can someone explain how to do this proof?

prove by contradiction that if n is a natural number then (n/n+1) > (n/n+2)

thanks

9/19/2006 5:18:51 PM

assume n is Natural and (n/n+1) <= (n/n+2) [ie you assume the antecedent and the negation of the consequent]

Then do a little algebra and you'll see a simple inequality that is ONLY true for non-positive numbers, thus it can't be true for natural numbers hence you have contradicted the assumption.

Try to figure it out from here, but if you need more help I can walk you through it.

[Edited on September 19, 2006 at 7:32 PM. Reason : .]

9/19/2006 7:29:23 PM

i put...

suppose there exist a n that is a natural number and (n/(n+1))<=(n/(n+2)) and since the larger the denominator is the smaller the value is. Then the only way that (n/(n+2)) is larger is if n is a negative number which is impossible since n is a natural number.

it feels like that is not correct though

9/19/2006 8:48:44 PM

you should do a little algebra to rearrange the thing to make it a little more obvious.

9/19/2006 9:12:42 PM

^agreed. The proof you gave is a little clumsy.

Quote :

"Then do a little algebra and you'll see a very simple inequality"

Then it'll be very clear.

9/19/2006 9:17:59 PM

would you just divide by n on both sides to get 1/n+1 <=1/n+2?

9/19/2006 9:41:39 PM

make the problem so you are no longer dealing with fractions. It will be real easy from there to derive a contradiction.

9/19/2006 9:48:15 PM

wow im an idiot....thanks a lot i got it now

i cant believe i didnt see that before

[Edited on September 19, 2006 at 9:52 PM. Reason : .]

9/19/2006 9:52:15 PM