Natural Numbers

• Feb 21st 2010, 11:51 AM
serious331
Natural Numbers
If x and y are natural numbers and x > y, then x-y is also a natural number.

how to prove this by induction?

is it becuase x-1 is a natural number,

and then for a number k, x - (k+1) = x-k-1=(x-k)-1 is also a natural number?
• Feb 21st 2010, 01:33 PM
emakarov
Please post the relevant details, in particular definitions. There are probably dozens of equivalent ways to define natural numbers, subtraction and >.
• Feb 21st 2010, 02:02 PM
serious331
Quote:

Originally Posted by emakarov
Please post the relevant details, in particular definitions. There are probably dozens of equivalent ways to define natural numbers, subtraction and >.

The reason my book states to show is that the set
S= {x E N| y E N or x-y E N} is inductive

I have no idea on how to work on the sets. The only thing I can think of is what I stated in the question
• Feb 21st 2010, 02:22 PM
emakarov
I don't understand the definition $\displaystyle S=\{x\in\mathbb{N}\mid y\in\mathbb{N}\lor x-y\in\mathbb{N}\}$. The variable $\displaystyle y$ is not bound (by a quantifier), so it is not clear whether $\displaystyle y\in\mathbb{N}\lor x_0-y\in\mathbb{N}$ is true or false for each particular $\displaystyle x_0$.
• Feb 21st 2010, 02:35 PM
serious331
Quote:

Originally Posted by emakarov
I don't understand the definition $\displaystyle S=\{x\in\mathbb{N}\mid y\in\mathbb{N}\lor x-y\in\mathbb{N}\}$. The variable $\displaystyle y$ is not bound (by a quantifier), so it is not clear whether $\displaystyle y\in\mathbb{N}\lor x_0-y\in\mathbb{N}$ is true or false for each particular $\displaystyle x_0$.

Sorry, I wrote the wrong thing. It is supposed to be
S(y) = { y E N | x E N, if x>y, x-y E N}
• Feb 21st 2010, 02:35 PM
emakarov
Quote:

If x and y are natural numbers and x > y, then x-y is also a natural number.
I guess, a simple answer is to do induction on x with the base case on x=y+1. Then if for x there is a number z (= x-y) such that y+z = x, then a number with the same property exists for x+1.
• Feb 21st 2010, 03:03 PM
serious331
Quote:

Originally Posted by emakarov
I guess, a simple answer is to do induction on x with the base case on x=y+1. Then if for x there is a number z (= x-y) such that y+z = x, then a number with the same property exists for x+1.

I have to do induction on y, so is it similar to what you have stated here?
• Feb 22nd 2010, 03:34 AM
HallsofIvy
S is the set of all natural numbers, y, such that "if x is a natural number and x> y then x-y is a natural number".

Your induction must start with y= 1. Show that if x is a natural number and x> 1 then x- 1 is a natural number.

But you still haven't done what emakarov asked you to do in his first reponse: what is your definition of ">" and what is your definition of "x- y" for natural numbers?