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tinynerdi January 5th, 2011 12:14 AM

addition is well-defined
 
define addition of natural number as follows:
a) for every x, define x+1=x'
b)for every x and y, defined x+y'=(x+y)'
show that this addition is well-defined. That is, show that for all x and w, the value of x+w is well-defined. Hint: you may want to used induction.

My question is what does it mean to be well-defined in this problem?

DLowry January 5th, 2011 02:13 AM

Re: addition is well-defined
 
One would expect this operation to work on all natural numbers, and for every natural number to appear as a result of this operation. Further, only natural numbers should appear.

dman315 January 5th, 2011 07:50 AM

Re: addition is well-defined
 
Unfortunately, induction will not work for this proof. It is a common mistake that I have seen tried a lot.

tinynerdi January 5th, 2011 12:49 PM

Re: addition is well-defined
 
Let M be the set of x.
can you set x=1 and w=y' ? so that
x=1 => x+1=x'=1' and x+(y')= (y')'= (x+y)' this implies that 1 belong to M.

dman315 January 5th, 2011 08:39 PM

Re: addition is well-defined
 
Not sure what you are trying to do.

Hooman January 10th, 2011 12:51 PM

Re: addition is well-defined
 
Dman is actually right. What you need here is recursion theorem on omega. In general, this theorem states that for any set A, an element a in A, and given f:A-->A, there exist a unique h:omega-->A s.t
h(0)=a
h(n')= f(h(n)) for all n in omega.
Then by recursion theorem we may define for all m in omega
Am:omega-->omega
Am(0)=m
Am(n')= Am(n)'
Then let
+={((m,n),p)| m and n are elements of omega ^ p= Am(n)}


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