The square root does not distribute over division in general. You should
try not to introduce the square root until you have the desired expression
already under the square root.

By the way, it looks like you are proving theorem quad. Depending on your
goals you could either use that theorem or use it as a template for your
theorem.

On Sat, Jun 29, 2019 at 7:22 PM Norman Megill <[email protected]> wrote:

> I, too, am puzzled by this.  My initial though was that maybe it could be
> done via complex exponentiation, which relates to the sqr function for all
> complex numbers via ~ cxpsqr.  However, ~ divcxp accepts only nonnegative
> real arguments for the ratio terms (although the exponent can be complex).
> Does anyone know what the issues are that require this limitation?
>
> Norm
>
> On Wednesday, June 26, 2019 at 1:55:03 PM UTC-4, Filip Cernatescu wrote:
>>
>> I have started from  (( ( A x. ( X ^ 2 ) ) + ( ( B x. X ) + C ) ) = 0
>> and  I have followed  the classical math proof  and
>> I have arrived  at:
>>
>> ( ( X + ( B / ( 2 x. A ) ) ) = ( sqr ` ( ( ( B ^ 2 ) - ( 4 x. ( A x. C )
>> ) ) / ( ( 2 x. A ) ^ 2 ) ) ) \/ ( X + ( B / ( 2 x. A ) ) ) = -u ( sqr ` ( (
>> ( B ^ 2 ) - ( 4 x.
>>
>> ( A x. C ) ) ) / ( ( 2 x. A ) ^ 2 ) ) ) )
>>
>> The problem is the distribution of square root over division(requires
>> real not complex):
>>
>> sqrdiv    ( ( ( A e. RR /\ 0 <_ A ) /\ B e. RR+ ) ->
>>                ( sqr ` ( A / B ) ) = ( ( sqr ` A ) / ( sqr ` B ) ) )
>>
>>
>> Thank you very much!!!
>>
>> Filip
>>
>>
>>
>>
>>
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