- #1
bomba923
- 760
- 0
Given that [itex] {\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3} \right\}} [/itex],
Can I list all the rational numbers in [itex] \left[ {a,b} \right] [/itex] as the sequence represented by
[tex] a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } [/tex]
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My notation is rather strange, but here
[tex] \left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } [/tex]
really just represents [itex] \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } \right\} [/itex]
(I used floor brackets b/c indexes of terms in a sequence must be whole numbers,
and to not let the last term exceed [itex] b [/itex] when multiplied by [itex] x [/itex] and added to [itex] a [/itex])
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For example, let [itex] a = 3 [/itex] and [itex] b = 5 [/itex]. Then, for [itex] x = 0.5 [/itex], the sequence will be [itex] \left\{ {3,3.5,4,4.5,5} \right\} [/itex].
Similarly, for [itex] x = 0.1 [/itex], the sequence will be [itex] \left\{ {3,3.1,3.2,3.3, \ldots ,5} \right\} [/itex].
And basically, the idea is to list all of the rationals within [itex] \left[ {a,b} \right] [/itex] as [itex] x [/itex] approaches zero (from the right!).
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Well then, (for [itex] {\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}} [/itex] ,
with [itex] \mathbb{Q} [/itex] being the set of all rationals, that is),
|Will this work?
[tex] {a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } [/tex]
(Btw, I expressed notation here that may be strange/non-standard.
How would I rewrite the expression to be more understandable--more standard??)
Can I list all the rational numbers in [itex] \left[ {a,b} \right] [/itex] as the sequence represented by
[tex] a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } [/tex]
----------------------------------------------------------------
My notation is rather strange, but here
[tex] \left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } [/tex]
really just represents [itex] \left\{ {0,1,2, \ldots ,\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } \right\} [/itex]
(I used floor brackets b/c indexes of terms in a sequence must be whole numbers,
and to not let the last term exceed [itex] b [/itex] when multiplied by [itex] x [/itex] and added to [itex] a [/itex])
----------------------------------------------------------------
For example, let [itex] a = 3 [/itex] and [itex] b = 5 [/itex]. Then, for [itex] x = 0.5 [/itex], the sequence will be [itex] \left\{ {3,3.5,4,4.5,5} \right\} [/itex].
Similarly, for [itex] x = 0.1 [/itex], the sequence will be [itex] \left\{ {3,3.1,3.2,3.3, \ldots ,5} \right\} [/itex].
And basically, the idea is to list all of the rationals within [itex] \left[ {a,b} \right] [/itex] as [itex] x [/itex] approaches zero (from the right!).
------------------------------------------------------------------
Well then, (for [itex] {\left\{ {\left( {a,b,x} \right)|0 < x < \left( {b - a} \right),\left( {a,b,x} \right) \in \mathbb{Q}^3 } \right\}} [/itex] ,
with [itex] \mathbb{Q} [/itex] being the set of all rationals, that is),
|Will this work?
[tex] {a + \mathop {\lim }\limits_{x \to 0} x\left\{ n \right\}_0^{\left\lfloor {\frac{{b - a}}{x}} \right\rfloor } [/tex]
(Btw, I expressed notation here that may be strange/non-standard.
How would I rewrite the expression to be more understandable--more standard??)
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