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Help me please $f(x)=x+a$, where $a$ is a constant.

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do you want to show that $f(x)$ is primitive recursive? –  Ram Jan 14 '13 at 9:28
    
Yes! Where f(x) is x+a –  user58222 Jan 14 '13 at 9:31
    
Welcome to Math.SE! You could have considered the addition of this link to your question: en.wikipedia.org/wiki/Primitive_recursive_function . In specific the paragraph "Addition" in the page above should answer your question. Is it correct? –  Giovanni De Gaetano Jan 14 '13 at 9:37

1 Answer 1

up vote 1 down vote accepted

First of all I think that your statement is wrong when $a\notin \mathbb{Z}$, and the function you define when $a\notin \mathbb{N}$ should have the domain restricted, because you cannot use the subtraction but only the limited subtraction.

So let us suppose $a\in \mathbb{N}$ for simplicity. If $a=0$ then $f(x)=x$ is the identity function, and this is known to be primitive recursive. Indeed $f(x) = P_1^1(x)$.

Now let us proceed by induction and suppose that $f_n(x)= x+n$ is primitive recursive. By $S$ we denote the successor function $S(k)=k+1$ which is axiomatically primitive recursive. Then $S(f_n(x))= (x+n)+1=f_{n+1}(x)$.

Since the composition of primitive recursive functions is primitive recursive we deduce that $f_{n+1}(x)=x+(n+1)$ is primitive recursive as well. By the induction hypothesis we are done.

Note: This has been brutally copied by Wikipedia: Primitive recursive function.

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Thank you! And there is my solution: (is it correct ?) $$f(x-a)=x-a+a=x$$ is the identity.....and primitive. $$f(x+1)=x+a+1=f(x)+1=S(f(x))$$ –  user58222 Jan 14 '13 at 10:50
    
You are welcome! The problem with your solution seems to me that $f(x-a)$ could be not well defined. In the sense that a primitive recursive function goes from $\mathbb{N}$ to $\mathbb{N}$ and if $x<a$ then $x-a\notin\mathbb{N}$. Moreover you should observe that the composition of primitive recursive functions is primitive recursive. If you want to be sure to have understood a good exercise could be to write $f_n(x)=x+n$ only in terms of projective and successor functions. Lastly, do you know how to accept an answer? –  Giovanni De Gaetano Jan 14 '13 at 11:02
    
i accepted. did i it right? –  user58222 Jan 14 '13 at 12:59
    
Perfect! ;) For your future participation in Math.SE remember that people do care about how many answers did you accept, so keep going! –  Giovanni De Gaetano Jan 14 '13 at 14:41

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