Is this proof okay?

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  • #1
asdf60
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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?
 

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  • #2
quasar987
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way to go!
 
  • #3
Werg22
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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
 
  • #4
quasar987
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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 ?
 

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