Is this proof okay?

asdf60 · Sep 14, 2005

I'm taking a course in rigorous calculus, using the famous calculus textbook by Tom Apostol. I'm required to prove that if x is real and satisfies 0 <= x < h, for all positive real h, then x = 0. Here is my 'proof':

if x >= 0, then either x > 0 or x= 0. i'm going to prove that x > 0 leads to contradiction. if x>0, then let h = x/2 > 0. then x-h = x-x/2 = x/2 > 0, and therefore, h < x...which contradicts the h > x for all h. So x=0.

Is this proof all right and sufficiently rigorous?

quasar987 · Sep 15, 2005

way to go!

Werg22 · Sep 15, 2005

Well h is always above 0. If you consider x=>0 either infinitly small or equal to zero, then it must be equal to zero to sastify the condition x<h. But there is a little contratiction. x=>0 means x is infinitly small but existant... there is a paradox in the condition 0<=x>h

quasar987 · Sep 15, 2005

What are you saying Werg22? Particularily, what do you mean by

"x=>0 means x is infinitly small but existant... "

and here...

"there is a paradox in the condition 0<=x>h"

did you mean 0<=x<h ?

Log in or Sign up

Is this proof okay?

asdf60

quasar987

Werg22

quasar987

Okay, just a little confusion. (Replies: 1)

Okay, I am down to only a few. (Replies: 4)

Is this okay? (Replies: 16)

Friction problem again Okay today is not my day (Replies: 2)

Proof ! (Replies: 0)