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(values are 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E, or F)
you are quoting a heck of a lot there.
[QUOTE]blah blah blah[/QUOTE] to reply to brian_dc.
Please remove excess text as not to re-post tons
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[QUOTE="brian_dc:349782"]whiskey_weed_and_women said:[QUOTE]ha prove it, PROVE IT !!!!! [/QUOTE] If we define two values on the domain [0, 4] (represent years, since before college I was not a math major and after college I will no longer be a math major) as x=my being a college student and y=my majoring in mathematics. We wish to prove that these two statements are uniformly continuous over some given statement of f(x) f: [0,4]--> R f(x)=x^2+x (note, by a theorem, the proof could end here with a statement of the fact that any polynomial is continuous, therefore the function is continuous. The theorem states that if the function is defined by continuous function over a compact interval (which [0,4] is since it is a closed and bounded set] it is uniformly continuous over that interval. I will utilize the episilon-delta (E, d) method of proving uniform continuity for exercise sake anyway. Let E>0. Assume x and y exist on [0,4]. Then |x-y|< d (from result below E/9) |f(x)-f(y)|=|(x^2+x)-(y^2+y)=|(x^2-y^2)+(x-y)|=|(x+y)(x-y)|+|x-y| at this point, consider that the greatest (x+y) can be equal to is if it is equal to the sum of the greatest upper bound of the domain = (4+4)=8 we can then say that |(x+y)(x-y)|+|x-y|<|(4+4)(x-y)|+|x-y|=8|(x-y)|+|x-y| set u=x-y 8|u|+|u| = 9|u| re-substitute --> 9|x-y| < E\9 so...there ya go...my being a math major in college is uniformly continuous over the domain of 0 to 4 years for the continuous function x^2+x you're probably asleep already[/QUOTE]
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