triclino Posted April 23, 2015 Share Posted April 23, 2015 Given the reals a<b find a, t such that : For all r>0 there exist an , x such that : 0<|x-t|<r and a<x<b Link to comment Share on other sites More sharing options...

imatfaal Posted April 23, 2015 Share Posted April 23, 2015 x=a + epsilon t=a epsilon is very small 1 Link to comment Share on other sites More sharing options...

triclino Posted April 23, 2015 Author Share Posted April 23, 2015 x=a + epsilon t=a epsilon is very small If you put t=a and x=a+ε ,then the inequality becomes: |a+ε-a|<ε => ε<ε, a contradiction By the way in the OP it is r and not ε Link to comment Share on other sites More sharing options...

imatfaal Posted April 24, 2015 Share Posted April 24, 2015 If you put t=a and x=a+ε ,then the inequality becomes: |a+ε-a|<ε => ε<ε, a contradiction By the way in the OP it is r and not ε "By the way in the OP it is r and not ε" - exactly and for the inquality to work epsilon is smaller than r (ie now matter hw small r is there is a smaller real that we designate epsilon) the inequality becomes 0<|a+ε-a|<r => 0<ε<r Link to comment Share on other sites More sharing options...

triclino Posted April 25, 2015 Author Share Posted April 25, 2015 "By the way in the OP it is r and not ε" - exactly and for the inquality to work epsilon is smaller than r (ie now matter hw small r is there is a smaller real that we designate epsilon) the inequality becomes 0<|a+ε-a|<r => 0<ε<r Suppose r=3b-a And since you clame the epsilon that you introduce in the problem to be ε<r then we can say that ε can be equal to 2b-a<3b-a=r Hence if we put x=a+ε=a+(2b-a)=2b Does this x satisfy also the relation a<x<b ? Furthermore when we introduce a new variable into a problem we have to quantify that variable to make our formula a well formed formula. And since you introduce ε how do you quantify it? Link to comment Share on other sites More sharing options...

imatfaal Posted April 25, 2015 Share Posted April 25, 2015 Suppose r=3b-a And since you clame the epsilon that you introduce in the problem to be ε<r then we can say that ε can be equal to 2b-a<3b-a=r Hence if we put x=a+ε=a+(2b-a)=2b Does this x satisfy also the relation a<x<b ? Furthermore when we introduce a new variable into a problem we have to quantify that variable to make our formula a well formed formula. And since you introduce ε how do you quantify it? No - if you checked my first response I said that epsilon was very small. Sorry I had thought you would know this terminology; epsilon is used to denote a very small addition of arbitrary smallness. What I am saying in my posts is that x is the tiniest bit bigger than a and that a is equal to t http://en.wikipedia.org/wiki/Epsilon In mathematics (particularly calculus), an arbitrarily small positive quantity is commonly denoted ε; see (ε, δ)-definition of limit. In reference to this, the late mathematician Paul Erdős also used the term "epsilons" to refer to children (Hoffman 1998, p. 4). Link to comment Share on other sites More sharing options...

triclino Posted April 25, 2015 Author Share Posted April 25, 2015 (edited) No - if you checked my first response I said that epsilon was very small. Sorry I had thought you would know this terminology; epsilon is used to denote a very small addition of arbitrary smallness. What I am saying in my posts is that x is the tiniest bit bigger than a and that a is equal to t http://en.wikipedia.org/wiki/Epsil In the whole of mathematics can you show me a single definition defining the exprssion " small epsilon"? In the definition of limits we say for all epsilon biger tha 0 ,there exists a delta biger than 0 such that: e.t.c e.t.c 99% of the problems in analysis stert with the expession . Let ε>0, or given ε>0 Yes you can use small or big epsilon,or delta . But if you do not clearly show their relatin to other variables of the problem or between them it is catastrophic. In the limits for example ,you have to clearly show the relatio9n between epsilon and delta, whether the epsilon and deltas are small or big,or whatever In your solution what will happen if for your " small epsion" we have a small : b-a?? Besides when you say : " for a small epsilon ",one may say ,ok But how small . Mathematics is relations between variables and certainly expressions or words of : small,big ,tiny e.t.c do not show relations Edited April 26, 2015 by triclino Link to comment Share on other sites More sharing options...

John Posted April 26, 2015 Share Posted April 26, 2015 But it's clear what the relationship must be. We must have [math]\varepsilon < \min(r, b - a)[/math]. If you want an equality, use the fact that [math]\min(x, y) = \frac{x + y - |x - y|}{2}[/math]. Link to comment Share on other sites More sharing options...

imatfaal Posted April 26, 2015 Share Posted April 26, 2015 In the whole of mathematics can you show me a single definition defining the exprssion " small epsilon"? In the definition of limits we say for all epsilon biger tha 0 ,there exists a delta biger than 0 such that: e.t.c e.t.c 99% of the problems in analysis stert with the expession . Let ε>0, or given ε>0 Yes you can use small or big epsilon,or delta . But if you do not clearly show their relatin to other variables of the problem or between them it is catastrophic. In the limits for example ,you have to clearly show the relatio9n between epsilon and delta, whether the epsilon and deltas are small or big,or whatever In your solution what will happen if for your " small epsion" we have a small : b-a?? Besides when you say : " for a small epsilon ",one may say ,ok But how small . Mathematics is relations between variables and certainly expressions or words of : small,big ,tiny e.t.c do not show relations http://mathworld.wolfram.com/Epsilon.html It is a little disingenuous to put "small epsilon" in quotation marks as if it is a phrase I have used - the only person to have used the phrase small epsilon is you. "99% of the problems..." really? "In your solution what will happen if for your " small epsion" we have a small : b-a??" It is arbitrarily small - ie it is smaller than... John even quantified the maximum above. Link to comment Share on other sites More sharing options...

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