Basics in integers

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July 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
July 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
July 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.
July 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
July 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 $a+b, ab \in \mathbb{Z}$ .
July 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)

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