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whats 1+1?

 Nov 25, 2019


whats 1+1?       


To solve this, we must count on our fingers.


Hold up a closed hand.

Say "one..."         and raise your index (first) finger.  

Say "plus one..."  and raise your middle (second) finger.


Examine your hand which has the first two fingers raised.


In a voice full of joyous discovery, shout "Bunny!"    


 Nov 26, 2019

I am hoping if you have the knowledge to type words on a computer you know that 1+1=2 

but here is an interesting proof I found for this.


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.

 Nov 26, 2019


I believed that 1+1=2 until I read this.  Now, I'm not so certain.  SMH.


Guest Nov 26, 2019

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