}[/tex])

Then [tex]\forall[/tex] r[tex]\in[/tex] Q, r[tex]\geq[/tex]1 ; P(r).

(PS: I apologize for my (probable) mistakes, because I'm neither an English speaker nor familiar with Latex.)