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 January 4th, 2011, 11:14 PM #1 Senior Member   Joined: Nov 2009 Posts: 129 Thanks: 0 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?
 January 5th, 2011, 01:13 AM #2 Senior Member   Joined: Nov 2010 Posts: 502 Thanks: 0 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.
 January 5th, 2011, 06:50 AM #3 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 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.
 January 5th, 2011, 11:49 AM #4 Senior Member   Joined: Nov 2009 Posts: 129 Thanks: 0 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.
 January 5th, 2011, 07:39 PM #5 Senior Member   Joined: Dec 2008 Posts: 306 Thanks: 0 Re: addition is well-defined Not sure what you are trying to do.
 January 10th, 2011, 11:51 AM #6 Senior Member   Joined: Jan 2011 Posts: 106 Thanks: 0 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 hmega-->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 Ammega-->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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### well defined problem and not well defined math problem

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