Question

Prove that 1+1=2.





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Answer 1

Hey,

Mate Here's the solution....

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:


Definition : 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





Or..


Note that while 1 + 1 = 2 can be false if you use

different axioms, 1 + 1 = 2 is not itself an axiom.

So using the Peano Axioms (which are usually

implied unless otherwise stated), 1 + 1 = 2 can

only be true since it necessarily follows from

the axioms.

Answer 2

Proved☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️☺️