\(2+2=4\)
I might be wrong here, I probably need one of the mods to check my work...
-Vinculum
Also,
"Given that all the numbers from 1 to N exist, the number N+1 is defined as -
N+1=N∪{N}N+1=N∪{N}
In other words we can say that the operator “+1” does nothing but unions the largest set smaller than the set we are operating upon with the set containing this largest set smaller than the set we were operating upon. (This is just the definition of the equation above :P )
We define 0 as an empty set,
0={}=Φ0={}=Φ
Now we recursively define the next numbers -
1=0∪{0}=Φ∪{0}={0}1=0∪{0}=Φ∪{0}={0}
2=1∪{1}={0}∪{1}={0,1}2=1∪{1}={0}∪{1}={0,1}
3=2∪{2}={0,1}∪{1}={0,1,2}3=2∪{2}={0,1}∪{1}={0,1,2}
4=3∪{3}={0,1,2}∪{3}={0,1,2,3}4=3∪{3}={0,1,2}∪{3}={0,1,2,3}
and so on…….
We can actually this way prove that - N={0,1,2,3,...,N−1}N={0,1,2,3,...,N−1}
Now N+2=N+(1+1)=(N+1)+1=(N+1)∪{(N+1)}={0,1,2,3...,N−1}∪{N}∪{(N+1)}={0,1,2,3...,N+1}N+2=N+(1+1)=(N+1)+1=(N+1)∪{(N+1)}={0,1,2,3...,N−1}∪{N}∪{(N+1)}={0,1,2,3...,N+1}
A good question would be why did the operator “+2” become equal to “+(1+1)”, this is simply because we defined 2 in that way. Also why did the brackets associate? Simply because we are dealing with sets(that too without any non-trivial operators), we can do that. ;)
Now we just plug in N=2 and Voila!!
2+2={0,1,2,3}=42+2={0,1,2,3}=4
PS. Sorry for being so long, but this is how we do it."
