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?

Printable View

- Feb 21st 2010, 11:51 AMserious331Natural 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 PMemakarov
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 PMserious331
- Feb 21st 2010, 02:22 PMemakarov
I don't understand the definition . The variable is not bound (by a quantifier), so it is not clear whether is true or false for each particular .

- Feb 21st 2010, 02:35 PMserious331
- Feb 21st 2010, 02:35 PMemakarovQuote:

If x and y are natural numbers and x > y, then x-y is also a natural number.

- Feb 21st 2010, 03:03 PMserious331
- Feb 22nd 2010, 03:34 AMHallsofIvy
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?