# Basics in integers

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• Jul 22nd 2007, 06:55 AM
Dili
Basics 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 AM
CaptainBlack
Quote:

Originally Posted by Dili
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

Where it belongs depends on what axioms you are using.

How are you defining the integers?

RonL
• Jul 22nd 2007, 09:10 AM
tukeywilliams
This is how I would do it:

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

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

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

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

Correct me if I am wrong.
• Jul 22nd 2007, 09:27 AM
Dili
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 AM
tukeywilliams
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 $a,b$ 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 $a+b, ab \in \mathbb{Z}$.
• Jul 22nd 2007, 10:08 AM
Dili
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)