Show that n+n=2n for every real number n.

Results 1 to 6 of 6

- September 22nd 2010, 08:40 PM #1

- September 22nd 2010, 08:46 PM #2

- September 22nd 2010, 09:18 PM #3

- September 22nd 2010, 09:30 PM #4
I don't follow. You say we are in reals, where 1+1 does equal 2. So either we're not actually in the reals, or you're suggesting making a proof that is not sound because it's based on a false premise. (In fact, by assuming a false premise one can logically "prove" any statement whatsoever.)

- September 22nd 2010, 11:05 PM #5
Principia Mathematica - Wikipedia, the free encyclopedia

Perhaps you should use their work as a guide.

(From your subsequent posts, your question almost certainly does not belong in the Pre-Algebra and Algebra subforum). It is impossible to know what level of proof you require. Unless you can make this crystal clear, I don't see the point in posting help with your question.

- September 23rd 2010, 01:48 AM #6

- Joined
- Apr 2005
- Posts
- 15,712
- Thanks
- 1472

Then you

**cannot**prove that n+ n= 2n for all n because it is not true for n= 1. Do you mean "Do not assume that 1+ 1= 2" (which is quite different from "assume that 1+ 1 is not equal to 2").

In Peano's axioms for the natural numbers we have the "successor function", s(n). Addition is defined by

a) n+ 1= s(n) and

b) if then there exist p such that m= s(p) and then n+ m= s(n+ p).

While 2 is**defined**as s(1) so it follows that 1+ 1= s(1)= 2.

If you don't like that then:

1) How are you**defining**"1"?

2) How are you**defining**"2"?

3) How are you**defining**"+"?