hello gurus in the house. I wonder why 1+1 would equals 2. though that's what we were thought. And i think this is just someone's idea, probably a whit man. can someone explain and prove to me that 1+1 = 11

Thanks for your response.

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«Home# Why Is 1+1 = 2,can Someone Prove 1+1=11?

hello gurus in the house. I wonder why 1+1 would equals 2. though that's what we were thought. And i think this is just someone's idea, probably a whit man. can someone explain and prove to me that 1+1 = 11

Thanks for your response.

3 answers

natural numbers existed even before the creation of the world and mankind, Man only discovers them as time goes on, In the ancient time peoples tried to keep reord of that they own. That was when people started counting with fingers and toes, even with marks on the wall. There was no education then bt there was something we call informal education. i.e if a man has 1 goat he will make 1 mark on the wall. If he has 2 he will make 2(i.e 1'1).

_____________________The proof starts from the Peano Postulates, which define the natural

numbers N. N is the smallest set satisfying these postulates:________________

P1. 1 is in N.

P2. If x is in N, then its "successor" x' is in N.

P3. There is no x such that x' = 1.

P4. If x isn't 1, then there is a y in N such that y' = x.

P5. If S is a subset of N, 1 is in S, and the implication

(x in S => x' in S) holds, then S = N.

Then you have to define addition recursively:

Def: Let a and b be in N. If b = 1, then define a + b = a'

(using P1 and P2). If b isn't 1, then let c' = b, with c in N

(using P4), and define a + b = (a + c)'.

Then you have to define 2:

Def: 2 = 1'

2 is in N by P1, P2, and the definition of 2.

Theorem: 1 + 1 = 2

Proof: Use the first part of the definition of + with a = b = 1.

Then 1 + 1 = 1' = 2 Q.E.D.

Note: There is an alternate formulation of the Peano Postulates which

replaces 1 with 0 in P1, P3, P4, and P5. Then you have to change the

definition of addition to this:

Def: Let a and b be in N. If b = 0, then define a + b = a.

If b isn't 0, then let c' = b, with c in N, and define

a + b = (a + c)'.

You also have to define 1 = 0', and 2 = 1'. Then the proof of the

Theorem above is a little different:

Proof: Use the second part of the definition of + first:

1 + 1 = (1 + 0)'

Now use the first part of the definition of + on the sum in

parentheses: 1 + 1 = (1)' = 1' = 2 Q.E.D.

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