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symbolic ring comparisons return strange results

asked 2017-10-04 09:34:08 -0500

DrNick gravatar image

Greetings,

bool(abs(-1+sqrt(2)) != abs(1-sqrt(2)))

erroneously returns True (I'm running SageMath-8.0 on a mac laptop, for what it's worth).

Is this a bug?

Thanks

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it looks like a bug to me, and also why it does this:

sage: x = SR.var('x')
sage: bool(abs(1-x) == abs(x-1))
True
sage: bool(abs(1-x) != abs(x-1))
True
mforets gravatar imagemforets ( 2017-10-05 22:41:00 -0500 )edit

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answered 2017-10-04 12:54:07 -0500

dan_fulea gravatar image

In order to establish if two expressions are equal, sage performs a "default wash machine program" and in case there is a proof for the True value, it returns this True. Else it returns False.

In such cases, one should help sage to find the truth.

In our case:

sage: expression = abs(-1+sqrt(2)) != abs(1-sqrt(2))
sage: bool(expression)
True
sage: bool(expression.simplify_full())
False

Just tell sage to perfom a simplification of the expression!

Note: One can find many "bugs" like the posted one, just try for instance:

sage: bool( abs(-1+sqrt(2)) != abs(1-sqrt(2)) )
True
sage: bool( abs(-1+sqrt(2)) == abs(1-sqrt(2)) )
True
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1

Yeah, I sort of understood the issue of returning False in case it can't really verify truth. But this is returning True when the statement is in fact False.

DrNick gravatar imageDrNick ( 2017-10-04 15:29:16 -0500 )edit

Most probably this is because a != b is first converted to not (a == b) before evaluation.

eric_g gravatar imageeric_g ( 2017-10-07 04:19:26 -0500 )edit

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Asked: 2017-10-04 09:34:08 -0500

Seen: 51 times

Last updated: Oct 04