sage: t = [t0,t1]
sage: t[0] is t0
True
sage: t[1] is t1
True

On 27 May 2014 12:07, David Joyner <[email protected]> wrote:
> On Mon, May 26, 2014 at 8:04 PM, msantopr <[email protected]> wrote:
>> Hi all,
>>
>> I am new to sage, so please forgive me if this is a trivial question.
>>
>> I tried to find a minimal example for my problem.
>> I can define two real interval fields:
>>
>> t0=RIF(2.9, 3.1)
>> t1=RIF(2.9, 3.1)
>>
>> and then define a matrix
>>
>> t=matrix([[t0],[t1]])
>>
>> Clearly
>>
>> parent(t0)
>>
>> Real Interval Field with 53 bits of precision
>>
>>
>> However,
>>
>>
>>  parent(t[0])
>>
>>
>> Vector space of dimension 1 over Real Interval Field with 53 bits of
>> precision
>>
>>
>> I would like to do some computations with t[0] and t[1] in which they should
>> be regarded as elements of a real interval fields. Is there any way to
>> convert from Vector space of dimension 1 over the real interval field to
>> real interval field? In my actual code I don't have t0 and t1 explicitly
>> since the matrix is obtained form a computation.
>>
>
>
> Does this help?
>
> sage: t0=RIF(2.9, 3.1)
> sage: t1=RIF(2.9, 3.1)
> sage: t=matrix([[t0],[t1]])
> sage: parent(t[0])
> Vector space of dimension 1 over Real Interval Field with 53 bits of precision
> sage: parent(t[0][0])
> Real Interval Field with 53 bits of precision
>
>>
>> Any help would be much appreciated!
>>
>>
>> Manuele
>>
>>
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