Thread: 1+1=2 ?

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

Originally Posted by Kaloda

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

I like your post title!

Under the reals we have the distributive property and also commuatitivity of multiplication, thus n+n=n(1+1)=n*2=2n.

No, assume that 1+1 is not equal to 2.

Originally Posted by Kaloda

No, assume that 1+1 is not equal to 2.

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.)

Originally Posted by Kaloda

No, assume that 1+1 is not equal to 2.

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.

Originally Posted by Kaloda

No, assume that 1+1 is not equal to 2.

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 "+"?