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Wolfram Alpha did not find the solution $$x=\frac{8}{7} , y=7$$

to the equations below. Why?

What can be done to expect more accurate results from Wolfram Alpha, in similar cases?

enter image description here

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closed as off-topic by dr.blochwave, LLlAMnYP, m_goldberg, Silvia, C. E. Sep 11 '15 at 13:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "The question is out of scope for this site. The answer to this question requires either advice from Wolfram support or the services of a professional consultant." – LLlAMnYP, C. E.
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ This is not the place to rant; if you have a serious question -- ask, if not send that as an email to support@wolfram.com $\endgroup$ – Sektor Sep 11 '15 at 6:26
  • $\begingroup$ @Sektor I completely disagree that this is a rant. One of the basic things that one expects from this software is the accuracy. I am not going to send any e-mail, and will leave this question here, to see what community thinks, I think it is useful for others to see, and for me to know the explanation. $\endgroup$ – VividD Sep 11 '15 at 6:33
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    $\begingroup$ Just leave the question as "Why?" and take out "This is a simple problem. This is disappointing". $\endgroup$ – alex.jordan Sep 11 '15 at 6:55
  • $\begingroup$ If I understood correctly, @Şektor is saying that if one has not-serious questions, they should be sent to support@wolfram.com?? Strange. $\endgroup$ – VividD Sep 11 '15 at 7:16
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    $\begingroup$ I'm voting to close this question as off-topic because it is about Wolfram Alpha, not Mathematica, as per the help center $\endgroup$ – dr.blochwave Sep 11 '15 at 7:48
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You can't expect a CAS to figure out everything for you. You need to give some hints:

Solve[x Floor[y] == 8 && y Floor[x] == 7 && x > y]

Better hint by indicating ranges of variables:

Solve[x Floor[y] == 8 && y Floor[x] == 7 && 1 <= x <= 10 && 
  1 <= y <= 10]

{{x -> 8/7, y -> 7}, {x -> 8/3, y -> 7/2}}

You can also do it by redefining $x$ as $x+d$ for some integer $x$ and $d \in [0, 1)$:

Reduce[(x + d) y == 8 && (y + e) x == 7 && x \[Element] Integers && 
 y \[Element] Integers && 0 <= d < 1 && 0 <= e < 1, {x, y, d, e}]
FindInstance[%, {x, y, d, e}, Reals, 100]

This returns all six solutions:

{{-8, -(7/8)}, {-4, -(7/4)}, {-(8/3), -(7/3)}, {-2, -(7/2)}, {8/7, 7}, {8/3, 7/2}}
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  • $\begingroup$ What does guarantee me that my hint will work? $\endgroup$ – VividD Sep 11 '15 at 7:28
  • $\begingroup$ Or, how to choose the hint that works? $\endgroup$ – VividD Sep 11 '15 at 7:34
  • $\begingroup$ Ok, thanks for the advice, I also added a subquestion to my question seeking such tips. $\endgroup$ – VividD Sep 11 '15 at 7:48
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    $\begingroup$ I've added a solution which gives every possible answer (I think). $\endgroup$ – Patrick Stevens Sep 11 '15 at 8:19
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    $\begingroup$ @PatrickStevens Beautiful! Hope it works so charmingly for similar cases, involving "floor" function. $\endgroup$ – VividD Sep 11 '15 at 8:30

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