5 ms·

Math person here. I'm curious to know if anyone used decimal numbers in their tests and if negative numbers were used. The rule is increasing real numbers and

Math person here. I'm curious to know if anyone used decimal numbers in their tests and if negative numbers were used. The rule is increasing real numbers and one can guess that the rule is increasing numbers without realizing this includes all real numbers and not just integers. In addition to getting it right did you use an exhaustive set of tests?

MalcolmPF 11y ago
I tried a negative series and a decimal series, just in case the rule was increasing natural numbers. I tried very large numbers to see if there was a limit to the rule, and a series that had large contrast in between each element. For fun I tried to see if the app would recognize "pi", "i", or "e", but, perhaps unsurprisingly, it did not.
- yequalsx 11y ago
  I would have checked complex numbers for fun but there is no reasonable ordering of the complex numbers in the way there is for real numbers. It's too bad it didn't recognize pi or e.
Infernal 11y ago
I did test with negatives, but I did not think to test with decimals. 15 yes's, 6 no's, and I successfully determined the rule.
pbnjay 11y ago
I got it right, using decimals - I was probably a little too exhaustive... http://imgur.com/6adtvqz http://imgur.com/6adtvqz
perlgeek 11y ago
Yes and yes. It wouldn't activate the "check" button with complex numbers.
- yequalsx 11y ago
  There isn't a way to check order for complex numbers in any reasonable way.
  - nandemo 11y ago
    But the problem statement doesn't imply the rule involves ordering. Likely, it doesn't accept complex numbers (or vectors or matrices) as input as part of the (implicit) spec: in common parlance, "number" tends to mean real number.
    - yequalsx 11y ago
      The tests with real numbers indicate that the rule is one based on order. So the problem doesn't state this but it is easily deducible when working with real numbers. Hence no need to check complex numbers. In common parlance in mathematical circles "number" tends not to mean real number unless context makes this obvious.
dorgo 11y ago
I wonder if somebody tested the rule to be deterministic by entering same number over and over. But I think the rule a<b<c is too simple to invite to use creative tests.
tjohns 11y ago
I tested with negative integers, but didn't think to test with real numbers unfortunately. (Still got it right, but I should've tested that.) That said, I did make sure to test all the edge cases I could think of. I was actually going to guess (n^1, n^2, n^3) at first, until it failed for (1, 1, 1). 10 "yes" tests, 6 "no" tests.
ddlutz 11y ago
I tested with real numbers to be curious, just tried 1.1, 1.2, and 1.3.
deleted 11y ago
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MalcolmPF 11y ago

I tried a negative series and a decimal series, just in case the rule was increasing natural numbers. I tried very large numbers to see if there was a limit to the rule, and a series that had large contrast in between each element. For fun I tried to see if the app would recognize "pi", "i", or "e", but, perhaps unsurprisingly, it did not.

yequalsx 11y ago
I would have checked complex numbers for fun but there is no reasonable ordering of the complex numbers in the way there is for real numbers. It's too bad it didn't recognize pi or e.

Infernal 11y ago

I did test with negatives, but I did not think to test with decimals. 15 yes's, 6 no's, and I successfully determined the rule.

pbnjay 11y ago

I got it right, using decimals - I was probably a little too exhaustive... http://imgur.com/6adtvqz http://imgur.com/6adtvqz

perlgeek 11y ago

Yes and yes. It wouldn't activate the "check" button with complex numbers.

yequalsx 11y ago
There isn't a way to check order for complex numbers in any reasonable way.
- nandemo 11y ago
  But the problem statement doesn't imply the rule involves ordering. Likely, it doesn't accept complex numbers (or vectors or matrices) as input as part of the (implicit) spec: in common parlance, "number" tends to mean real number.
  - yequalsx 11y ago
    The tests with real numbers indicate that the rule is one based on order. So the problem doesn't state this but it is easily deducible when working with real numbers. Hence no need to check complex numbers. In common parlance in mathematical circles "number" tends not to mean real number unless context makes this obvious.

dorgo 11y ago

I wonder if somebody tested the rule to be deterministic by entering same number over and over. But I think the rule a<b<c is too simple to invite to use creative tests.

tjohns 11y ago

I tested with negative integers, but didn't think to test with real numbers unfortunately. (Still got it right, but I should've tested that.) That said, I did make sure to test all the edge cases I could think of. I was actually going to guess (n^1, n^2, n^3) at first, until it failed for (1, 1, 1). 10 "yes" tests, 6 "no" tests.

ddlutz 11y ago

I tested with real numbers to be curious, just tried 1.1, 1.2, and 1.3.

deleted 11y ago

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