Question

Prove that 1+1=2 (mathematically).

Answer 1

Answer:

there is no proof of this mate

bcoz we r studying this from childhood and it has no proof

1+1=2

stop putting this type of question here

Answer 2

Hey mate!!

here's your answer...

»The proof starts from the Peano Postulates, which define the natural

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.

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.

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:

Then you have to define addition recursively: Def: Let a and b be in N. If b = 1, then define a + b = a'

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

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

•Then you have to define 2:

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.

Proof: Use the first part of the definition of + with a = b = 1. Then 1 + 1 = 1' = 2 Q.E.D.

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

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

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:

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.

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

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

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:

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

Now use the first part of the definition of + on the sum in » parentheses: 1 + 1 = (1)' = 1' = 2

$<marquee>★★Hope it will be helpful for you ☺★★$