Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Bladito****Member**- Registered: 2009-10-07
- Posts: 1

Theorem: All positive integers are equal.

Proof: Sufficient to show that for any two positive integers, A and B, A = B.

Further, it is sufficient to show that for all N > 0, if A and B (positive integers) satisfy (MAX(A, B) = N) then A = B.

Proceed by induction.

If N = 1, then A and B, being positive integers, must both be 1. So A = B.

Assume that the theorem is true for some value k. Take A and B with MAX(A, B) = k+1. Then MAX((A-1), (B-1)) = k. And hence (A-1) = (B-1). Consequently, A = B.

Anyone got a solution? Cuz I have no idea

Offline

**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi Bladito;

Bladito wrote:

Assume that the theorem is true for some value k.

One problem is right there, you can't assume that. You can't find any k where that is true to start the inductive process. For any value of k > 1, I will always have the counterexample Max( A = k, B = k-1) = k for k >1, clearly A > B. This holds for all k > 1. In other words the theorem doesn't hold for 2,3,4,5 ... Max(A,B) = k does not imply A = B = k.

*Last edited by bobbym (2009-10-07 03:54:13)*

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

Offline

**Avon****Member**- Registered: 2007-06-28
- Posts: 80

bobbym wrote:

Bladito wrote:Assume that the theorem is true for some value k.

One problem is right there, you can't assume that. You can't find any k where that is true to start the inductive process.

The statement is true when k=1 as Bladito has stated. You seem to agree.

Bladito, the problem is that A-1 and B-1 are not necessarily both positive integers.

Offline

Pages: **1**