Hello,

Can you help me with proving for integer a,b :

a+b integer

a.b integer

I posted this on number theory but apparently it belongs to set theory

thank you

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- Jul 22nd 2007, 06:55 AMDiliBasics in integers
Hello,

Can you help me with proving for integer a,b :

a+b integer

a.b integer

I posted this on number theory but apparently it belongs to set theory

thank you - Jul 22nd 2007, 08:04 AMCaptainBlack
- Jul 22nd 2007, 09:10 AMtukeywilliams
This is how I would do it:

The successor function is defined as $\displaystyle s: \mathbb{Z}^{+} \rightarrow \mathbb{Z}^{+} $ as $\displaystyle s(n) = n+1 $ for $\displaystyle n \in \mathbb{Z}^{+} $ (for positive integers).

So by the Peano postulates, $\displaystyle a+b $ and $\displaystyle ab $ should be integers.

$\displaystyle a+1 = s(a) $, and $\displaystyle a + s(k) = s(a+k) $.

$\displaystyle a \times 1 = a $ and $\displaystyle a \times s(k) = a \times k +a $.

Correct me if I am wrong. - Jul 22nd 2007, 09:27 AMDili
The integer was defined using the set theory (real number inductive set properties)

I was cofused because it was given with questions related to prime numbers

Can it be proven in either way?

thank you - Jul 22nd 2007, 09:51 AMtukeywilliams
I think so. I believe you have to use the fundamental theorem of arithmetic for other other method of proof. So write out the unique prime factors of $\displaystyle a,b $ and then add them and multiply them. If you get a unique set of prime factors for each operation then you will have proved that $\displaystyle a+b, ab \in \mathbb{Z} $.

- Jul 22nd 2007, 10:08 AMDili
Thank you tukeywilliams. Any replies from others are welcomed even using the set theory properties. (but peano postulates were not defined in the class work we have done)